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Table of contents
First published on Saturday, Jul 6, 2024 and last modified on Thursday, Apr 10, 2025 by François Chaplais.
Mathedu SAS
1 Introduction
In that last text of the course, you will learn deeply whta complex numbers are and how to manipulate then, as well as their use to calculate the roots of real quadratic polynomials having with negative discriminant.
You will then define the matrices and column vectors with complex elements, and learn to manipulate them.
And you will see that this allows the diagonalisation of the real matrices having no real eigenvalues.
2 The Field of Complex Numbers \( (\mathbb{C},+,\times)\)
The field \( (\mathbb{C},+,\times)\) is an extention of the field \( (\mathbb{R},+,\times)\) .
2.1 Reminder: the Addition and Subtraction of Complex Numbers
The complex numbers where introduced in the section 8, where they where considered as vectors.
Definition 1
Assume that \( (x_{1},y_{1},x_{2},y_{2})\in\mathbb{R}^{4}\) are real numbers, and consider the complex numbers \( z_1=x_1+iy_1\) and \( z_2=x_2+iy_2\) .
Then we define the sum and the difference of \( z_{1}\) and \( z_{2}\) as:
\( z_{1}+z_{2}=(x_{1}+x_{2})+i(y_{1}+y_{2})\) ,
and \( z_{1}-z_{2}=(x_{1}-x_{2})+i(y_{1}-y_{2})\) .
Theorem 1
Assume that \( (z_{1},z_{2}\in\mathbb{C}^{2}\) are complex numbers with \( z_2=x_2+iy_2\) , wher \( (x_{2},y_{2})\in\mathbb{R}^{2}\) are real numbers.
Consider the opposite of of \( z_{2}\) , \( -z_2=(-x_2)+i(-y_2)\) .
Then \( z_{1}-z_{2}=z_{1}+(-z_{2})\) .
This is because, for any real numbers \( ((a,b)\in\mathbb{R}^{2}\) , \( a-b=a+(-b)\) .
2.2 The Commutative Group \( (\mathbb{C},+)\)
\( (\mathbb{C},+)\) is a commutative group because of the following theorem.
Theorem 2
Assume that \( (z_1,z_2,z_3)\in\mathbb{C}^3\) are complex numbers.
Then the following properties hold for the addition.
Commutativity: \( z_1+z_2=z_2+z_1\) .
Associativity: \( (z_1+z_2)+z_3=z_1+(z_2+z_3)\) .
\( 0=0+0i\) is neutral: \( z_1+0=0+z_1=z_1\)
\( z_1+(-z_1)=(-z_1)+z_1=0\) , which justifies the fact that we call \( -z_{1}\) the opposite of \( z_{1}\) .
Proof
Assume that \( (x_{1},y_{1},x_{2},y_{2},x_{3},y_{3})\in\mathbb{R}^{6}\) are real numbers, and consider the complex numbers \( z_1=x_1+iy_1\) , \( z_2=x_2+iy_2\) and \( z_3=x_3+iy_3\) .
Then the following calculations may be performed.
As the addition is commutative in \( \mathbb{R}\) , we have:
\( z_1+z_2 =(x_{1}+x_{2})+i(y_{1}+y_{2}) =(x_{2}+x_{1})+i(y_{2}+y_{1}) =z_2+z_1\)
As the addition is associative in \( \mathbb{R}\) , we have:
\( (z_1+z_2)+z_3 =((x_{1}+x_{2})+i(y_{1}+y_{2}))+z_3 =(((x_{1}+x_{2})+x_{3})+i((y_{1}+y_{2})+y_{3})) =((x_{1}+(x_{2}+x_{3}))+i(y_{1}+(y_{2})+y_{3}) =z_1+(x_{2}+x_{3})+i(y_{2}+y_{3}) =z_1+(z_2+z_3)\)
As \( 0=0+0i\) and \( 0\) is neutral for the addition in \( \mathbb{R}\) , we have:
\( z_1+0 =(x_{1}+0)+i(y_{1}+0) =x_1+iy_1 =z_1\) .
\( 0+z_1 =(0+x_{1})+i(0+y_{1}) =x_1+iy_1 =z_1\) .
As for any real number \( a\in\mathbb{R}\) , \( a+(-a)=(-a)+a=0\) , we have:
\( z_1+(-z_1) =(x_{1}+(-x_{1}))+i(y_{1}+(-y_{1})) =0+0i =0\) .
\( (-z_1)+z_1 =((-x_{1})+x_{1})+i((-y_{1})+y_{1}) =0+0i =0\) .
2.3 Reminder: the Multiplication and Division of Complex Numbers
Definition 2
Assume that \( (x_{1},y_{1},x_{2},y_{2})\in\mathbb{R}^{4}\) are real numbers, and consider the complex numbers \( z_1=x_1+iy_1\) and \( z_2=x_2+iy_2\) .
Then we define the product and the quotient (if \( z_{2}\ne 0\) ) of \( z_{1}\) and \( z_{2}\) as:
\( z_1z_2=(x_1x_2-y_1y_2)+i(x_1y_2+y_1x_2)\) ,
and, if \( z_{2}\ne 0\) , \( \frac{z_1}{z_2}=\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2} +i\;\frac{-x_1y_2+x_2y_1}{x_2^2+y_2^2}\) .
Theorem 3
Assume that \( (z_{1},z_{2})\in\mathbb{C}\times\mathbb{C}^{*}\) are complex numbers such that \( z_{2}\ne 0\) . with \( z_2=x_2+iy_2\) , where \( (x_{2},y_{2})\in\mathbb{R}^{2}\) are real numbers.
Consider the inverse of of \( \frac{1}{z_2}=\frac{x_2}{x_2^2+y_2^2}+i\;\frac{-y_2}{x_2^2+y_2^2}\) .
Then \( \frac{z_1}{z_2}=z_1\frac{1}{z_2}\) .
Proof
Assume that \( (z_{1},z_{2})\in\mathbb{C}\times\mathbb{C}^{*}\) are complex numbers such that \( z_{2}\ne 0\) . with \( z_1=x_1+iy_1\) and \( z_2=x_2+iy_2\) , where \( (x_{1},y_{2},x_{1},y_{2})\in\mathbb{R}^{4}\) are real numbers.
Then the real part and imaginary part of \( z_1\frac{1}{z_2}\) are the following.
\( \Re(z_1\frac{1}{z_2}) =x_{1}\frac{x_2}{x_2^2+y_2^2}-y_{1}\frac{-y_2}{x_2^2+y_2^2} =\frac{x_{1}x_2}{x_2^2+y_2^2}+\frac{y_{1}y_2}{x_2^2+y_2^2} =\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2} =\Re(\frac{z_1}{z_2})\)
\( \Im(z_1\frac{1}{z_2}) =x_{1}\frac{-y_2}{x_2^2+y_2^2}+y_{1}\frac{x_2}{x_2^2+y_2^2} =\frac{-x_{1}y_2}{x_2^2+y_2^2}+\frac{y_{1}x_2}{x_2^2+y_2^2} =\frac{-x_1y_2+y_1x_2}{x_2^2+y_2^2} =\Im(\frac{z_1}{z_2})\)
Consequently, \( z_1\frac{1}{z_2}=\frac{z_1}{z_2}\) .
Theorem 4
Assume that \( (z_1,z_2)\in(\mathbb{C}^*)^2\) are non-zero complex numbers, and assume that \( z_1=R_1e^{i\theta_1}\) and \( z_2=R_2e^{i\theta_2}\) , with \( (R_{1},R_{2})\in(\mathbb{R}^{*})^{2}\) the modules of \( z_{1}\) and \( z_{2}\) respectively, and \( (\theta_1,\theta_2)\in\mathbb{R}^{2}\) are the arguments of \( z_{1}\) and \( z_{2}\) respectively.
Then \( z_1z_2=R_1R_2e^{i(\theta_1+\theta_2)}\) and \( \frac{z_{1}}{z_{2}}=\frac{R_{1}}{R_{2}}e^{i(\theta_{1}-\theta_{2})}\) .
Proof
Assume that \( (z_1,z_2)\in(\mathbb{C}^*)^2\) are non-zero complex numbers, and assume that \( z_1=R_1e^{i\theta_1}\) and \( z_2=R_2e^{i\theta_2}\) , where \( (R_{1},R_{2})\in(\mathbb{R}^{*})^{2}\) are the modules of \( z_{1}\) and \( z_{2}\) respectively, and \( (\theta_1,\theta_2)\in\mathbb{R}^{2}\) the arguments of \( z_{1}\) and \( z_{2}\) respectively.
Then the following assertions hold:
\( z_{1}=x_{1}+iy_{1}\) , with \( x_{1}=R_{1}\cos(\theta_{1})\) and \( y_{1}=R_{1}\sin(\theta_{1})\) .
\( z_{2}=x_{2}+iy_{2}\) , with \( x_{2}=R_{2}\cos(\theta_{2})\) and \( y_{2}=R_{2}\sin(\theta_{2})\) .
Then we may use the definition of the multiplication of complex numbers and the trigonometric formulae to perform the following calculations.
\( z_{1}z_{2}=(x_{1}x_{2}-y_{1}y_{2})+i\;(x_{1}y_{2}+y_{1}x_{2})\)
\( =(R_{1}\cos(\theta_{1})R_{2}\cos(\theta_{2})-R_{1}\sin(\theta_{1})R_{2}\sin(\theta_{2})) +i(\;R_{1}\cos(\theta_{1}R_{2}\sin(\theta_{2})+R_{1}\sin(\theta_{1})R_{2}\cos(\theta_{2}))\)
\( =R_{1}R_{2}(\cos(\theta_{1})\cos(\theta_{2})-\sin(\theta_{1})\sin(\theta_{2})) +i\;R_{1}R_{2}(\cos(\theta_{1}\sin(\theta_{2})+\sin(\theta_{1})\cos(\theta_{2}))\)
\( =R_{1}R_{2}\cos(\theta_{1}+\theta_{2})+i\;R_{1}R_{2}\sin(\theta_{1}+\theta_{2})\)
Consequently, the module of \( z_{1}z_{2}\) is
\( |z_{1}z_{2}|=R_{1}R_{2}\sqrt{\cos^{2}(\theta_1+\theta_2)+\sin^{2}(\theta_1+\theta_2)}=R_{1}R_{2}\)
and its argument is \( \theta_{1}+\theta_{2}\) modulo \( 2\pi\) .
So that \( z_1z_2=R_1R_2e^{i(\theta_1+\theta_2)}\) .
Moreover, the real part and the imaginary part of \( \frac{z_{1}}{z_{2}}\) are the following.
\( \Re(\frac{z_{1}}{z_{2}})=\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2}\)
\( =\frac{R_{1}\cos(\theta_{1})R_{2}\cos(\theta_{2})+R_{2}\sin(\theta_{2})R_{2}\sin(\theta_{2})}{R_2^2}\)
\( =\frac{R_{1}}{R_{2}}(\cos(\theta_{1})R_{2}\cos(\theta_{2})+\sin(\theta_{2})R_{2}\sin(\theta_{2})\)
\( =\frac{R_{1}}{R_{2}}\cos(\theta_{1}-\theta_{2})\) .
\( \Im(\frac{z_{1}}{z_{2}})=\frac{-x_1y_2+y_1x_2}{x_2^2+y_2^2}\)
\( =\frac{-R_{1}\cos(\theta_{1})R_{2}\sin(\theta_{2})+R_{2}\sin(\theta_{2})R_{2}\cos(\theta_{2})}{R_2^2}\)
\( =\frac{R_{1}}{R_{2}}(-\cos(\theta_{1})R_{2}\sin(\theta_{2})+\sin(\theta_{2})R_{2}\cos(\theta_{2})\)
\( =\frac{R_{1}}{R_{2}}\sin(\theta_{1}-\theta_{2})\) .
Consequently, \( \frac{z_{1}}{z_{2}}=\frac{R_{1}}{R_{2}}e^{i(\theta_{1}-\theta_{2})}\) .
2.4 The Commutative Group \( (\mathbb{C}^{*},\times)\)
\( (\mathbb{C}^{*},\times)\) is a commutative group because of the following theorem.
Theorem 5
Assume that \( (z_1,z_2,z_3)\in(\mathbb{C})^{*})^3\) are non zero complex numbers.
Then the following properties hold for the multiplication.
Commutativity: \( z_1z_2=z_2z_1\) .
Associativity: \( (z_1z_2)z_3=z_1(z_2z_3)\) .
\( 1=1e^{i0}i\) is neutral: \( 1\times z_1=z_1\times 1=z_1\)
\( z_1\frac{1}{z_1}=\frac{1}{z_1}\times z_1=1\) , which justifies the fact that we call \( \frac{1}{z_1}\) the inverse of \( z_{1}\) .
Proof
Assume that \( (z_1,z_2,z_3)\in(\mathbb{C})^{*})^3\) are non zero complex numbers, with \( z_1=R_1e^{i\theta_1}\) , \( z_2=R_2e^{i\theta_2}\) and \( z_3=R_3e^{i\theta_3}\) , where \( (R_{1},R_{2},R_{3})\in(\mathbb{R}^{*})^{2}\) are the modules of \( z_{1}\) , \( z_{2}\) and \( z_{3}\) respectively, and \( (\theta_1,\theta_2,\theta_3)\in\mathbb{R}^{2}\) are the arguments of \( z_{1}\) , \( z_{2}\) and \( z_{3}\) respectively.
Then the following calculations may be performed.
Because the multiplication and the addition are commutative in \( \mathbb{R}\) , we have:
\( z_1z_2 =R_1R_2e^{i(\theta_1+\theta_2)} =R_2R_1e^{i(\theta_2+\theta_1)} =z_2z_1\)
Because the multiplication and the addition are associative in \( \mathbb{R}\) , we have:
\( (z_1z_2)z3 =(R_1R_2e^{i(\theta_1+\theta_2)})z_{3} =((R_1R_2)R_{3})e^{i((\theta_1+\theta_2)+\theta_3)})\)
\( =(R_1(R_2R_{3}))e^{i(\theta_1+(\theta_2+\theta_3))}) =z_{1}((R_2R_3)e^{i(\theta_2+\theta_3)}) =z_1(z_2z_{3})\)
Because \( 1\) is neutral for the multiplication in \( \mathbb{R}\) , \( 0\) is neutral for the addition in \( \mathbb{R}\) , and \( 1=1e^{i0}\) we have:
\( 1\times z_1 =(1\times R_1)e^{i(0+\theta_1)} =R_1e^{i\theta_1} =z_1\)
\( z_1\times 1 =(R_1\times 1)e^{i(\theta_1+0)} =R_1e^{i\theta_1} =z_1\)
Because of the theorems 3 and 4, we have:
\( z_1\frac{1}{z_1} =\frac{z_1}{z_1} =\frac{R_1}{R_1}e^{i(\theta_1-\theta_1)} =1e^{i0} =1\)
And because the multiplication is commutative,
\( \frac{1}{z_1}\times z_1=z_1\frac{1}{z_1}=1\)
2.5 The Unitary Commutative Monoid \( (\mathbb{C},\times)\)
\( (\mathbb{C},\times)\) is a unitary commutative monoid because of the following theorem.
Theorem 6
Assume that \( (z_1,z_2,z_3)\in\mathbb{C}^3\) are complex numbers.
Then the following properties hold for the multiplication.
Commutativity: \( z_1z_2=z_2z_1\) .
Associativity: \( (z_1z_2)z_3=z_1(z_2z_3)\) .
\( 1=1+0i\) is neutral: \( 1\times z_1=z_1\times 1=z_1\) .
Lemma 1
Assume that \( z\in\mathbb{C}\) is a complex numbers.
Then \( 0=0+0i\) is absorbent for the multiplication, that is:
\( 0\times z=z\times 0=0\) .
Proof (of the lemma 1)
Assume that \( z\in\mathbb{C}\) is a complex numbers, with \( z=x+yi\) , \( (x,y)\in\mathbb{R}^{2}\) .
Then the following calculations may be performed, with \( 0=0+0i\) .
\( 0\times z=(0\times x-0\times y)+i(0\times y + 0\times x)=0+0i=0\) ,
and \( z\times 0=(0x\times 0-y\times 0)+i(x\times 0 + y\times 0)=0+0i=0\) .
Proof (of the theorem 6)
Assume that \( (z_1,z_2,z_3)\in\mathbb{C}^3\) are complex numbers.
Then the following cases may be distinguished.
Let’s prove that \( z_1z_2=z_2z_1\) in any cases.
Assume that \( z_{1}=0\) .
Then, as \( 0\) is absorbent for the multiplication in \( \mathbb{Z}\) , we have:
\( z_{1}z_{2}=0\times z_{2}=0\) , and \( z_{2}z_{1}=z_{2}\times 0=0\) .
Consequently, \( z_1z_2=z_2z_1\) .
Assume that \( z_{2}=0\) .
Then, as \( 0\) is absorbent for the multiplication in \( \mathbb{Z}\) , we have:
\( z_{1}z_{2}=z_{1}\times 0=0\) , and \( z_{2}z_{1}=0\times z_{1}=0\) .
Consequently, \( z_1z_2=z_2z_1\) .
Assume that \( z_{1}\ne 0\) and \( z_{2}\ne 0\) .
Then, as the multiplication is commutative in \( \mathbb{C}^{*}\) , \( z_1z_2=z_2z_1\) .
Let’s prove that \( (z_1z_2)z_3=z_1(z_2z_3)\) in any cases.
Assume that \( z_{1}=0\) .
Then, as \( 0\) is absorbent for the multiplication in \( \mathbb{Z}\) , we have:
\( (z_{1}z_{2})z_{3}=(0\times z_{2})z_{3}=0\times z_{3}=0\) , and \( z_{1}(z_{2}z_{3})=0\times (z_{2}z_{3})=0\) .
Consequently, \( (z_1z_2)z_3=z_1(z_2z_3)\) .
Assume that \( z_{2}=0\) .
Then, as \( 0\) is absorbent for the multiplication in \( \mathbb{Z}\) , we have:
\( (z_{1}z_{2})z_{3}=(z_{1}\times 0)z_{3}=0\times z_{3}=0\) , and \( z_{1}(z_{2}z_{3})=z_{1}(0\times z_{3})=z_{1}\times 0=0\) .
Consequently, \( (z_1z_2)z_3=z_1(z_2z_3)\) .
Assume that \( z_{3}=0\) .
Then, as \( 0\) is absorbent for the multiplication in \( \mathbb{Z}\) , we have:
\( (z_{1}z_{2})z_{3}=(z_{1}z_{2})\times 0=0\) , and \( z_{1}(z_{2}z_{3})=z_{1}(z_{2}\times 0)=z_{1}\times 0=0\) .
Consequently, \( (z_1z_2)z_3=z_1(z_2z_3)\) .
Assume that \( z_{1}\ne 0\) , \( z_{2}\ne 0\) and \( z_{3}\ne 0\) .
Then, as the multiplication is associative in \( \mathbb{C}^{*}\) , \( (z_1z_2)z_3=z_1(z_2z_3)\) .
Let’s prove that \( 1\times z_1=z_1\times 1=z_1\) in any cases.
Assume that \( z_{1}=0\) .
Then, as \( 0\) is absorbent for the multiplication in \( \mathbb{Z}\) , we have:
\( 1\times z_1=1\times 0=0=z_{1}\) ,
and \( z_{1}\times 1=0\times 1=0=z_{1}\) .
Assume that \( z_{1}\ne 0\) .
Then, as \( 1\) is neutral for the multiplication in \( \mathbb{C}^{*}\) , \( 1\times z_1=z_1\times 1=z_1\) .
2.6 The Distributivity Laws in \( \mathbb{C}\)
The multiplication is distributive to the left and to the right on the addition in \( \mathbb{C}\) because of the following theorem.
Theorem 7
Assume that \( (z_0,z_1,z_2)\in\mathbb{C}^3\) are any complex numbers.
Then the following properties hold for the addition and multiplication.
Distributivity to the left of the multiplication on the addition: \( z_0(z_1+z_2)=z_0z_1+z_0z_2\) .
Distributivity to the right of the multiplication on the addition: \( (z_1+z_2)z_0=z_1z_0+z_2z_0\)
Proof
Assume that \( (z_0,z_1,z_2)\in\mathbb{C}^3\) are any complex numbers.
Assume that \( z_{0}=x_{0}+iy_{0}\) , \( z_{1}=x_{1}+iy_{1}\) and \( z_{2}=x_{2}+iy_{2}\) , with \( (x_{0},y_{0},x_{1},y_{1},x_{2},y_{2})\in\mathbb{R}^{6}\) real numbers.
Then the following calculations may be performed.
Because the addition in \( \mathbb{R}\) is associative and commutative, and the multiplication in \( \mathbb{R}\) is distributive to the left on the addition, we have:
\( \Re(z_0(z_1+z_2)) =x_{0}(x_{1}+x_{2})-y_{0}(y_{1}+y_{2}) =x_{0}x_{1}+x_{0}x_{2}-y_{0}y_{1}-y_{0}y_{2} =(x_{0}x_{1}-y_{0}y_{1})+(x_{0}x_{2}-y_{0}y_{2}) =\Re(z_0z_1+z_0z_2)\)
\( \Im(z_0(z_1+z_2)) =x_{0}(y_{1}+y_{2})+y_{0}(x_{1}+x_{2}) =x_{0}y_{1}+x_{0}y_{2}+y_{0}x_{1}+y_{0}x_{2} =(x_{0}y_{1}+y_{0}x_{1})+(x_{0}y_{2}+y_{0}x_{2}) =\Im(z_0z_1+z_0z_2)\)
Because the multiplication in \( \mathbb{C}\) is commutative, we have: \( (z_1+z_2)z_0 =z_0(z_1+z_2) =z_0z_1+z_0z_2 =z_1z_0+z_2z_0\)
2.7 The Commutative Field \( (\mathbb{C},+,\times)\)
\( (\mathbb{C},+,\times)\) is a commutative field because of the following properties.
\( (\mathbb{C},+)\) is a commutative group.
\( (\mathbb{C}^*,\times)\) is a commutative group.
\( (\mathbb{C},\times)\) is a unitary commutative monoid.
The multiplication is distributive to the left on the addition.
Assume that \( x\in\mathbb{R}\) is a real number, and consider the complex number \( z_{x}=x+i0\) with a null imaginary part.
Then we may assimilate \( z_{x}\) to \( x\) , in such a way that the following properties hold.
\( \mathbb{R}\) is a subset of \( \mathbb{C}\) .
The addition in \( \mathbb{R}\) is compatible with the addition in \( \mathbb{C}\) .
And the multiplication in \( \mathbb{R}\) is compatible with the addition in \( \mathbb{C}\) .
Consequently, the commutative field \( (\mathbb{R},+,\times)\) is a sub-field of the commutative field \( (\mathbb{C},+,\times)\) .
3 The Conjugate Complex Numbers
The conjugate complex numbers have the same real parts and opposite imaginary parts. They are quite important in solving quadratic equations with real coefficients, such as the caracteristic polynomial of real matrices.
3.1 The Complex Conjugate of a Complex Number
Definition 3
Assume that \( z\in\mathbb{C}\) is a complex number, with \( z=x+iy\) , where \( (x,y\in\mathbb{R}^{2}\) are real numbers.
Then the complex conjugate of is defined as \( \overline{z}=x-iy\)
3.2 Properties of the Complex Conjugates
Theorem 8
Assume that \( (z_1,z_2)\in\mathbb{C}^2\) are complex numbers, and that \( x\in\mathbb{R}\) is a real number, considered as the complex number \( x+0i\) .
Then the following equalities hold.
\( \overline{x}=x\) .
\( \overline{xz_1}=x\overline{z_1}\) .
\( \overline{\overline{z_1}}=z_1\) , so that we may speak of a pair of conjugate complex numbers.
\( \overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\) .
\( \overline{-z_1}=-\overline{z_1}\) .
\( \overline{z_1-z_2}=\overline{z_1}-\overline{z_2}\) .
\( \overline{z_1z_2}=\overline{z_1}\;\overline{z_2}\)
If \( z_1\ne 0\) , then \( \overline{z_1}\ne 0\) and \( \overline{\left(\frac{1}{z_1}\right)}=\frac{1}{\overline{z_1}}\) .
If \( z_2\ne 0\) , then \( \overline{z_2}\ne 0\) and \( \overline{\left(\frac{z_{1}}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}\) .
\( z_{1}\overline{z_{1}}\) is a positive or zero real number, and the module of \( z_{1}\) is equal to \( |z_{1}|=\sqrt{z_{1}\overline{z_{1}}}\) .
If \( z_1\ne 0\) and \( z_{1}=Re^{i\theta}\) , with \( R=|z_{1}|\) and \( \theta=\text{arg}(z_{1})\) , then \( \overline{z_1}=Re^{-i\theta}\) .
Proof
Assume that \( (z_1,z_2)\in\mathbb{C}^2\) are complex numbers, and that \( x\in\mathbb{R}\) is a real number, considered as the complex number \( x+0i\) .
Assume that \( z_{1}=x_{1}+iy_{1}\) and \( z_{2}=x_{2}+iy_{2}\) , with \( (x_{1},y_{1},x_{2},y_{2})\in\mathbb{R}^{4}\) real numbers.
Then the following calculations may be performed.
\( \overline{x}=\overline{x+0i}=x-0i=x\) .
As the multiplicaiton is distributive in teh addition in \( \mathbb{C}\) , we have:
\( \overline{xz_1} =\overline{x(x_1+iy_{1})} =\overline{x_1+i(xy_{1})} =xx_1-i(xy_{1}) =x(x_1-iy_{1}) =x\overline{z_1}\) .
\( \overline{\overline{z_1}} =\overline(x_{1}-iy_{1}) =x_{1}+iy_{1} =z_1\) .
As the addition is assosiative and commutative in \( \mathbb{C}\) , we have:
\( \overline{z_1+z_2} =\overline{(x_{1}+x_{2})+i(y_{1}+y_{2})} =((x_{1}+x_{2})-i(y_{1}+y_{2})\)
\( =(x_{1}-iy_{1})+(x_{2}-iy_{2}) =\overline{z_1}+\overline{z_2}\) .
\( \overline{-z_1} =\overline{-x_{1}-iy_{1}} =-x_{1}+iy_{1} =-(x_{1}-iy_{1} =-\overline{z_1}\) .
As the addition is assosiative and commutative in \( \mathbb{C}\) , we have:
\( \overline{z_1-z_2} =\overline{(x_{1}-x_{2})+i(y_{1}-y_{2})} =(x_{1}-x_{2})-i(y_{1}-y_{2})\)
\( =(x_{1}-iy_{1})-(-x_{2}-iy_{2}) =(x_{1}-iy_{1})-(x_{2}-iy_{2}) =\overline{z_1}-\overline{z_2}\) .
\( \overline{z_1z_2} =\overline{(x_{1}x_{2}-y_{1}y_{2})+i(x_{1}y_{2}+y_{1}x_{2})} =(x_{1}x_{2}-y_{1}y_{2})-i(x_{1}y_{2}+y_{1}x_{2})\)
And \( \overline{z_1}\;\overline{z_2} =(x_{1}-iy_{1})(x_{2}-iy_{2})\)
\( =(x_{1}x_{2}-(-y_{1})(-y_{2}))+i(x_{1}(-y_{2})+(-y_{1})x_{2}) =(x_{1}x_{2}-y_{1}y_{2})-i(x_{1}y_{2}+y_{1}x_{2})\) .
Consequently, \( \overline{z_1z_2}=\overline{z_1}\;\overline{z_2}\) .
Assume that \( z_1\ne 0\) .
Then \( x_{1}\) and \( y_{1}\) are not zero together, so that \( \overline{z_1}=x_{1}-iy_{1}\ne 0\) .
Moreover, we have:
\( \overline{\left(\frac{1}{z_1}\right)} =\overline{\left(\frac{x_{1}}{x_{1}^{2}+y_{1}^{2}}+i\frac{-y_{1}}{x_{1}^{2}+y_{1}^{2}}\right)} =\frac{x_{1}}{x_{1}^{2}+y_{1}^{2}}+i\frac{y_{1}}{x_{1}^{2}+y_{1}^{2}}\) ,
and \( \frac{1}{\overline{z_1}} =\frac{1}{x_{1}-iy_{1}} =\overline{\left(\frac{x_{1}}{x_{1}^{2}+(y_{1})^{2}}+i\frac{-(-y_{1})}{x_{1}^{2}+(-y_{1})^{2}}\right)} =\frac{x_{1}}{x_{1}^{2}+y_{1}^{2}}+i\frac{y_{1}}{x_{1}^{2}+y_{1}^{2}}\) .
Consequently, \( \overline{\left(\frac{1}{z_1}\right)}=\frac{1}{\overline{z_1}}\) .
Assume that \( z_2\ne 0\) .
Then \( x_{1}\) and \( y_{1}\) are not zero together, so that \( \overline{z_1}=x_{1}-iy_{1}\ne 0\) .
Moreover, we have:
\( \overline{\left(\frac{z_{1}}{z_2}\right)} =\overline{\left(\frac{x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}+i\frac{-x_{1}y_{2}+y_{1}x_{2}}{x_{2}^{2}+y_{2}^{2}}\right)} =\frac{x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}-i\frac{-x_{1}y_{2}+y_{1}x_{2}}{x_{2}^{2}+y_{2}^{2}}\)
\( =\frac{x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}+i\frac{x_{1}y_{2}-y_{1}x_{2}}{x_{2}^{2}+y_{2}^{2}}\) ,
and \( \frac{\overline{z_1}}{\overline{z_2}} =\frac{x_{1}-iy_{1}}{x_{2}-iy_{2}} =\frac{x_{1}x_{2}+(-y_{1})(-y_{2})}{x_{2}^{2}+(-y_{2})^{2}}+i\frac{-x_{1}(-y_{2})+(-y_{1}x_{2})}{x_{2}^{2}+(-y_{2})^{2}}\)
\( =\frac{x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}+i\frac{x_{1}y_{2}-y_{1}x_{2}}{x_{2}^{2}+y_{2}^{2}}\) .
Consequently, \( \overline{\left(\frac{z_{1}}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}\) .
\( z_{1}\overline{z_{1}} =(x_{1}+iy_{1})(x_{1}-iy_{1}) =(x_{1}^{2}-y_{1}(-y_{1})+i(x_{1}(-y_{1}+y_{1}x_{1}) =x_{1}^{2}+y_{1}^{2}=|z_{1}|^{2}\) .
Consequently, \( z_{1}\overline{z_{1}}\ge 0\) and \( |z_{1}|=\sqrt{z_{1}\overline{z_{1}}}\) .
Assume that \( z_1\ne 0\) and \( z_{1}=Re^{i\theta}\) , with \( R=|z_{1}|\) and \( \theta=\text{arg}(z_{1})\) .
Then \( z_{1}=R\cos(\theta)+iR\sin(\theta)\) , so that \( \overline{z_1}=R\cos(\theta)-iR\sin(\theta) =R(\cos(-\theta)+i\sin(-\theta))=Re^{-i\theta}\) .
4 Resolution in \( \mathbb{C}\) of quadratic equations with real coefficients
4.1 The Complex Square Roots of the Real Numbers
Theorem 9
Assume that \( a\in\mathbb{R}\) is a real number of any sign.
Then there exists at least one complex square root of \( a\) , a complex number \( z\in\mathbb{C}\) such that \( z^{2}=a\) .
Namely, these are the following complex numbers.
If \( a=0\) , it has a unique complex square root \( z_{0}=0\) .
If \( a>0\) , it has two complex square roots, that are the real numbers \( z_1=\sqrt{a}\) and \( z_2=-\sqrt{a}\) .
If \( a<0\) , it has two complex square roots, that are the conjugate complex numbers \( z_1=i\sqrt{-a}\) and \( z_2=-i\sqrt{-a}\) .
Proof
Assume that \( a\in\mathbb{R}\) is a real number of any sign.
Then the following calculations may be performed.
To begin with it, let’s consider that the equation \( z^{2}=a\) in \( \mathbb{C}\) is equivalent to the quadratic equation \( z^{2}-a=0\) .
Assume that \( a=0\) .
Then \( z^{2}-a=0\) if and only if \( z^{2}=0\) , that is equivalent to \( z=0\) .
Assume that \( a>0\) and consider the candidate square roots \( z_1=\sqrt{a}\) and \( z_2=-\sqrt{a}\) .
Then we may calculate:
\( (z-z_{1})(z-z_{2})=(z-\sqrt{a})(z+\sqrt{a})=z^{2}-\sqrt{a}z+z\sqrt{a}-\sqrt{a}^{2}=z^{2}-a\) .
Consequently, \( z^{2}-a=0\) if and onmy if \( z=z_{1}=\sqrt{a}\) or \( z=z_{2}=-\sqrt{a}\) .
Assume that \( a<0\) and consider the candidate square roots \( z_1=i\sqrt{-a}\) and \( z_2=-i\sqrt{-a}\) .
Then, as \( i^{2}=-1\) we may calculate:
\( (z-z_{1})(z-z_{2})=(z-i\sqrt{-a})(z+i\sqrt{-a})\)
\( =z^{2}-i\sqrt{-a}z+z\times i\sqrt{-a}-i^{2}\sqrt{-a}^{2}=z^{2}-a\) .
Consequently, \( z^{2}-a=0\) if and onmy if \( z=z_{1}=i\sqrt{-a}\)
or \( z=z_{2}=-i\sqrt{-a}\) .
4.2 Solve a Quadratic Equation with Real Coefficients in \( \mathbb{C}\)
Assume that \( (a,b,c)\in\mathbb{R}^*\times\mathbb{R}\times\mathbb{R}\) are real numbers such that \( a\ne 0\) .
Let’s try to solve in \( \mathbb{C}\) the quadratic equation with real coefficients: \( ax^{2}+bx+c=0\) .
To do that, let’s perform the following calculations.
\( ax^{2}+bx+c=a\left(x^{2}+\frac{b}{a}x+\frac{c}{a}\right)\) .
But because \( \left( x+\frac{b}{2a} \right)^{2}=x^{2}+\frac{b}{a}x+\left( \frac{b}{2a} \right)^{2}\) , we have:
\( ax^{2}+bx+c=a\left(\left( x+\frac{b}{2a} \right)^{2}-\left( \frac{b}{2a} \right)^{2}+\frac{c}{a}\right) =a\left(\left( x+\frac{b}{2a} \right)^{2}-\left( \frac{b^{2}-4ac}{4a^{2}} \right)\right)\) .
We find the discriminant of the quadratic equation: \( \Delta=b^2-4ac\) , so that \( ax^{2}+bx+c=0\) if and only if
\( \left( x+\frac{b}{2a} \right)^{2}=\left( \frac{\Delta}{4a^{2}} \right)\) ,
so that \( x+\frac{b}{2a}\) is a square root of \( \frac{\Delta}{4a^{2}}\) , that has the same signe as \( \Delta\) .
Then we may distinguish the following cases.
Assume that \( \Delta=0\) .
Then the quadratic equation has a unique solution \( x_{0}\) , that is such that:
\( x_{0}+\frac{b}{2a}=0\) , i.e. \( x_{0}=-\frac{b}{2a}\) .
Assume that \( \Delta>0\) .
Then the quadratic equation has two real solutions \( x_{1}\) and \( x_{2}\) , that are such that:
\( x_{1}+\frac{b}{2a}=\frac{\sqrt{\Delta}}{2a}\) , i.e. \( x_1=\frac{-b+\sqrt{\Delta}}{2a}\) .
\( x_{2}+\frac{b}{2a}=-\frac{\sqrt{\Delta}}{2a}\) , i.e. \( x_1=\frac{-b-\sqrt{\Delta}}{2a}\) .
Assume that \( \Delta<0\) .
Then the quadratic equation has two conjugate complex solutions \( x_{1}\) and \( x_{2}\) , that are such that:
\( x_{1}+\frac{b}{2a}=i\frac{\sqrt{-\Delta}}{2a}\) , i.e. \( x_1=\frac{-b+i\sqrt{-\Delta}}{2a}\) .
\( x_{2}+\frac{b}{2a}=-i\frac{\sqrt{-\Delta}}{2a}\) , i.e. \( x_2=\frac{-b-i\sqrt{-\Delta}}{2a}\) .
5 Matrices and Column Vectors with Complex Elements
5.1 \( 2\times 2\) Matrices with Complex Elements
Definition 4
Assume that \( (a,b,c,d)\in\mathbb{C}^4\) are complex numbers.
Then the \( 2\times 2\) matrix with complex elements \( (a,b,c,d)\) is:
\( A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\)
If \( (a,b,c,d)\in\mathbb{R}^4\) are real numbers, is is also a \( 2\times 2\) matrix with real elements.
5.2 Add and Subtract \( 2\times 2\) Matrices with Complex Elements
Definition 5
Assume that \( (a,b,c,d,e,f,g,h)\in\mathbb{C}^8\) are complex numbers.
Consider the matrices with complex elements \( A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) and \( B=\begin{bmatrix}e&f\\ g&h\end{bmatrix}\) .
Then the sum \( S=A+B\) of \( A\) and \( B\) is defined as the \( 2\times 2\) matrix with complex elements:
\( S=\begin{bmatrix}a+e&b+f\\ c+g&d+h\end{bmatrix}\)
And the difference \( S=A-B\) of \( A\) and \( B\) is defined as the \( 2\times 2\) matrix with complex elements:
\( D=\begin{bmatrix}a-e&b-f\\ c-g&d-h\end{bmatrix}\)
Note that these are the same formulae as for \( 2\times 2\) matrices with real elements.
5.3 Properties of the Addition of \( 2\times 2\) Matrices with Complex Elements
Let’s denote \( \mathbb{M}_{22}^{\mathbb{C}}\) the set of \( 2\times 2\) matrices with complex elements.
Theorem 10
Assume that \( (A,B,C)\in(\mathbb{M}_{22}^{\mathbb{C}})^3\) are \( 2\times 2\) matrices with complex elements, with \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) .
Then \( (\mathbb{M}_{22}^{\mathbb{C}}\;,+)\) is a commutative group because of the following properties of the addition of \( 2\times 2\) matrices with complex elements.
Commutativity: \( A+B=B+A\) .
Associativity: \( (A+B)+C=A+(B+C)\) .
The null matrix \( O_{22}=\begin{bmatrix}0&0\\ 0&0\end{bmatrix}\) is neutral: \( A+O_{22}=O_{22}+A=A\) .
If we define the opposite \( -A\) of \( A\) as \( -A=\begin{bmatrix}-a_{11}&-a_{12}\\ -a_{21}&-a_{22}\end{bmatrix}\) , then \( A+(-A)=(-A)+A=O_{22}\) .
The last formula justifies the fact that we call \( -A\) the opposite of \( A\) .
Proof
Assume that \( (A,B,C)\in(\mathbb{M}_{22}^{\mathbb{C}})^3\) are \( 2\times 2\) matrices with complex elements, with \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) , \( B=\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix}\) , and \( C=\begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix}\) .
Then the following calculations may be performed.
As the addition is commutative in \( \mathbb{C}\) , we have:
\( A+B =\begin{bmatrix}a_{11}+b_{11}&a_{12}+b_{12}\\ a_{21}+b_{21}&a_{22}+b_{22}\end{bmatrix} =\begin{bmatrix}b_{11}+a_{11}&b_{12}+a_{12}\\ b_{21}+a_{21}&b_{22}+a_{22}\end{bmatrix} =B+A\)
As the addition is associative in \( \mathbb{C}\) , we have:
\( (A+B)+C =\begin{bmatrix}a_{11}+b_{11}&a_{12}+b_{12}\\ a_{21}+b_{21}&a_{22}+b_{22}\end{bmatrix} +\begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}(a_{11}+b_{11})+c_{11}&(a_{12}+b_{12})+c_{12}\\ (a_{21}+b_{21})+c_{21}&(a_{22}+b_{22})+c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}+(b_{11}+c_{11})&a_{12}+(b_{12}+c_{12})\\ a_{21}+(b_{21}+c_{21})&a_{22}+(b_{22})+c_{22})\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} +\begin{bmatrix}b_{11}+c_{11}&b_{12}+c_{12}\\ b_{21}+c_{21}&b_{22})+c_{22}\end{bmatrix} =A+(B+C)\) .
As \( 0\) is neutral for the addition in \( \mathbb{C}\) , we have:
\( A+O_{22} =\begin{bmatrix}a_{11}+0&a_{12}+0\\ a_{21}+0&a_{22}+0\end{bmatrix} =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} =A\) .
\( O_{22}+A =\begin{bmatrix}0+a_{11}&0+a_{12}\\ 0+a_{21}&0+a_{22}\end{bmatrix} =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} =A\) .
As for any complex number \( z\in\mathbb{C}\) , \( z+(-z)=(-z)+z=0\) , we have:
\( A+(-A) =\begin{bmatrix}a_{11}+(-a_{11})&a_{12}+(-a_{12})\\ a_{21}+(-a_{21})&a_{22}+(-a_{22})\end{bmatrix} =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} =O_{22}\) .
\( (-A)+A =\begin{bmatrix}(-a_{11})+a_{11}&(-a_{12})+a_{12}\\ (-a_{21})+a_{21}&(-a_{22})+a_{22}\end{bmatrix} =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} =O_{22}\) .
5.4 Properties of the Subtraction of \( 2\times 2\) Matrices with Complex Elements
Theorem 11
Assume that \( A\in\mathbb{M}_{22}^{\mathbb{C}}\) is a \( 2\times 2\) matrix with complex elements.
Then the following assertions hold for the subtraction.
\( A-O_{22}=A\) .
\( O_{22}-A=-A\) .
\( A-A=O_{22}\) .
Lemma 2
Assume that \( z\in\mathbb{C}\) is a complex number.
Then the following assertions hold for the subtraction.
\( z-0=z\) .
\( 0-z=-z\) .
\( z-z=0\) .
Proof (of the lemma 2)
Assume that \( z\in\mathbb{C}\) is a complex number, with \( z=x+iy\) , where \( (x,y)\in\mathbb{R}^{2}\) are real numbers.
Then the following calculations may be performed.
As \( a-0=a\) for any real number \( a\in\mathbb{R}\) , we have:
\( z-0=(x-0)+i(y-0)=x+iy=z\) .
As \( 0-a=-a\) for any real number \( a\in\mathbb{R}\) , we have:
\( 0-z=(0-x)+i(0-y)=-x-iy=-z\) .
As \( a-a=0\) for any real number \( a\in\mathbb{R}\) , we have:
\( z-z=(x-x)+i(y-y)=0+0i=0\) .
Proof (of the theorem 11)
Assume that \( A\in\mathbb{M}_{22}^{\mathbb{C}}\) is a \( 2\times 2\) matrix with complex elements, with \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) .
Then because of the lemma 2, the following calculations may be performed.
\( A-O_{22} =\begin{bmatrix}a_{11}-0&a_{12}-0\\ a_{21}-0&a_{22}-0\end{bmatrix} =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} =A\) .
\( O_{22}-A =\begin{bmatrix}0-a_{11}&0-a_{12}\\ 0-a_{21}&0-a_{22}\end{bmatrix} =\begin{bmatrix}-a_{11}&-a_{12}\\ -a_{21}&-a_{22}\end{bmatrix} =-A\) .
\( A-A =\begin{bmatrix}a_{11}-a_{11}&a_{12}-a_{12}\\ a_{21}-a_{21}&a_{22}-a_{22}\end{bmatrix} =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} =O_{22}\) .
5.5 Mutual Properties of the Addition and the Subtraction of \( 2\times 2\) Matrices with Complex Elements
Theorem 12
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})^2\) are \( 2\times 2\) matrices with complex elements. Then the following assertions hold for the addition and the subtraction.
Subtract a matrix is adding its opposite: \( A-B=A+(-B)\) .
The addition and subtraction of matrices are mutually reciprocal:
\( (A+B)-B=A\) .
\( (A-B)+B=A\) .
Lemma 3
Assume that \( (z_{1},z_{2})\in(\mathbb{C})^2\) are complex numbers.
Then the following assertions hold for the addition and the subtraction.
Subtract a complex number is adding its opposite: \( z_{1}-z_{2}=z_{1}+(-z_{2})\) .
The addition and subtraction of complex numbers are mutually reciprocal:
\( (z_{1}+z_{2})-z_{2}=z_{1}\) .
\( (z_{1}-z_{2})+z_{2}=z_{1}\) .
Proof (of the lemma 3)
Assume that \( (z_{1},z_{2})\in(\mathbb{C})^2\) are complex numbers, with \( z_{1}=x_{1}+iy_{1}\) and \( z_{2}=x_{2}+iy_{2}\) , where \( (x_{1},y_{1},x_{2},y_{2})\in\mathbb{R}^{4}\) are real nubers.
Then the following calculations may be performed.
As subtract a real number is adding its opposite, we have:
\( z_{1}-z_{2} =(x_{1}-x_{2})+i(y_{1}-y_{2}) =(x_{1}+(-x_{2}))+i(y_{1}+(-y_{2})) =z_{1}+(-z_{2})\) .
As the addition and subtraction of real numbers are mutually reciprocal, we have:
\( (z_{1}+z_{2})-z_{2} =((x_{1}+x_{2}-x_{2})+i((y_{1}+y_{2})-y_{2}) =x_{1}+iy_{1} =z_{1}\) .
\( (z_{1}-z_{2})+z_{2} =((x_{1}-x_{2}+x_{2})+i((y_{1}-y_{2})+y_{2}) =x_{1}+iy_{1} =z_{1}\) .
Proof (of the theorem 12)
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})^2\) are \( 2\times 2\) matrices with complex elements, with \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) and \( B=\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix}\) .
Then because of the lemma 3, the following calculations may be performed.
\( A-B =\begin{bmatrix}a_{11}-b_{11}&a_{12}-b_{12}\\ a_{21}-b_{21}&a_{22}-b_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}+(-b_{11})&a_{12}+(-b_{12})\\ a_{21}+(-b_{21})&a_{22}+(-b_{22})\end{bmatrix} =A+(-B)\) .
Moreover:
\( (A+B)-B =\begin{bmatrix}(a_{11}+b_{11})-b_{11}&(a_{12}+b_{12})-b_{12}\\ (a_{21}+b_{21})-b_{21}&(a_{22}+b_{22})-b_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} =A\) .
\( (A-B)+B =\begin{bmatrix}(a_{11}-b_{11})+b_{11}&(a_{12}-b_{12})+b_{12}\\ (a_{21}-b_{21})+b_{21}&(a_{22}-b_{22})+b_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} =A\) .
5.6 Multiply a \( 2\times 2\) Matrix with Complex Elements by a Complex Scalar
Definition 6
Assume that \( (a,b,c,d,k)\in\mathbb{C}^5\) are complex numbers.
Consider the \( 2\times 2\) matrix with complex elements \( A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) .
Then the product \( P=kA\) of \( A\) by the complex scalar \( k\) is defined as the \( 2\times 2\) matrix with complex elements:
\( P=\begin{bmatrix}ka&kb\\ kc&kd\end{bmatrix}\)
It is the same formula as for \( 2\times 2\) matrices with real elements and real scalars.
5.7 Properties of the Scalar Multiplication of \( 2\times 2\) Matrices with Complex Elements
Theorem 13
Assume that \( A\in\mathbb{M}_{22}^{\mathbb{C}}\) is a \( 2\times 2\) matrix with complex elements and that \( k\in\mathbb{C}\) is a complex scalar.
Then the following properties hold for the scalar multiplication.
\( 1\times A=A\) .
\( 0\times A=O_{22}\) .
\( k\times O_{22}=O_{22}\)
\( k(-A)=-kA\)
\( (-k)A=-kA\)
Lemma 4
Assume that \( (z_{1},z_{2})\in(\mathbb{C})^2\) are complex numbers.
Then the following assertions hold for the addition and the subtraction.
\( z_{1}(-z_{2})=-z_{1}z_{2}\) .
\( (-z_{1})z_{2}=-z_{1}z_{2}\) .
Proof (of the lemma 4)
Assume that \( (z_{1},z_{2})\in(\mathbb{C})^2\) are complex numbers, with \( z_{1}=x_{1}+iy_{1}\) and \( z_{2}=x_{2}+iy_{2}\) , where \( (x_{1},y_{1},x_{2},y_{2})\in\mathbb{R}^{4}\) are real nubers.
Then the following calculations may be performed.
\( z_{1}(-z_{2}) =(x_{1}+iy_{1})((-x_{2})+i(-y_{2})) =(x_{1}(-x_{2})-y_{1}(-y_{2})+i(x_{1}(-y_{2})+y_{1}(-x_{2})) ==(-(x_{1}x_{2})-y_{1}y_{2})+i(-(x_{1}y_{2})+y_{1}x_{2})) =-z_{1}z_{2}\) .
As the multiplication is commutative in \( \mathbb{C}\) ,and because of the item (I), we have:
\( (-z_{1})z_{2}=z_{2}(-z_{1})=-z_{2}z_{1}=-z_{1}z_{2}\) .
Proof (of the theorem 13)
Assume that \( A\in\mathbb{M}_{22}^{\mathbb{C}}\) is a \( 2\times 2\) matrix with complex elements, with \( A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) , and that \( k\in\mathbb{C}\) is a complex scalar.
Then the following properties hold for the scalar multiplication.
As \( 1\) is neutral for the multiplication in \( \mathbb{C}\) , we have:
\( 1\times A =\begin{bmatrix}1\times a&1\times b\\ 1\times c&1\times d\end{bmatrix} =\begin{bmatrix}a&b\\ c&d\end{bmatrix} =A\) .
As \( 0\) is absorbent for the multiplication in \( \mathbb{C}\) , we have:
\( 0\times A =\begin{bmatrix}0\times a&0\times b\\ 0\times c&0\times d\end{bmatrix} =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} =O_{22}\) .
As \( 0\) is absorbent for the multiplication in \( \mathbb{C}\) , we have:
\( k\times O_{22} =\begin{bmatrix}k\times 0&k\times 0\\ k\times 0&k\times 0\end{bmatrix} =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} =O_{22}\)
Because of the item (I) of the lemma 4, we have:
\( k(-A) =\begin{bmatrix}k(-a)&k(-b)\\ k(-c)&k(-d)\end{bmatrix} =\begin{bmatrix}-ka&-kb\\ -kc&-kd\end{bmatrix} =-kA\)
Because of the item (II) of the lemma 4, we have:
\( (-k)A =\begin{bmatrix}(-k)a&(-k)b\\ (-k)c&(-k)d\end{bmatrix} =\begin{bmatrix}-ka&-kb\\ -kc&-kd\end{bmatrix} =-kA\)
5.8 Mutual Properties of the Addition and the Scalar Multiplication of \( 2\times 2\) Matrices with Complex Elements
Theorem 14
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})^2\) are \( 2\times 2\) matrices with complex elements and that \( (k,l)\in\mathbb{C}^2\) are complex scalars.
Then \( (\mathbb{M}_{22}^{\mathbb{C}}\;,+,\cdot)\) is a vector space on \( \mathbb{C}\) because of the following properties of the addition and the scalar multiplication.
\( (\mathbb{M}_{22}^{\mathbb{C}}\;,+)\) is a commutative group.
First distributivity law: \( k(A+B)=kA+kB\) .
Second distributivity law: \( (k+l)A=kA+lA\) .
Associativity law: \( k(lA)=(kl)A\) .
Proof
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})^2\) are \( 2\times 2\) matrices with complex elements, with \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) and \( B=\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix}\) , and that \( (k,l)\in\mathbb{C}^2\) are complex scalars.
Then the following calculations may be performed.
As stated by the theorem 10, \( (\mathbb{M}_{22}^{\mathbb{C}}\;,+)\) is a commutative group.
Because the multiplication is distributive to the left on the addition in \( \mathbb{C}\) , we have:
\( k(A+B) =\begin{bmatrix}k(a_{11}+b_{11})&k(a_{12}+b_{12})\\ k(a_{21}+b_{21})&k(a_{22}+b_{22})\end{bmatrix}\)
\( =\begin{bmatrix}ka_{11}+kb_{11}&ka_{12}+kb_{12}\\ ka_{21}+kb_{21}&ka_{22}+kb_{22}\end{bmatrix} =kA+kB\) .
Because the multiplication is distributive to the left on the addition in \( \mathbb{C}\) , we have:
\( (k+l)A =\begin{bmatrix}(k+l)a_{11}&(k+l)a_{12}\\ (k+l)a_{21}&(k+l)a_{22}\end{bmatrix}\)
\( =\begin{bmatrix}ka_{11}+la_{11}&ka_{12}+la_{12}\\ ka_{21}+la_{21}&ka_{22}+la_{22}\end{bmatrix} =kA+lA\) .
Because the multiplication is associative in \( \mathbb{C}\) , we have:
\( k(lA) =\begin{bmatrix}k(la_{11})&k(la_{12})\\ k(la_{21})&k(la_{22})\end{bmatrix} =\begin{bmatrix}(kl)a_{11}&(kl)a_{12}\\ (kl)a_{21}&(kl)a_{22}\end{bmatrix} =(kl)A\) .
5.9 Matrix Multiplication of \( 2\times 2\) Matrices with Complex Elements
Definition 7
Assume that \( (a,b,c,d,e,f,g,h)\in\mathbb{C}^8\) are complex numbers.
Consider the matrices with complex elements \( A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) and \( B=\begin{bmatrix}e&f\\ g&h\end{bmatrix}\) .
Then the sum result of the matrix multiplication \( M=AB\) of \( A\) and \( B\) is defined as the \( 2\times 2\) matrix with complex elements:
\( M=\begin{bmatrix}ae+bg&af+bh\\ ce+dg&cf+dh\end{bmatrix}\)
Note that it is the same formula as for \( 2\times 2\) matrices with real elements.
5.10 Properties of the Matrix Multiplication of \( 2\times 2\) Matrices with Complex Elements
Theorem 15
Assume that \( (A,B,C)\in(\mathbb{M}_{22}^{\mathbb{C}})^3\) are \( 2\times 2\) matrices with complex elements.
Then \( (\mathbb{M}_{22}^{\mathbb{C}}\;,\times)\) is a unitary monoid because of the following properties of the addition of \( 2\times 2\) matrices with complex elements.
Associativity: \( (AB)C=A(BC)\) .
The identity matrix \( I=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}\) is neutral: \( IA=AI=A\) .
Moreover, the following property holds.
The null matrix \( O_{22}=\begin{bmatrix}0&0\\ 0&0\end{bmatrix}\) is absorbent: \( AO_{22}=O_{22}A=O_{22}\) .
Proof
Assume that \( (A,B,C)\in(\mathbb{M}_{22}^{\mathbb{C}})^3\) are \( 2\times 2\) matrices with complex elements, with \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) , \( B=\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix}\) , and \( C=\begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix}\) .
Then the following calculations may be performed.
As the multiplication is distributive to the right and to the left on the addition, and the addition is associative and commutative in \( \mathbb{C}\) , we have:
\( (AB)C =\left( \begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix} \right) \begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix} \begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}(a_{11}b_{11}+a_{21}b_{21})c_{11}+(a_{11}b_{12}+a_{12}b_{22})c_{21}& (a_{11}b_{11}+a_{12}b_{21})c_{12}+(a_{11}b_{12}+a_{12}b_{22})c_{22}\\ (a_{21}b_{11}+a_{22}b_{21})c_{11}+(a_{21}b_{12}+a_{22}b_{22})c_{21}& (a_{21}b_{11}+a_{22}b_{21})c_{12}+(a_{21}b_{12}+a_{22}b_{22})c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}b_{11}c_{11}+a_{21}b_{21}c_{11}+a_{11}b_{12}c_{21}+a_{12}b_{22}c_{21}& a_{11}b_{11}c_{12}+a_{12}b_{21}c_{12}+a_{11}b_{12}c_{22}+a_{12}b_{22}c_{22}\\ a_{21}b_{11}c_{11}+a_{22}b_{21}c_{11}+a_{21}b_{12}c_{21}+a_{22}b_{22}c_{21}& a_{21}b_{11}c_{12}+a_{22}b_{21}c_{12}+a_{21}b_{12}c_{22}+a_{22}b_{22}c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}(b_{11}c_{11}+b_{12}c_{21})+a_{12}(b_{21}c_{11}+b_{12}c_{21})& a_{11}(b_{11}c_{12}+b_{12}c_{22})+a_{12}(b_{21}c_{12}+b_{22}c_{22})\\ a_{21}(b_{11}c_{11}+b_{12}c_{21})+a_{22}(b_{21}c_{11}+b_{12}c_{21})& a_{21}(b_{11}c_{12}+b_{12}c_{22})+a_{22}(b_{21}c_{12}+b_{22}c_{21})\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}b_{11}c_{11}+b_{12}c_{21}& b_{11}c_{12}+b_{12}c_{22}\\ b_{21}c_{11}+b_{22}c_{21}& b_{21}c_{12}+b_{22}c_{22}\end{bmatrix}\)
\( =A\;\left( \begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix} \begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix} \right)\)
\( =A(BC)\)
As \( 1\) is neutral and \( 0\) is absorbent for the multiplication in \( \mathbb{C}\) , we have:
\( IA =\begin{bmatrix}1&0\\ 0&1\end{bmatrix} \begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} =\begin{bmatrix}1\times a_{11}+0\times a_{21}& 1\times a_{12}+0\times a_{22}\\ 0\times a_{11}+1\times a_{21}& 0\times a_{12}+1\times a_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} =A\)
\( AI =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}1&0\\ 0&1\end{bmatrix} =\begin{bmatrix}a_{11}\times 1+a_{12}\times 0& a_{11}\times 0+a_{12}\times 1\\ a_{21}\times 1+a_{22}\times 0& a_{21}\times 0+a_{22}\times 1\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} =A\)
Moreover, as \( 0\) is absorbent for the multiplication in \( \mathbb{C}\) , we have:
\( AO_{22} =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}0&0\\ 0&0\end{bmatrix} =\begin{bmatrix}a_{11}\times 0+a_{12}\times 0& a_{11}\times 0+a_{12}\times 0\\ a_{21}\times 0+a_{22}\times 0& a_{21}\times 0+a_{22}\times 0\end{bmatrix}\)
\( =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} =O_{22}\)
\( O_{22}A =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} \begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} =\begin{bmatrix}0\times a_{11}+0\times a_{21}& 0\times a_{12}+0\times a_{22}\\ 0\times a_{11}+0\times a_{21}& 0\times a_{12}+0\times a_{22}\end{bmatrix}\)
\( =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} =O_{22}\)
5.11 Mutual Properties of the Matrix Multiplication and the Addition of \( 2\times 2\) Matrices with Complex Elements
Theorem 16
Assume that \( (A,B,C)\in(\mathbb{M}_{22}^{\mathbb{C}})^3\) are \( 2\times 2\) matrices with complex elements.
Then \( (\mathbb{M}_{22}^{\mathbb{C}}\;,+,\times)\) is a unitary ring because of the following properties of the matrix multiplication and the addition of \( 2\times 2\) matrices with complex elements.
\( (\mathbb{M}_{22}^{\mathbb{C}}\;,+)\) is a commutative group.
\( (\mathbb{M}_{22}^{\mathbb{C}}\;,\times)\) is a unitary monoid.
Distributivity to the left: \( A(B+C)=AB+AC\) .
Distributivity to the right: \( (A+B)C=AC+BC\) .
Proof
Assume that \( (A,B,C)\in(\mathbb{M}_{22}^{\mathbb{C}})^3\) are \( 2\times 2\) matrices with complex elements, with \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) , \( B=\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix}\) , and \( C=\begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix}\) .
Then the following calculations may be performed.
\( (\mathbb{M}_{22}^{\mathbb{C}}\;,+)\) is a commutative group because of the theorem 10.
\( (\mathbb{M}_{22}^{\mathbb{C}}\;,\times)\) is a unitary monoid because of the theorem 15.
As the multiplication is distributive to the left on the addition, and the addition is associative and commutative in \( \mathbb{C}\) , we have:
\( A(B+C) =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}b_{11}+c_{11}&b_{12}+c_{12}\\ b_{21}+c_{21}&b_{22}+c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}(b_{11}+c_{11})+a_{12}(b_{21}+c_{21})& a_{11}(b_{12}+c_{12})+a_{12}(b_{22}+c_{22}\\ a_{21}(b_{11}+c_{11})+a_{22}(b_{21}+c_{21})& a_{21}(b_{12}+c_{12})+a_{22}(b_{22}+c_{22})\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}b_{11}+a_{11}c_{11}+a_{12}b_{21}+a_{12}c_{21}& a_{11}b_{12}+a_{11}c_{12}+a_{12}b_{22}+a_{12}c_{22}\\ a_{21}b_{11}+a_{21}c_{11}+a_{22}b_{21}+a_{22}c_{21}& a_{21}b_{12}+a_{21}c_{12}+a_{22}b_{22}+a_{22}c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}& a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}& a_{21}b_{12}+a_{22}b_{22}\end{bmatrix} +\begin{bmatrix}+a_{11}c_{11}+a_{12}c_{21}& a_{11}c_{12}+a_{12}c_{22}\\ a_{21}c_{11}+a_{22}c_{21}& a_{21}c_{12}+a_{22}c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix} +\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix}\)
\( =AB+AC\) .
As the multiplication is distributive to the right on the addition, and the addition is associative and commutative in \( \mathbb{C}\) , we have:
\( (A+B)C \begin{bmatrix}a_{11}+b_{11}&a_{12}+b_{12}\\ a_{21}+b_{21}&a_{22}+b_{22}\end{bmatrix} \begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix} \)
\( =\begin{bmatrix}(a_{11}+b_{11})c_{11}+(a_{12}+b_{12})c_{21}& (a_{11}+b_{11})c_{12}+(a_{12}+b_{12})c_{22}\\ (a_{21}+b_{21})c_{11}+(a_{22}+b_{22})c_{21}& (a_{21}+b_{21}c_{12}+(a_{22}+b_{22})c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}c_{11}+b_{11}c_{11}+a_{12}c_{21}+b_{12}c_{21}& a_{11}c_{12}+b_{11}c_{12}+a_{12}c_{22}+b_{12}c_{22}\\ a_{21}c_{11}+b_{21}c_{11}+a_{22}c_{21}+b_{22}c_{21}& a_{21}c_{12}+b_{21}c_{12}+a_{22}c_{22}+b_{22}c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}c_{11}+a_{12}c_{21}& a_{11}c_{12}+a_{12}c_{22}\\ a_{21}c_{11}+a_{22}c_{21}& a_{21}c_{12}+a_{22}c_{22}\end{bmatrix} +\begin{bmatrix}a_{11}c_{11}+b_{11}c_{11}+a_{12}c_{21}+b_{12}c_{21}& b_{11}c_{12}+b_{12}c_{22}\\ b_{21}c_{11}+b_{22}c_{21}& b_{21}c_{12}+b_{22}c_{22}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix} +\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix} \begin{bmatrix}c_{11}&c_{12}\\ c_{21}&c_{22}\end{bmatrix} \)
\( =AC+BC\) .
5.12 Mutual Properties of the Matrix Multiplication, the Addition and the Scalar Multiplication of \( 2\times 2\) Matrices with Complex Elements
Theorem 17
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})\) are \( 2\times 2\) matrices with complex elements and that \( (k,l)\in\mathbb{C}^2\) are complex scalars.
Then \( (\mathbb{M}_{22}^{\mathbb{C}}\;,+,\times,\cdot)\) is a unitary algebra because of the following properties of the matrix multiplication, the addition and the scalar multiplication of \( 2\times 2\) matrices with complex elements.
\( (\mathbb{M}_{22}^{\mathbb{C}}\;,+,\cdot)\) is a vector space on \( \mathbb{C}\) .
\( (\mathbb{M}_{22}^{\mathbb{C}}\;,+,\times)\) is a unitary ring.
Associativity law: \( (kA)(lB)=(kl)(AB)\) .
Proof
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})\) are \( 2\times 2\) matrices with complex elements and that \( (k,l)\in\mathbb{C}^2\) are complex scalars.
Assume that \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) and \( B=\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix}\) .
Then the following calculations may be performed.
\( (\mathbb{M}_{22}^{\mathbb{C}}\;,+,\cdot)\) is a vector space on \( \mathbb{C}\) because of the theorem 21.
\( (\mathbb{M}_{22}^{\mathbb{C}}\;,+,\times)\) is a unitary ring because of the theorem 16.
As the multiplication is distributive to the left on the addition, and the addition is associative and commutative in \( \mathbb{C}\) , we have:
\( (kA)(lB) =\begin{bmatrix}ka_{11}&ka_{12}\\ ka_{21}&ka_{22}\end{bmatrix} \begin{bmatrix}lb_{11}&lb_{12}\\ lb_{21}&lb_{22}\end{bmatrix}\)
\( =\begin{bmatrix}(ka_{11})(lb_{11})+(ka_{12})(lb_{21})& (ka_{21})(lb_{11})+(ka_{22})(lb_{21})\\ (ka_{21})(lb_{11})+(ka_{22})(lb_{21})& (ka_{21})(lb_{21})+(ka_{22})(lb_{22})\end{bmatrix}\)
\( =\begin{bmatrix}(kl)(a_{11}b_{11})+(kl)(a_{12}b_{21})& (kl)(a_{21}b_{11})+(kl)(a_{22}b_{21})\\ (kl)(a_{21}b_{11})+(kl)(a_{22}b_{21})& (kl)(a_{21}b_{21})+(kl)(a_{22}b_{22})\end{bmatrix}\)
\( =\begin{bmatrix}(kl)(a_{11}b_{11}+a_{12}b_{21})& (kl)(a_{21}b_{11}+a_{22}b_{21})\\ (kl)(a_{21}b_{11})+a_{22}b_{21})& (kl)(a_{21}b_{21}+a_{22}b_{22})\end{bmatrix}\)
\( =(kl)\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}& a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}& a_{21}b_{12}+a_{22}b_{22}\end{bmatrix}\)
\( =(kl)\left( \begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix} \right)\)
\( =(kl)(AB)\) .
5.13 Determinant and Inverse of a \( 2\times 2\) Matrix with Complex Elements
Definition 8
Assume that \( (a,b,c,d)\in\mathbb{C}^4\) are complex numbers.
Consider the \( 2\times 2\) matrix with complex elements \( A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) .
Then the determinant of the matrix \( A\) is defined as the complex number \( \det(A)=ad-bc\) .
And if \( \det(A)\ne 0\) , then \( A\) is said to be invertible and its inverse is defined as \( A^{-1}=\frac{1}{ad-cb}\begin{bmatrix}d&-b\\ -c&a\end{bmatrix}\) .
Note that these are the same formulae as for \( 2\times 2\) matrices with real elements.
5.14 Properties of the Inverses of \( 2\times 2\) Matrices with Complex Elements
Theorem 18
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})\) are \( 2\times 2\) matrices with complex elements such that \( \det(A)\ne 0\) and \( \det(B)\ne 0\) .
Then the following assertions hold for the inverses of matrices.
\( A^{-1}A=AA^{-1}=I\) , the identity matrix.
\( \det(AB)\ne 0\) and \( (AB)^{-1}=B^{-1}A^{-1}\) .
Lemma 5
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})\) are \( 2\times 2\) matrices with complex elements. Then \( \det(AB)=\det(A)\det(B)\) .
Proof (of the lemma 5)
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})\) are \( 2\times 2\) matrices with complex elements, with \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) and \( B=\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix}\) .
Then the following calculations may be performed, using the fact that the multiplication is distributive on the addition, and the addition and the multoplication are associative and commutative in \( \mathbb{C}\) .
Then the following formulae hold.
\( AB=\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix}\)
\( \det(A)=a_{11}a_{22}-a_{12}a_{21}\)
\( \det(B)=b_{11}b_{22}-b_{12}b_{21}\)
Consequently, as the multiplication is distributive to the left and to the right on the addition, and the addition and the multiplication are associative and commutative in \( \mathbb{C}\) , we have:
\( \det(AB)=(a_{11}b_{11}+a_{12}b_{21})(a_{21}b_{12}+a_{22}b_{22})\)
\( -(a_{11}b_{12}+a_{12}b_{21})(a_{21}b_{11}+a_{22}b_{22})\)
\( =a_{11}b_{11}a_{21}b_{12}+a_{11}b_{11}a_{22}b_{22}+a_{12}b_{21}a_{21}b_{12}+a_{12}b_{21}a_{22}b_{22}\)
\( -a_{11}b_{12}a_{21}b_{11}-a_{11}b_{12}a_{22}b_{21}-a_{12}b_{22}a_{21}b_{11}-a_{12}b_{21}a_{22}b_{22}\)
\( =a_{11}a_{12}b_{11}(b_{12}-b_{12})+a_{11}a_{22}(b_{11}b_{22}-b_{12}b_{21})\)
\( +a_{12}a_{21}(b_{21}b_{12}-b_{22}b_{11})+a_{12}a_{22}b_{22}(b_{21}-b_{21})\)
\( =a_{11}a_{22}(b_{11}b_{22}-b_{12}b_{21})-a_{21}a_{22}(b_{11}b_{22}-b_{12}b_{21})\)
\( =(a_{11}a_{22}-a_{12}a_{21})(b_{11}b_{22}-b_{12}b_{21})\)
\( =\det(A)\det(B)\)
Proof (of the theorem 18)
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})\) are \( 2\times 2\) matrices with complex elements such that \( \det(A)\ne 0\) and \( \det(B)\ne 0\) .
Then \( A\) and \( B\) are invertible.
Assume that \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) and \( B=\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix}\) .
Consider the product \( C\) of \( A\) and \( B\) :
\( C=AB=\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix}\) .
Because of the lemma 5, we have \( \det(C)\ne 0\) and \( C\) is invertible.
Consider the following matrices, the “adjoint” of \( A\) and \( B\) respectively:
\( A^{*}=\begin{bmatrix}a_{22}&-a_{12}\\ -a_{21}&a_{11}\end{bmatrix}\) ,
\( B^{*}=\begin{bmatrix}b_{22}&-b_{12}\\ -b_{21}&b_{11}\end{bmatrix}\) ,
and \( C^{*}=\begin{bmatrix}a_{21}b_{12}+a_{22}b_{22}&-(a_{11}b_{12}+a_{12}b_{22})\\ -(a_{21}b_{11}+a_{22}b_{21})&a_{11}b_{11}+a_{12}b_{21}\end{bmatrix}\) .
Then by definition of the inverses of matrices, we have:
\( A^{-1}=\frac{1}{\det(A)}A^{*}\) ,
\( B^{-1}=\frac{1}{\det(B)}B^{*}\) ,
and \( C^{-1}=\frac{1}{\det(C)}C^{*}\) .
Consequently, because of the associativity law of the theorem 17, we have:
\( B^{-1}A^{-1}=\frac{1}{\det(A)\det(B)}B^{*}A^{*}\) .
As the lemma 5 says that \( \det(A)\det(B)=\det(AB)=\det(C)\) , and because \( C^{-1}=\frac{1}{\det(C)}C^{*}\) , we have to prove that \( C^{*}=B^{*}A^{*}\) .
But, as the addtion and the multiplication are commutative in \( \mathbb{C}\) , we have:
\( B^{*}A^{*} =\begin{bmatrix}b_{22}&-b_{12}\\ -b_{21}&b_{11}\end{bmatrix} \begin{bmatrix}a_{22}&-a_{12}\\ -a_{21}&a_{11}\end{bmatrix}\)
\( =\begin{bmatrix}b_{22}a_{22}+(-b_{12})(-a_{21})& b_{22}(-a_{12})+(-b_{12})a_{11}\\ (-b_{21})a_{22}+b_{11}(-a_{21})& (-b_{21})(-a_{12})+b_{11}a_{11}\end{bmatrix}\)
\( =\begin{bmatrix}a_{21}b_{12}+a_{22}b_{22}&-(a_{11}b_{12}+a_{12}b_{22})\\ -(a_{21}b_{11}+a_{22}b_{21})&a_{11}b_{11}+a_{12}b_{21}\end{bmatrix}\)
\( =C^{*}\)
5.15 Column Vectors with \( 2\) Complex Elements
Definition 9
Assume that \( (e,f)\in\mathbb{C}^2\) are complex numbers.
Then the column vector with \( 2\) complex elements \( (e,f)\) is:
\( U=\begin{bmatrix}e\\ f\end{bmatrix}\) .
If \( (e,f)\in\mathbb{R}^2\) are real numbers, it is also a column vector with \( 2\) real elements.
We note \( \mathbb{M}_{21}^{\mathbb{C}}\) the set of column vectors with \( 2\) complex elements.
5.16 Add and Subtract Column Vectors with \( 2\) Complex Elements
Definition 10
Assume that \( (e,f,g,h)\in\mathbb{C}^4\) are complex numbers.
Then the sum \( W_{+}=U+V\) of \( U\) and \( V\) is defined as the column vector with \( 2\) complex elements:
\( W_+=\begin{bmatrix}e+g\\ f+h\end{bmatrix}\)
And the difference \( W_{-}=U-V\) of \( U\) and \( V\) is defined as the column vector with \( 2\) complex elements:
\( W_-=\begin{bmatrix}e-g\\ f-h\end{bmatrix}\)
These are the same formulae as for column vectors with \( 2\) real elements.
5.17 Properties of the Addition of Column Vectors with \( 2\) Complex Elements
Theorem 19
Assume that \( (U,V,W)\in(\mathbb{M}_{21}^{\mathbb{C}})^3\) are column vectors with \( 2\) complex elements, with \( U=\begin{bmatrix}e\\ f\end{bmatrix}\) .
Then \( (\mathbb{M}_{21}^{\mathbb{C}}\;,+)\) is a commutative group because of the following properties of the addition of column vectors with \( 2\) complex elements.
Commutativity: \( U+V=V+U\) .
Associativity: \( (U+V)+W=U+(V+W)\)
The null column vector \( O_{21}=\begin{bmatrix}0\\ 0\end{bmatrix}\) is neutral: \( U+O_{21}=O_{21}+U=U\) .
If we define the opposite \( -U\) of \( U\) as \( -U=\begin{bmatrix}-e\\ -f\end{bmatrix}\) , then \( U+(-U)=(-U)+U=O_{21}\) .
The last formula justifies the fact that we call \( -U\) the opposite of \( U\) .
Proof
Assume that \( (U,V,W)\in(\mathbb{M}_{21}^{\mathbb{C}})^3\) are column vectors with \( 2\) complex elements, with \( U=\begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix}\) , \( V=\begin{bmatrix}v_{1}\\ v_{2}\end{bmatrix}\) , and \( W=\begin{bmatrix}w_{1}\\ w_{2}\end{bmatrix}\) .
Then the following calculations may be performed.
As the addition is commutative in \( \mathbb{C}\) , we have:
\( U+V =\begin{bmatrix}u_{1}+v_{1}\\ u_{2}+v_{2}\end{bmatrix} =\begin{bmatrix}v_{1}+u_{1}\\ v_{2}+u_{2}\end{bmatrix} =V+U\)
As the addition is associative in \( \mathbb{C}\) , we have:
\( ((U+V)+W =\begin{bmatrix}u_{1}+v_{1}\\ u_{2}+v_{2}\end{bmatrix} +\begin{bmatrix}w_{1}\\ w_{2}\end{bmatrix}\)
\( =\begin{bmatrix}(u_{1}+v_{1})+w_{1}\\ (u_{2}+v_{2})+w_{2}\end{bmatrix}\)
\( =\begin{bmatrix}u_{1}+(v_{1}+w_{1})\\ u_{2}+(v_{2}+w_{2})\end{bmatrix}\)
\( =\begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix} +\begin{bmatrix}v_{1}+w_{1}\\ v_{2}+w_{2}\end{bmatrix}\)
\( =U+(V+W)\) .
As \( 0\) is neutral for the addition in \( \mathbb{C}\) , we have:
\( U+O_{21} =\begin{bmatrix}u_{1}+0\\ u_{2}+0\end{bmatrix} =\begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix} =U\) .
\( O_{21}+U =\begin{bmatrix}0+u_{1}\\ 0+u_{2}\end{bmatrix} =\begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix} =U\) .
As for any complex number \( z\in\mathbb{C}\) , \( z+(-z)=(-z)+z=0\) , we have:
\( U+(-U) =\begin{bmatrix}u_{1}+(-u_{1})\\ y_{2}+(-u_{2}\end{bmatrix} =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} =O_{21}\) .
\( (-U)+U =\begin{bmatrix}(-a_{11})+a_{11}&(-a_{12})+a_{12}\\ (-a_{21})+a_{21}&(-a_{22})+a_{22}\end{bmatrix} =\begin{bmatrix}0\\ 0\end{bmatrix} =O_{21}\) .
5.18 Multiply a Column Vector with \( 2\) Complex Elements by a Complex Scalar
Definition 11
Assume that \( (e,f,k)\in\mathbb{C}^3\) are complex numbers.
Consider the column vector with \( 2\) complex elements \( U=\begin{bmatrix}e\\ f\end{bmatrix}\) .
Then the product \( T=kU\) of \( U\) by the complex scalar \( k\) is defined as the column vector with \( 2\) complex elements:
\( T=\begin{bmatrix}ke\\ kf\end{bmatrix}\) .
It is the same formula as for columns vectors with \( 2\) real elements and real scalars.
5.19 Properties of the Scalar Multiplication of Column Vectors with \( 2\) Complex Elements
Theorem 20
Assume that \( U\in\mathbb{M}_{21}^{\mathbb{C}}\) is a column vector with \( 2\) complex elements and that \( k\in\mathbb{C}\) is a complex scalar.
Then the following properties hold for the scalar multiplication.
\( 1\times U=U\) .
\( 0\times U=O_{21}\) .
\( k\times O_{21}=O_{21}\)
\( k(-A)=-kA\)
\( (-k)A=-kA\)
Proof
Assume that \( U\in\mathbb{M}_{21}^{\mathbb{C}}\) is a column vector with \( 2\) complex elements, with \( U=\begin{bmatrix}e\\ f\end{bmatrix}\) , and that \( k\in\mathbb{C}\) is a complex scalar.
Then the following properties hold for the scalar multiplication.
As \( 1\) is neutral for the multiplication in \( \mathbb{C}\) , we have:
\( 1\times U =\begin{bmatrix}1\times e\\ 1\times f\end{bmatrix} =\begin{bmatrix}e\\ f\end{bmatrix} =U\) .
As \( 0\) is absorbent for the multiplication in \( \mathbb{C}\) , we have:
\( 0\times U =\begin{bmatrix}0\times e\\ 0\times f\end{bmatrix} =\begin{bmatrix}0\\ 0\end{bmatrix} =O_{21}\) .
As \( 0\) is absorbent for the multiplication in \( \mathbb{C}\) , we have:
\( k\times O_{21} =\begin{bmatrix}k\times 0\\ k\times 0\end{bmatrix} =\begin{bmatrix}0\\ 0\end{bmatrix} =O_{21}\)
Because of the item (I) of the lemma 4, we have:
\( k(-U) =\begin{bmatrix}k(-e)\\ k(-cf)\end{bmatrix} =\begin{bmatrix}-ke\\ -kf\end{bmatrix} =-kU\)
Because of the item (II) of the lemma 4, we have:
\( (-k)U =\begin{bmatrix}(-k)e\\ (-k)f\end{bmatrix} =\begin{bmatrix}-ke\\ -kf\end{bmatrix} =-kU\)
5.20 Mutual Properties of the Addition and the Scalar Multiplication of Column Vectors with \( 2\) Complex Elements
Theorem 21
Assume that \( (U,V)\in(\mathbb{M}_{21}^{\mathbb{C}})^2\) are column vectors with \( 2\) complex elements and that \( (k,l)\in\mathbb{C}^2\) are complex scalars.
Then \( (\mathbb{M}_{21}^{\mathbb{C}}\;,+,\cdot)\) is a vector space on \( \mathbb{C}\) because of the following properties of the addition and the scalar multiplication.
\( (\mathbb{M}_{21}^{\mathbb{C}}\;,+)\) is a commutative group.
First distributivity law: \( k(U+V)=kU+kV\) .
Second distributivity law: \( (k+l)U=kU+lU\) .
Associativity law: \( k(lU)=(kl)U\) .
Proof
Assume that \( (U,V)\in(\mathbb{M}_{21}^{\mathbb{C}})^2\) are column vectors with \( 2\) complex elements, with \( U=\begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix}\) and \( V=\begin{bmatrix}v_{1}\\ v_{2}\end{bmatrix}\) , and that \( (k,l)\in\mathbb{C}^2\) are complex scalars.
Then the following calculations may be performed.
As stated by the theorem 19, \( (\mathbb{M}_{21}^{\mathbb{C}}\;,+)\) is a commutative group.
Because the multiplication is distributive to the left on the addition in \( \mathbb{C}\) , we have:
\( k(U+V) =\begin{bmatrix}k(u_{1}+v_{1})\\ k(u_{2}+v_{2})\end{bmatrix}\)
\( =\begin{bmatrix}ku_{1}+kv_{1}\\ ku_{2}+kv_{2}\end{bmatrix} =kU+kV\) .
Because the multiplication is distributive to the right on the addition in \( \mathbb{C}\) , we have:
\( (k+l)U =\begin{bmatrix}(k+l)u_{1}\\ (k+l)au_{2}\end{bmatrix}\)
\( =\begin{bmatrix}ku_{1}+lu_{1}\\ ku_{2}+lu_{2}\end{bmatrix} =kU+lU\) .
Because the multiplication is associative in \( \mathbb{C}\) , we have:
\( k(lU) =\begin{bmatrix}k(lu_{1})\\ k(lu_{2})\end{bmatrix} =\begin{bmatrix}(kl)u_{1}\\ (kl)u_{2}\end{bmatrix} =(kl)U\) .
5.21 Matrix Multiplication of a \( 2\times 2\) Matrix with Complex Elements and a Column Vector with \( 2\) Complex Elements
Definition 12
Assume that \( (a,b,c,d,e,f)\in\mathbb{C}^8\) are complex numbers.
Consider the \( 2\times 2\) matrix with complex elements \( A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) and the column vector with \( 2\) complex elements \( U=\begin{bmatrix}e\\ f\end{bmatrix}\) .
Then the result of the matrix multiplication \( H=AU\) is defined as the column vector with complex elements:
\( H=\begin{bmatrix}ae+bf\\ ce+df\end{bmatrix}\) .
It is the same formula as for \( 2\times 2\) matrices with real elements and column vectors with \( 2\) real elements.
5.22 Properties of the Matrix Multiplication of a \( 2\times 2\) Matrix with Complex Elements and a Column Vector with \( 2\) Complex Elements
Theorem 22
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})^2\) are \( 2\times 2\) matrices with complex elements and that \( U\in\mathbb{M}_{21}^{\mathbb{C}}\) is a column vector with \( 2\) complex elements.
Then the following properties hold.
Associativity: \( (AB)U=A(BU)\) .
The identity matrix \( I=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}\) is neutral: \( IU=U\) .
The null matrix \( O_{22}=\begin{bmatrix}0&0\\ 0&0\end{bmatrix}\) is absorbent: \( O_{22}U=O_{21}\) .
The null column vector \( O_{21}=\begin{bmatrix}0\\ 0\end{bmatrix}\) is absorbent: \( AO_{21}=O_{21}\) .
Proof
Assume that \( (A,B)\in(\mathbb{M}_{22}^{\mathbb{C}})^2\) are \( 2\times 2\) matrices with complex elements, with \( A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}\) and \( B=\begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix}\) .
And assume that \( U\in\mathbb{M}_{21}^{\mathbb{C}}\) is a column vector with \( 2\) complex elements, with \( U=\begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix}\) .
Then the following calculations may be performed.
As the multiplication is distributive to the right and to the left on the addition, and the addition is associative and commutative in \( \mathbb{C}\) , we have:
\( (AB)U =\left( \begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix} \right) \begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix} \begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix}\)
\( =\begin{bmatrix}(a_{11}b_{11}+a_{21}b_{21})u_{1}+(a_{11}b_{12}+a_{12}b_{22})u_{2}\\ (a_{21}b_{11}+a_{22}b_{21})u_{1}+(a_{21}b_{12}+a_{22}b_{22})u_{2}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}b_{11}u_{1}+a_{21}b_{21}u_{1}+a_{11}b_{12}u_{2}+a_{12}b_{22}u_{2}\\ a_{21}b_{11}u_{1}+a_{22}b_{21}u_{1}+a_{21}b_{12}u_{2}+a_{22}b_{22}u_{2}\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}(b_{11}u_{1}+b_{12}u_{2})+a_{12}(b_{21}u_{1}+b_{12}u_{2})\\ a_{21}(b_{11}u_{1}+b_{12}u_{2})+a_{22}(b_{21}u_{1}+b_{12}u_{2})\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}b_{11}u_{1}+b_{12}u_{2}\\ b_{21}u_{1}+b_{22}u_{2}\end{bmatrix}\)
\( =A\;\left( \begin{bmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{bmatrix} \begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix} \right)\)
\( =A(BU)\)
As \( 1\) is neutral and \( 0\) is absorbent for the multiplication in \( \mathbb{C}\) , we have:
\( IU=\begin{bmatrix}1&0\\ 0&1\end{bmatrix} \begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix} =\begin{bmatrix}1\times u_{1}+0\times u_{2}\\ 0\times u_{1}+1\times u_{2}\end{bmatrix} =\begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix} =U\)
As \( 0\) is absorbent for the multiplication in \( \mathbb{C}\) , we have
\( O_{22}U =\begin{bmatrix}0&0\\ 0&0\end{bmatrix} \begin{bmatrix}u_{1}\\ u_{2}\end{bmatrix} =\begin{bmatrix}0\times u_{1}+0\times u_{2}\\ 0\times u_{1}+0\times u_{2}\end{bmatrix} =\begin{bmatrix}0\\ 0\end{bmatrix} =O_{21}\)
\( AO_{21} =\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}0\\ 0\end{bmatrix} =\begin{bmatrix}a_{11}\times 0+a_{12}\times 0\\ a_{21}\times 0+a_{22}\times 0\end{bmatrix} =\begin{bmatrix}0\\ 0\end{bmatrix} =O_{21}\)
6 \( 2\times 2\) Linear Systems in \( \mathbb{C}\)
6.1 Define a \( 2\times 2\) Linear Systems in \( \mathbb{C}\)
Definition 13
Assume that \( (a,b,c,d,e,f)\in\mathbb{C}^6\) are complex numbers.
Then solving the system of \( 2\) linear equations of \( 2\) complex variables:
(1)
is finding the couple(s) of complex numbers \( (x,y)\in\mathbb{C}^2\) that fulfil both equations of the system together.
6.2 Solve a \( 2\times 2\) Linear System in \( \mathbb{C}\) with a Non Zero Determinant
Let’s try to solve the linear system 1.
6.2.1 First case: \( a=0\)
If \( a=0\) , then \( b\neq 0\) , \( c\) and \( d\) are not zero together, and the system becomes:
(2)
As \( b\neq 0\) , the first equation \( by=e\) has a unique solution in , and that solution is:
(3)
Then we may substitute \( y\) with its value \( y=\frac{e}{b}\) in the second equation \( y=\frac{e}{b}\) .
The variable \( y\) is eliminated, and we obtain the equation in \( x\) :
(4)
The equation 4 is equivalent to the equation in \( x\) \( cx=f-\frac{de}{b}=\frac{fb}{b}-\frac{de}{b}=\frac{bf-de}{b} \)
That equation has a unique solution in \( x\) if and only if \( c\neq 0\) , and that solution is:
(5)
Consequently, the \( 2\times 2\) linear system 2 has a unique solution in \( (x,y)\in\mathbb{R}^2\) if and only if \( c\neq 0\) , that is equivalent to \( bc\neq 0\) (because \( b\neq 0\) ), and that solution is:
(6)
6.2.2 Alternate case: \( a\neq 0\)
If \( a\neq 0\) , then we may solve the first equation \( ax+b=e\) in \( x\) .
It becomes: \( ax=e-by\) , leading to \( x=\frac{e-by}{a}\) .
The solution depends on and is:
(7)
Then we may substitute \( x\) with its value \( x=\frac{e}{a}-\frac{b}{a}y\) in the second equation \( cx+dy=f\) .
The variable \( x\) is eliminated, and we obtain the equation in \( y\) :
(8)
Then we may distribute \( c\) and put \( y\) into factor in the first member of the equation 8, to obtain:
(9)
If we multiply both members of the equality by \( a\) , that is non zero, the equation 9 is equivalent to:
(10)
That equation has a unique solution in \( y\in\mathbb{R}\) if and only if \( ad-bc\neq 0\) , and that solution is:
(11)
Then we may substitute with its value in equation 11, in the equation 7 goving the value of \( x\) as a function of \( x\) :
(12)
The solution in \( x\) of the system 1 is the following (where we find again the condition \( ad-bc\neq 0\) ):
(13)
Consequently, when \( a\neq 0\) the \( 2\times 2\) linear system 1 has a unique solution in \( (x,y)\in\mathbb{R}^2\) if and only if \( ad-bc\neq 0\) , and that solution is:
(14)
6.2.3 Revisit the case \( a=0\)
If \( a=0\) the following facts hold with the solution in the case \( a\neq 0\) :
The condition \( ad-bc\neq 0\) becomes \( bc\neq0\) .
It is the condition to have a unique solution when \( a=0\) .
The solution when \( ad-bc\neq 0\) , that is \( (x,y)=\left(\frac{de-bf}{ad-bc},\;\;\frac{af-ce}{ad-bc}\right)\) becomes \( (x,y)=\left(\frac{bf-de}{bc},\frac{e}{b}\right)\) .
It is the solution we found in the case \( a=0\) when \( bc\neq 0\) .
Consequently, we may get rid of the condition about \( a\) , and set the theorem below.
6.2.4 The generic \( 2\times 2\) Linear System in \( \mathbb{C}\) with a Non Zero Determinant
Theorem 23
Assume \( (a,b,c,d,e,f)\in\mathbb{C}^6\) are real numbers such that \( a\) and \( b\) are not zero together, and \( c\) and \( d\) are not zero together. Then the system of 2 linear equations of 2 real variables:
has a unique solution in \( (x,y)\in\mathbb{R}^2\) if and only if \( ad-bc\neq 0\) , and that solution is:
6.3 Solutions of a \( 2\times 2\) Linear System in \( \mathbb{C}\) with a Zero Determinant
Assume that the coefficients of the linear system 1 are such that \( ad-bc=0\) .
Let’s try to solve the system under that hypothesis.
6.3.1 First case: \( a=b=c=d=0\)
In that case, the equations of the system are equivalent to \( 0=e\) and \( 0=f\) respectively.
Consequently, the system has no solution unless \( e=f=0\) , in which case every \( (x,y)\in\mathbb{R}^{2}\) are solutions of the system, and the system has an infinity of solutions.
6.3.2 Second Case: \( a\ne 0\)
In that case, the first equation of the system is equivalent to \( x=\frac{e-by}{a}\) .
And if we substitute the value of \( x\) as a function of \( y\) in the second equation, we obtain:
\( \left(c\frac{e-by}{a}+dy\right)=f\) .
If we multiply both members of the equality by \( a\ne 0\) , we obtain the equivalent equation:
\( (ad-bc)y+ce=af\) ,
that is equivalent to:
\( (ad-bc)y=af-ce\) ,
i.e. \( 0=af-ce\) because \( ad-bc=0\) .
Consequently, the system has no solution unless \( af-ce=0\) , in which case the system is equivalent to the first equation \( ax+by=e\) , that is the equation of a straight line, and the system has an infinity of solutions.
6.3.3 Third Case Case: \( b\ne 0\)
In that case, the first equation of the system is equivalent to \( y=\frac{e-ax}{b}\) .
And if we substitute the value of \( x\) as a function of \( y\) in the second equation, we obtain:
\( \left(cx+d\frac{e-ax}{b}\right)=f\) .
If we multiply both members of the equality by \( b\ne 0\) , we obtain the equivalent equation:
\( (bc-ad)x+de=bf\) ,
that is equivalent to:
\( (ad-bc)y=de-bf\) ,
i.e. \( 0=de-bf\) because \( ad-bc=0\) .
Consequently, the system has no solution unless \( de-bf=0\) , in which case the system is equivalent to the first equation \( ax+by=e\) , that is the equation of a straight line, and the system has an infinity of solutions.
6.3.4 Fourth Case: \( c\ne 0\)
In that case, the second equation of the system is equivalent to \( x=\frac{f-dy}{c}\) .
And if we substitute the value of \( x\) as a function of \( y\) in the first equation, we obtain:
\( \left(a\frac{f-dy}{c}+by\right)=e\) .
If we multiply both members of the equality by \( c\ne 0\) , we obtain the equivalent equation:
\( (bc-ad)y+af=ce\) ,
that is equivalent to:
\( (ad-bc)y=af-ce\) ,
i.e. \( 0=af-ce\) because \( ad-bc=0\) .
Consequently, the system has no solution unless \( af-ce=0\) , in which case the system is equivalent to the first equation \( ax+by=e\) , that is the equation of a straight line, and the system has an infinity of solutions.
6.3.5 Fifth Case Case: \( d\ne 0\)
In that case, the second equation of the system is equivalent to \( y=\frac{f-cx}{d}\) .
And if we substitute the value of \( x\) as a function of \( y\) in the first equation, we obtain:
\( \left(ax+b\frac{f-cx}{d}\right)=e\) .
If we multiply both members of the equality by \( d\ne 0\) , we obtain the equivalent equation:
\( (ad-bc)x+bf=de\) ,
that is equivalent to:
\( (ad-bc)y=de-bf\) ,
i.e. \( 0=de-bf\) because \( ad-bc=0\) .
Consequently, the system has no solution unless \( de-bf=0\) , in which case the system is equivalent to the first equation \( ax+by=e\) , that is the equation of a straight line, and the system has an infinity of solutions.
6.3.6 The generic \( 2\times 2\) Linear System in \( \mathbb{C}\) with a Non Zero Determinant
Consequently, the theorem below holds.
Theorem 24
Assume \( (a,b,c,d,e,f)\in\mathbb{C}^6\) are real numbers such that
\( ad-bc=0\) . Then the set of solutions of the system of 2 linear equations of 2 real variables:
is either an infinite set, or the void set \( \emptyset\) .
6.4 The Matrix View of a \( 2\times 2\) Linear System in \( \mathbb{C}\)
Theorem 25
Assume that \( (a,b,c,d,e,f)\in\mathbb{C}^6\) are complex numbers, and consider:
the \( 2\times 2\) matrix with complex elements \( A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) ,
and the column vector with \( 2\) complex elements \( B=\begin{bmatrix}e\\ f\end{bmatrix}\) .
Then the \( 2\times 2\) linear system in the complex variables \( (x,y)\in\mathbb{C}^{2}\) :
is equivalent to the matrix equation \( AX=B\) in the unknown column
vector \( X=\begin{bmatrix}x\\ y\end{bmatrix}\) .
This is due to the fact that \( AX=\begin{bmatrix}a&b\\ c&d\end{bmatrix} \begin{bmatrix}x\\ y\end{bmatrix} =\begin{bmatrix}ax+by\\ cx+dy\end{bmatrix}\) .
6.5 Solve a Matrix Equation with a Non Zero Determinant in \( \mathbb{C}\)
Theorem 26
Assume that \( A\in\mathbb{M}_{22}^{\mathbb{C}}\) is a \( 2\times 2\) matrix with complex elements such that \( \det(A)\ne 0\) , and that \( B\in\mathbb{M}_{21}^{\mathbb{C}}\) is a column vector with \( 2\) complex elements.
Then the matrix equation in the column vector with \( 2\) complex elements \( X\in\mathbb{M}_{21}^{\mathbb{C}}\) :
has a unique solution:
This is due to the fact that \( X=IX=(A^{-1}A)X=A^{-1}(AX)=A^{-1}B\) .
6.6 Solve an Homogenous Matrix Equation in \( \mathbb{C}\)
Theorem 27
Assume that \( (a,b,c,d)\in\mathbb{C}^4\) are complex numbers, and consider the matrix \( A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) and the null column vector \( O_{21}=\begin{bmatrix}0\\ 0\end{bmatrix}\) .
Then the solution(s) of the matrix equation \( AX=O_{21}\) is (are) the following.
If \( \det(A)\ne 0\) , then the matrix equation has a unique solution, and it is \( X=O_{21}=\begin{bmatrix}0\\ 0\end{bmatrix}\) .
If \( A\ne O_{22}\) and \( \det(A) = 0\) , then the solutions are the column vectors that are the scalar multiples of \( \begin{bmatrix}-b\\ a\end{bmatrix}\) , unless \( a=b=0\) , in which case the solutions are the column vectors that are the scalar multiples of \( \begin{bmatrix}-d\\ c\end{bmatrix}\) .
And if \( A=O_{22}\) the solutions are all the column vectors \( X=\begin{bmatrix}x\\ y\end{bmatrix}\) , for any \( (x,y)\in\mathbb{C}^2\) .
This is a simple transcription of the following lemma.
Lemma 6
Assume that \( (a,b,c,d)\in\mathbb{R}^4\) are real numbers and consider the homogeneous system:
Then the solution(s) of the homogeneous system is (are) the following.
If \( ad-bc\ne 0\) , then the system has a unique solution, and it is
\( (x,y)=(0,0)\) .
If \( a\) , \( b\) , \( c\) and \( d\) are not all zeros and \( ad-bc=0\) , then the solutions are the couples of complex numbers that are proprtionnal to \( (-b,a)\) , unless \( a=b=0\) , in which case the solutions are couples of complex numbers that are \( (-d,c)\) .
And if \( a=b=c=d=0\) the solutions are all the elements of \( \mathbb{R}^2\) .
Proof (of the lemma 6)
Assume that \( (a,b,c,d)\in\mathbb{C}^4\) are complex numbers and consider the homogeneous system in the complex variables \( (x,y)\in\mathbb{C}^{2}\) :
Then the following calculations may be performed.
If \( ad-bc\ne 0\) , then the determinant of the system is non zero and the system has a unique solution. And as \( (x,y)=(0,0)\) is a solution, then it is the unique solution of the system.
Assume that \( a\) , \( b\) , \( c\) and \( d\) are not all zeros and \( ad-bc=0\) .
Assume that \( a\ne 0\) .
Then the first equation of the system is equivealent to \( x=-\frac{b}{a}y\) .
And if we substitute the value of \( x\) as a function of \( y\) in the second equation, we obtain:
\( \left(-c\frac{b}{a}+d\right)y=0\) .
That is equivalent to:
\( \frac{ad-bc}{a}y=0\) ,
i.e. \( 0=0\) because \( ad-bc=0\) .
As \( 0=0\) is always true, then the system is equivalent to the first equation \( ax+by=0\) .
The solutions are thus the couples of complex numbers that are \( (-b,a)\) .
Assume that \( b\ne 0\) .
Then the first equation of the system is equivalent to \( y=-\frac{a}{b}x\) .
And if we substitute the value of \( y\) as a function of \( x\) in the second equation, we obtain:
\( \left(c-d\frac{a}{b}\right)x=0\) .
That is equivalent to:
\( -\frac{ad-bc}{a}x=0\) ,
i.e. \( 0=0\) because \( ad-bc=0\) .
As \( 0=0\) is always true, then the system is equivalent to the first equation \( ax+by=0\) .
The solutions are thus the couples of complex numbers that are \( (-b,a)\) .
Assume that \( c\ne 0\) .
Then the second equation of the system is equivalent to \( x=-\frac{d}{c}y\) .
And if we substitute the value of \( x\) as a function of \( y\) in the first equation, we obtain:
\( \left(-a\frac{d}{c}+b\right)y=0\) .
That is equivalent to:
\( -\frac{ad-bc}{a}y=0\) ,
i.e. \( 0=0\) because \( ad-bc=0\) .
As \( 0=0\) is always true, then the system is equivalent to the second equation \( cx+dy=0\) .
The solutions are thus the couples of complex numbers that are \( (-d,c)\) .
Assume that \( d\ne 0\) .
Then the second equation of the system is equivalent to \( y=-\frac{c}{d}x\) .
And if we substitute the value of \( x\) as a function of \( y\) in the first equation, we obtain:
\( \left(a-b\frac{c}{d}\right)x=0\) .
That is equivalent to:
\( \frac{ad-bc}{a}x=0\) ,
i.e. \( 0=0\) because \( ad-bc=0\) .
As \( 0=0\) is always true, then the system is equivalent to the second equation \( cx+dy=0\) .
The solutions are thus the couples of complex numbers that are \( (-d,c)\) .
Assume that \( a=b=c=d=0\) .
Then both equations of the system are equivalent to \( 0=0\) , so that every \( (x,y)\in\mathbb{R}^{2}\) are solutions.
7 Diagonalisation in \( \mathbb{C}\) of Real Matrices
The \( 2\times 2\) matrices with real elements that have no real eigenvalues are diagonalisable in \( \mathbb{C}\) .
7.1 Complex Eigenvalues for a \( 2\times 2\) Real Matrix with no real eigenvalue
A \( 2\times 2\) real matrix is the matrix of a linear mapping with no real eigenvalue if and only if the discriminant of its characteristic polynomial is negative.
But if the discriminant of its characteristic polynomial is negative, the characteristic polynomial has two conjugate complex roots.
Consequently, if a \( 2\times 2\) real matrix is the matrix of a linear mapping with no real eigenvalue, it has two conjugate complex eigenvalues.
This is because for a root \( \lambda\in\mathbb{C}\) of the characteristic polynomial of \( A\in\mathbb{M}_{22}^{\mathbb{R}}\) , \( \chi(\lambda)=\det(\lambda I-A)=0\) and thus there exists a column vector \( V\in\mathbb{M}_{21}^{\mathbb{C}}\) such that \( AV=\lambda V\) .
7.2 Complex Eigenvalues and Eigenvectors of a Real Matrix with no Real Eigenvalues
Assume that \( A=\begin{bmatrix} a&b\\ c&d \end{bmatrix}\) is a \( 2\times 2\) matrix with real elements, such that it is the matrix in the canonical basis of a linear mapping with no real eigenvalues.
Consider the characteristic polynomial \( \chi(\lambda)=\lambda^2-(a+d)\lambda+ad-bc\) of \( A\) .
Then the following assertions hold.
The discriminant \( \Delta=(a+d)^2-4(ad-bc)\) of \( \chi(\lambda)\) is negative.
Consequently, \( \chi(\lambda)\) has two conjugate complex roots:
\( \lambda_1=\frac{a+d-i\sqrt{-\Delta}}{2}\) and \( \lambda_2=\overline{\lambda_1}=\frac{a+d+i\sqrt{-\Delta}}{2}\) .
Because \( \chi(\lambda_1)=\det(\lambda_1 I-A)=0\) , then there exists a non zero column vector with \( 2\) complex elements \( V_{1}=\begin{bmatrix} v_{11}\\ v_{21} \end{bmatrix}\) such that \( AV_1=\lambda_1 V_1\) .
This is the extension to the complex numbers of the notions of eigenvalue and eigenvector.
So that we say that \( \lambda_1\) is a complex eigenvalue of \( A\) , and \( V_{1}\) is a complex eigenvector of \( A\) for the eigenvalue \( \lambda_1\) .
And as \( A\) is a matrix with real elements and that \( \lambda_2=\overline{\lambda_1}\) , if we set \( V_2=\begin{bmatrix} \overline{v_{11}}\\ \overline{v_{21}} \end{bmatrix}\) , then \( V_{2}\) is a non zero vector such that \( AV_2=\lambda_2 V_2\) .
So that \( \lambda_2\) is another complex eigenvalue of \( A\) , and \( V_{2}\) is a complex eigenvector of \( A\) for the eigenvalue \( \lambda_2\) .
7.3 Diagonalisation of a Real Matrix with no Real Eigenvalues in \( \mathbb{C}\)
With the notations of the previous peregraph, let’s consider the matrix \( P=\begin{bmatrix} v_{11}&\overline{v_{11}}\\ v_{21}&\overline{v_{21}} \end{bmatrix}\) with columns the column vectors \( V_{1}\) and \( V_{2}\) .
Then the following assertions hold.
The determinant of is equal to
\( \det{P}=v_{11}\overline{v_{21}}-v_{21}\overline{v_{11}}=v_{11}\overline{v_{21}}-\overline{v_{11}\overline{v_{21}}} =2i\;\Im(v_{11}\overline{v_{21}})\) , the imaginary part of \( v_{11}\overline{v_{21}}\) .
But we may choose \( v_{11}=b\) , that is real, and \( v_{21}=\lambda_1-a\) , that is not real, so that \( v_{11}\overline{v_{21}}\) is not real and has a non zero imaginary part.
Consequently, \( \det(P)\ne 0\) , an \( P\) is invertible.
On the other hand, \( AP=\begin{bmatrix} av_{11}+bv_{21}&a\overline{v_{11}}+b\overline{v_{21}} \\ cv_{11}+dv_{21}&c\overline{v_{11}}+d\overline{v_{21}} \end{bmatrix}\) , with the columns \( AV_{1}\) and \( AV_{2}\) .
Let’s consider the diagonal matrix \( D=\begin{bmatrix} \lambda_1&0\\ 0&\lambda_2 \end{bmatrix}\) .
Then \( PD=\begin{bmatrix} \lambda_1 v_{11}& \lambda_2 \overline{v_{11}}\\ \lambda_1 v_{21}& \lambda_2 \overline{v_{21}} \end{bmatrix}\) with the columns \( \lambda_1V_1\) and \( \lambda_2V_2\) .
Consequently, as \( AV_1=\lambda_1 V_1\) and \( AV_2=\lambda_2 V_2\) , we have \( AP=PD\) , and thus \( P^{-1}AP=D\) , that is a diagonal matrix.
We have thus diagonalised the matrix \( A\) in \( \mathbb{C}\) .
8 Conclusion
In that last section of the course of Liear Algebra in the Euclidean Plane, we learned more information about the complex number field \( (\mathbb{C},+,\times)\) , which allowed us to diagonalize in \( \mathbb{C}\) the real matrices with no real eigenvalues.
It is to denote that in Python, the determination of eigenvalues and eigenvectors of matrices is done in \( \mathbb{C}\) .