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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
Now you shall discover the nice properties of some important kinds of matrices.
The symmetric matrices, that are symmetric along their main diagonal.
The skew symmetric matrices, that are anti-symmetric along their main diagonal.
And the unit matrices, the matrices of the linear mappings that preserve the norms of vectors.
2 The Transpose of a Matrix
The transposition of a matrix, consisting in exchanging its rows and columns, will appear as a major tool to define interesting special matrices.
2.1 Definition of the Transpose of a Matrix
Definition 1
Assume that \( (a,b,c,d)\in\mathbb{R}^4\) are real numbers, and consider the matrix \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) .
Then the transpose of the matrix \( A\) is the matrix \( A^T=\begin{bmatrix} a&c\\b&d \end{bmatrix}\) , where the rows and columns of \( A\) are exchanged.
2.2 Properties of the Transposition of Matrices
Theorem 1
Assume that \( A\) and \( B\) are \( 2\times 2\) matrices with real elements, and that \( \lambda\in\mathbb{R}\) is a scalar.
Then the following assertions hold.
The transpose of the transpose of \( A\) is equal to \( A\) : \( (A^T)^T=A\) .
The transpose of the sum of \( A\) and \( B\) is the sum of the transposes of \( A\) and \( B\) : \( (A+B)^T=A^T+B^T\) .
The transpose of the opposite of \( A\) is the opposite of the transpose of \( A\) : \( (-A)^T=-A^T\) .
The transpose of the scalar product of \( A\) by \( \lambda\) is the scalar product of the transpose of \( A\) by \( \lambda\) : \( (\lambda A)^T=\lambda A^T\) .
The transpose of the matrix product \( AB\) is the matrix product \( B^T A^T\) : \( (AB)^T=B^T A^T\) .
If \( A\) is invertible, then its transpose \( A^{T}\) is invertible and its inverse is equal to the transpose of the inverse of \( A\) : \( (A^T)^{-1}=(A^{-1})^T\) .
Proof
Assume that \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) and \( B=\begin{bmatrix} e&f\\g&h \end{bmatrix}\) are \( 2\times 2\) matrices with real elements, and that \( \lambda\in\mathbb{R}\) is a scalar.
Then the following calculations may be done.
\( A^T=\begin{bmatrix} a&c\\b&d \end{bmatrix}\) and \( (A^T)^T=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) = \( A\) .
\( A+B=\begin{bmatrix} a+e&b+f\\c+g&d+h \end{bmatrix}\) ,
so that \( (A+B)^{T}=\begin{bmatrix} a+e&c+g\\b+f&d+h \end{bmatrix} =\begin{bmatrix} a&c\\b&d \end{bmatrix}+\begin{bmatrix} e&g\\f&h \end{bmatrix} =A^{T}+B^{T}\) .
\( -A=\begin{bmatrix} -a&-b\\{-c}&-d \end{bmatrix}\) , so that \( (-A)^T=\begin{bmatrix} -a&-c\\{-b}&-d \end{bmatrix} =-\begin{bmatrix} a&c\\b&d \end{bmatrix}=-A^{T}\) .
\( \lambda A=\begin{bmatrix} \lambda a&\lambda b\\\lambda c&\lambda d \end{bmatrix}\) , so that \( (\lambda A)^T = \begin{bmatrix} \lambda a&\lambda c\\\lambda b&\lambda d \end{bmatrix} =\lambda \begin{bmatrix} a&c\\b&d \end{bmatrix} =\lambda A^T\) .
\( AB=\begin{bmatrix} a&b\\c&d \end{bmatrix}\begin{bmatrix} e&f\\g&h \end{bmatrix} =\begin{bmatrix} ae+bg&af+bh\\ce+dg&cf+dh \end{bmatrix}\) , so that
\( (AB)^T=\begin{bmatrix} ae+bg&ce+dg\\af+bh&cf+dh \end{bmatrix}\) .
And \( B^T A^T=\begin{bmatrix} e&g\\f&h \end{bmatrix}\begin{bmatrix} a&c\\b&d \end{bmatrix} =\begin{bmatrix} ea+gb&ec+gd\\fa+hb&fc+hd \end{bmatrix} =(AB)^T\) .
Assume that \( A\) is invertible. Then \( \det(A)=ad-bc\ne 0\) .
But \( \det(A^{T})=ad-cb=\det(A)\) , that is non zero, so that \( A^{T}\) is invertible as well.
Moreover, \( A^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d&-b\\{-c}&a \end{bmatrix}\) , so that, because of item (I),
\( (A^{-1})^T=\frac{1}{ad-bc}\begin{bmatrix} d&-c\\{-b}&a \end{bmatrix} =\frac{1}{ad-cb}\begin{bmatrix} d&-c\\{-b}&a \end{bmatrix}=(A^T)^{-1}\) .
3 Symmetric and Skew Symmetric Matrices
The symmetric and skew symmetric matrices exhibit symmetry and anti-symmetry along their diagonal.
3.1 Example of a Symmetric Matrix
Assume that \( (a,b)\in\mathbb{R}^2\) are real numbers that are not zero together, and consider the symmetry \( \sigma_{ab}\) along the line \( (D_{ab})\) of equation \( ax+by=0\) in the canonical basis.
Then the matrix of \( \sigma_{ab}\) in the canonical basis is equal to \( S_{ab}=\begin{bmatrix} \frac{a^2-b^2}{a^2+b^2}&\frac{2ab}{a^2+b^2}\\ \frac{2ab}{a^2+b^2}&\frac{b^2-a^2}{a^2+b^2}\end{bmatrix}\) .
We observe that the matrix \( S_{ab}\) is symmetric along its main diagonal: we say that it is a symmetric matrix.
3.2 Definition of a Symmetric Matrix
Definition 2
A symmetric \( 2\times 2\) matrix with real elements is a matrix of the type \( A=\begin{bmatrix} a&b\\b&d \end{bmatrix}\) , where \( (a,b,d)\in\mathbb{R}^3\) are any real numbers.
Theorem 2
Assume that \( A\) is a \( 2\times 2\) matrix with real elements.
Then the following assertions are equivalent:
- \( A\) is a symmetric matrix.
- \( A\) is symmetric along its main diagonal.
- \( A\) is equal to its transpose: \( A^{T}=A\) .
The proof relies on the fact that, if \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) , then it is symmetric if and only if \( b=c\) .
3.3 Other Examples of Symmetric Matrices
The following matrices are symmetric.
- The null matrix: \( O_{22}=\begin{bmatrix} 0&0\\0&0\end{bmatrix}\) .
- The identity matrix: \( I=\begin{bmatrix} 1&0\\0&1\end{bmatrix}\) .
- Any diagonal matrix \( D_{ab}=\begin{bmatrix} a&0\\0&b\end{bmatrix}\) , where \( (a,b)\in\mathbb{R}^2\) are any real numbers.
- In particular, any scalar matrix \( D_{aa}=\begin{bmatrix} a&0\\0&a\end{bmatrix}\) , where \( a\in\mathbb{R}\) is any real number.
- Any matrices of the form \( A^TA\) or \( AA^T\) , where \( A\) is any \( 2\times 2\) matrix with real elements.
All these matrices are clearly symmetric, except for the last one, that uses the theorem 1 in the following calculations:
\( (A^TA)^T=A^T(A^T)^T=A^TA\) and \( (AA^T)^T=(A^T)^TA^T=AA^T\) .
3.4 Properties of Symmetric Matrices
Theorem 3
Assume that \( A\) and \( B\) are \( 2\times 2\) symmetric matrices with real elements, and that \( \lambda\in\mathbb{R}\) is a scalar.
Then the following assertions hold.
- \( A^{T}\) is symmetric.
- \( A+B\) is symmetric.
- \( -A\) is symmetric.
- \( \lambda A\) is symmetric.
- If \( A\) is invertible, then \( A^{-1}\) is symmetric.
The proof relies on the theorem 1 and on the characteristic property of a symmetric matrix that says that it is equal to its transpose.
3.5 Example of a Skew Symmetric Matrix
Consider the rotation \( \rho_{\pi/2}\) of angle \( \frac{\pi}{2}\) .
Then the matrix of \( \rho_{\pi/2}\) in the canonical basis is equal to \( R_{\pi/2}=\begin{bmatrix}0&-1\\1&0 \end{bmatrix}\) .
We observe that the elements on the diagonal of \( R_{\pi/2}\) are \( 0\) and that the element below the main diagonal of \( R_{\pi/2}\) is the opposite of the element above its main diagonal.
We say that \( R_{\pi/2}\) is a skew symmetric matrix.
3.6 Definition of a Skew Symmetric Matrix
Definition 3
A skew symmetric \( 2\times 2\) matrix with real elements is a matrix of the type \( A=\begin{bmatrix} 0&a\\{-a}&0 \end{bmatrix}\) , where \( a\in\mathbb{R}\) is any real number.
Theorem 4
Assume that \( A\) is a \( 2\times 2\) matrix with real elements.
Then the following assertions are equivalent:
\( A\) is a skew symmetric matrix.
The elements on the main diagonal of \( A\) are \( 0\) and the element below the main diagonal of \( A\) is the opposite of the element above its main diagonal.
The transpose of \( A\) is equal to the opposite of \( A\) : \( A^{T}=-A\) .
The proof relies on the fact that, if \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) , then it is skew symmetric if and only if \( a=d=0\) and \( c=-b\) .
3.7 Other Examples of Skew Symmetric Matrices
The following matrices are skew symmetric.
The null matrix: \( O_{22}=\begin{bmatrix} 0&0\\0&0\end{bmatrix}\) .
The matrix of the rotation \( \rho_{-\pi/2}\) of angle \( -\frac{\pi}{2}\) : \( R_{-\pi/2}=\begin{bmatrix}0&1\\{-1}&0 \end{bmatrix}\) .
3.8 Properties of Skew Symmetric Matrices
Theorem 5
Assume that \( A\) is a \( 2\times 2\) skew symmetric matrices with real elements, and that \( \lambda\in\mathbb{R}\) is a scalar.
Then the following assertions hold.
- \( A^{T}\) is skew symmetric.
- \( A+B\) is skew symmetric.
- \( -A\) is symmetric.
- \( \lambda A\) is skew symmetric.
- If \( A\) is invertible, then \( A^{-1}\) is skew symmetric.
The proof relies on the theorem 1 and on the characteristic property of a symmetric matrix that says that it is equal to the opposite of its transpose.
4 The Transpose of a Column Vector
The transpose of a column vector is a line vector, and the transpose of a line vector is a column vector.
4.1 The Line Vectors with 2 Real Elements
Assume that \( (v_1,v_2)\in\mathbb{R}^2\) are real numbers.
- Then the line vector with elements \( (v_1,v_2)\) is denoted \( V=\begin{bmatrix}v_1&v_2\end{bmatrix}\) .
Assume that \( X=\begin{bmatrix}x_1\\x_2\end{bmatrix}\) is a column vector with 2 real elements.
- Then we may multiply the column vector \( X\) to the left by the line vector to obtain a scalar: \( VX=\begin{bmatrix}v_1&v_2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=v_1x_1+v_2x_2\) .
Assume that \( A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\) is a matrix with real elements.
- Then we may multiply the matrix \( A\) to the left by the line vector \( V\) to obtain a line vector: \( VA=\begin{bmatrix}v_1&v_2\end{bmatrix} \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} =\begin{bmatrix}v_1a_{11}+v_2a_{21}&v_1a_{12}+v_2a_{22}\end{bmatrix}\) .
4.2 Contextual Line and Column Vectors in Python
In ‘numpy’ scientific library, all vectors are basically line vectors.
If a \( 2\times 2\) matrix is multiplied to the left by a vector with 2 elements (operator @), it is considered as a line vector, and the result is a line vector.
If a \( 2\times 2\) matrix is multiplied to the right by a vector with 2 elements (operator @), it is considered as a column vector, and the result is a column vector.
If two vectors with 2 elements are multiplied together with the operator @, the first vector is seen as a line vector, the second vector is seen as a column vector, and the result is their dot product.
4.3 Extended Associativity Laws
Theorem 6
Assume that \( V\) is a line vector with 2 real elements.
Assume that \( X\) is a column vector with 2 real elements.
Assume that \( A\) and \( B\) are \( 2\times 2\) matrices with real elements.
Then the following assertions hold.
- \( (VA)X=V(AX)\) and it is a scalar that may be written \( VAX\) .
- \( (VA)B=V(AB)\) and it is a line vector that may be written \( VAB\) .
- \( (AB)X=A(BX)\) and it is a column vector that may be written \( ABX\) .
- We may define the scalar \( VABX\) as:
- the product of a line vector by a column vector: \( (VA)(BX)\) ,
- or a matrix multiplied by a line vector to the left and by a column vector to the right: \( V(AB)X\) .
Proof
Assume that \( V=\begin{bmatrix}v_1&v_2\end{bmatrix}\) is a line vector with 2 real elements.
Assume that \( X=\begin{bmatrix}x_1\\x_2\end{bmatrix}\) is a column vector with 2 real elements.
Assume that \( A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\) and \( B=\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}\) are \( 2\times 2\) matrices with real elements.
Then we may perform the following calculations.
\( VA=\begin{bmatrix}v_1&v_2\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} =\begin{bmatrix}v_1a_{11}+v_2a_{21}&v_1a_{12}+v_2a_{22}\end{bmatrix}\) ,
so that \( (VA)X=\begin{bmatrix}v_1a_{11}+v_2a_{21}&v_1a_{12}+v_2a_{22}\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}\)
\( =(v_1a_{11}+v_2a_{21})x_{1}+(v_1a_{12}+v_2a_{22})x_{2}\)
\( =v_1a_{11}x_{1}+v_2a_{21}x_{1}+v_1a_{12}x_{2}+v_2a_{22}x_{2}\) .
and \( AX=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} =\begin{bmatrix}a_{11}x_{1}+a_{12}x_{2}\\a_{21}x_{1}+a_{22}x_2\end{bmatrix}\) ,
so that \( V(AX)=\begin{bmatrix}v_1&v_2\end{bmatrix}\begin{bmatrix}a_{11}x_{1}+a_{12}x_{2}\\a_{21}x_{1}+a_{22}x_2\end{bmatrix}\)
\( =v_{1}(a_{11}x_{1}+a_{12}x_{2})+v_{2}(a_{21}x_{1}+a_{22}x_2)\)
\( =v_{1}a_{11}x_{1}+v_{1}a_{12}x_{2}+v_{2}a_{21}x_{1}+v_{2}a_{22}x_2 =(VA)X\) .
\( VA=\begin{bmatrix}v_1&v_2\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} =\begin{bmatrix}v_1a_{11}+v_2a_{21}&v_1a_{12}+v_2a_{22}\end{bmatrix}\) ,
so that \( (VA)B=\begin{bmatrix}v_1a_{11}+v_2a_{21}&v_1a_{12}+v_2a_{22}\end{bmatrix} \begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}\)
\( =\begin{bmatrix}(v_1a_{11}+v_2a_{21})b_{11}+(v_1a_{12}+v_2a_{22})b_{21}& (v_1a_{11}+v_2a_{21})b_{12}+(v_1a_{12}+v_2a_{22})b_{22}\end{bmatrix}\)
\( =\begin{bmatrix}v_1a_{11}b_{11}+v_2a_{21}b_{11}+v_1a_{12}b_{21}+v_2a_{22}b_{21}& v_1a_{11}b_{12}+v_2a_{21}b_{12}+v_1a_{12}b_{22}+v_2a_{22}b_{22}\end{bmatrix}\) .
And \( AB=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix} =\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix}\) ,
so that \( V(AB)=\begin{bmatrix}v_1&v_2\end{bmatrix} \begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix}\)
\( =\begin{bmatrix}v_1(a_{11}b_{11}+a_{12}b_{21})+v_2(a_{21}b_{11}+a_{22}b_{21})& v_1(a_{11}b_{12}+a_{12}b_{22})+v_2(a_{21}b_{12}+a_{22}b_{22})&\end{bmatrix}\)
\( =\begin{bmatrix}v_1a_{11}b_{11}+v_1a_{12}b_{21}+v_2a_{21}b_{11}+v_2a_{22}b_{21}& v_1a_{11}b_{12}+v_1a_{12}b_{22}+v_2a_{21}b_{12}+v_2a_{22}b_{22}&\end{bmatrix}\)
\( =(VA)B\) .
\( AB=\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix}\) ,
so that\( (AB)X=\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\ a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}\)
\( =\begin{bmatrix}(a_{11}b_{11}+a_{12}b_{21})x_1+(a_{11}b_{12}+a_{12}b_{22})x_2\\ (a_{21}b_{11}+a_{22}b_{21})x_1+(a_{21}b_{12}+a_{22}b_{22})x_2\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}b_{11}x_1+a_{12}b_{21}x_1+a_{11}b_{12}x_2+a_{12}b_{22}x_2\\ a_{21}b_{11}x_1+a_{22}b_{21}x_1+a_{21}b_{12}x_2+a_{22}b_{22}x_2\end{bmatrix}\) .
And \( BX=\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} =\begin{bmatrix}b_{11}x_1+b_{12}x_2\\b_{21}x_1+b_{22}x_2\end{bmatrix}\) ,
so that \( A(BX)=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} \begin{bmatrix}b_{11}x_1+b_{12}x_2\\b_{21}x_1+b_{22}x_2\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}(b_{11}x_1+b_{12}x_2)+a_{12}(b_{21}x_1+b_{22}x_2)\\ a_{21}(b_{11}x_1+b_{12}x_2)+a_{22}(b_{21}x_1+b_{22}x_2)\end{bmatrix}\)
\( =\begin{bmatrix}a_{11}b_{11}x_1+a_{11}b_{12}x_2+a_{12}b_{21}x_1+a_{12}b_{22}x_2\\ a_{21}b_{11}x_1+a_{21}b_{12}x_2+a_{22}b_{21}x_1+a_{22}b_{22}x_2\end{bmatrix}\)
\( =(AB)X\) .
Because of the previous items, we may get rid of the parentheses or put parentheses where we wish in any valid expression with matrices and with or without a line vector to the left, and with or without a column vector to the right.
Consequently, \( VABX=(VA)(BX)=V(AB)X\) .
4.4 Transposes of Column Vectors and of Line Vectors
Definition 4
Assume that \( X=\begin{bmatrix}x_1\\x_2\end{bmatrix}\) is a column vector with 2 real elements.
Then we define the transpose of \( X\) as the line vector with elements \( (x_{1},x_{2})\) :
\( X^T=\begin{bmatrix}x_1&x_2\end{bmatrix}\) .
Definition 5
Assume that \( V=\begin{bmatrix}v_1&v_2\end{bmatrix}\) is a line vector with 2 real elements.
Then we define the transpose of \( V\) as the column vector with elements \( (v_{1},v_{2})\) :
\( V^T=\begin{bmatrix}v_1\\v_2\end{bmatrix}\) .
4.5 Properties of theTransposes of Column and Line Vectors
4.5.1 The transpose of the Transpose of a line or column vector
Theorem 7
Assume that \( V\) is a line vector with 2 real elements.
Assume that \( X\) is a column vector with 2 real elements.
Then the following assertions hold.
The transpose of the transpose of \( V\) is equal to \( V\) : \( (V^{T})^{T}=V\) .
The transpose of the transpose of \( X\) is equal to \( X\) : \( (X^{T})^{T}=X\) .
Proof
Assume that \( V=\begin{bmatrix}v_1&v_2\end{bmatrix}\) is a line vector with 2 real elements.
Then \( V^T=\begin{bmatrix}v_1\\v_2\end{bmatrix}\) , so that \( (V^{T})^{T}=\begin{bmatrix}v_1&v_2\end{bmatrix}=V\) .
Assume that \( X=\begin{bmatrix}x_1\\x_2\end{bmatrix}\) is a column vector with 2 real elements.
Then \( X^T=\begin{bmatrix}x_1&x_2\end{bmatrix}\) , so that \( (X^{T})^{T}=\begin{bmatrix}x_1\\x_2\end{bmatrix}=X\) .
4.5.2 Transpose of the Product of a Line or Column Vector by a Matrix
Theorem 8
Assume that \( V\) is a line vector with 2 real elements.
Assume that \( X\) is a column vector with 2 real elements.
Assume that \( A\) is a \( 2\times 2\) matrix with real elements.
Then the following assertions hold.
The transpose of the product \( VA\) is equal to: \( (VA)^T=A^TV^T\) .
The transpose of the product \( AX\) is equal to: \( (AX)^T=X^T A^T\) .
Proof
Assume that \( A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\) is a \( 2\times 2\) matrix with real elements.
Assume that \( V=\begin{bmatrix}v_1&v_2\end{bmatrix}\) is a line vector with 2 real elements.
Then \( VA=\begin{bmatrix}v_1&v_2\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} =\begin{bmatrix}v_1a_{11}+v_2a_{21}&v_1a_{12}+v_2a_{22}\end{bmatrix}\) ,
so that \( (VA)^{T}=\begin{bmatrix}v_1a_{11}+v_2a_{21}\\v_1a_{12}+v_2a_{22}\end{bmatrix}\) .
And \( A^{T}V^{T}=\begin{bmatrix}a_{11}&a_{21}\\a_{12}&a_{22}\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} =\begin{bmatrix}a_{11}v_1+a_{21}v_2\\a_{12}v_1+a_{22}v_2\end{bmatrix} =(VA)^{T}\) .
Assume now that \( X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}\) is a column vector with 2 real elements.
Then \( AX=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix} =\begin{bmatrix} a_{11}x_{1}+a_{12}x_{2}\\a_{21}x_{1}+a_{22}x_{2} \end{bmatrix}\) ,
so that \( (AX)^{T}=\begin{bmatrix} a_{11}x_{1}+a_{12}x_{2}&a_{21}x_{1}+a_{22}x_{2} \end{bmatrix}\) .
And \( X^{T}A^{T}=\begin{bmatrix} x_{1}&x_{2} \end{bmatrix} \begin{bmatrix}a_{11}v_1+a_{21}v_2\\a_{12}v_1+a_{22}v_2\end{bmatrix}\)
\( =\begin{bmatrix} x_{1}a_{11}+x_{2}a_{12}&x_{1}a_{21}+x_{2}a_{22} \end{bmatrix} =(AX)^{T}\)
4.5.3 Matrix Form of the Dot Product and of the Square of the Norm
Theorem 9
Assume that \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^2\) are vectors such that:
the column vector of coordinates of \( \overrightarrow{u}\) in the canonical basis \( B_{0}\) is equal to \( [\overrightarrow{u}]_{B_0}=X\) ,
and the column vector of coordinates of \( \overrightarrow{v}\) in the canonical basis \( B_{0}\) is equal to \( [\overrightarrow{v}]_{B_0}=Y\) .
Then the following assertions hold.
The dot product of \( \overrightarrow{u}\) and \( \overrightarrow{v}\) is equal to: \( \overrightarrow{u}\cdot\overrightarrow{v}=X^T Y\) .
And consequently, the square of the norm of \( \overrightarrow{u}\) is equal to: \( \left\| \overrightarrow{u} \right\|^2=X^T X\)
Proof
Assume that \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^2\) are vectors such that:
the column vector of coordinates of \( \overrightarrow{u}\) in the canonical basis \( B_{0}\) is equal to \( [\overrightarrow{u}]_{B_0}=X=\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}\) ,
and the column vector of coordinates of \( \overrightarrow{v}\) in the canonical basis \( B_{0}\) is equal to \( [\overrightarrow{v}]_{B_0}=Y=\begin{bmatrix}y_{1}\\y_{2}\end{bmatrix}\) .
Then \( X^{T}Y=\begin{bmatrix}x_{1}&x_{2}\end{bmatrix}\begin{bmatrix}y_{1}\\y_{2}\end{bmatrix} =x_{1}y_{1}+x_{2}y_{2}=\overrightarrow{u}\cdot\overrightarrow{v}\) .
And \( \left\| \overrightarrow{u} \right\|^2=\overrightarrow{u}\cdot\overrightarrow{u}=X^T X\) .
That theorem justifies the fact that, in Python, u@v is the dot product of the (line) vector u and the (column) vector v.
4.5.4 Dot Product of Images of Vectors by a Linear Mapping
Theorem 10
Assume that \( f:\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping in the plane with matrix \( A\) in the canonical basis.
Assume that \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^2\) are vectors such that:
the column vector of coordinates of \( \overrightarrow{u}\) in the canonical basis \( B_{0}\) is equal to \( [\overrightarrow{u}]_{B_0}=X\) ,
and the column vector of coordinates of \( \overrightarrow{v}\) in the canonical basis \( B_{0}\) is equal to \( [\overrightarrow{v}]_{B_0}=Y\) .
Then the following assertions hold.
The dot product of \( f(\overrightarrow{u})\) and \( f(\overrightarrow{v})\) is equal to: \( f(\overrightarrow{u})\cdot f(\overrightarrow{v})=X^TA^{T}A Y\) .
And consequently, the square of the norm of \( \overrightarrow{u}\) is equal to:
\( \left\| f(\overrightarrow{u}) \right\|^2=X^TA^{T}A X\)
Proof
Assume that \( f:\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping in the plane with matrix \( A\) in the canonical basis.
Assume that \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^2\) are vectors such that:
the column vector of coordinates of \( \overrightarrow{u}\) in the canonical basis \( B_{0}\) is equal to \( [\overrightarrow{u}]_{B_0}=X\) ,
and the column vector of coordinates of \( \overrightarrow{v}\) in the canonical basis \( B_{0}\) is equal to \( [\overrightarrow{v}]_{B_0}=Y\) .
Then \( [f(\overrightarrow{u})]_{B_0}=AX\) and \( [f(\overrightarrow{v})]_{B_0}=AY\) , so that
\( f(\overrightarrow{u})\cdot f(\overrightarrow{v})=(AX)^{T}(AY)=(X^TA^{T})(A Y)=X^TA^{T}A Y\) .
And \( \left\| f(\overrightarrow{u}) \right\|^2=\overrightarrow{u}\cdot\overrightarrow{u}=X^TA^{T}A X\) .
5 The Unitary Matrices
The unitary matrices are the matrices of the linear mappings that preserve the norm of the vectors.
5.1 Definition of the Unitary Matrices
Definition 6
Assume that \( A\) is a \( 2\times 2\) matrix with real elements.
Assume that \( A\) is the matrix of the linear mapping \( f:\mathbb{P}\rightarrow\mathbb{P}\) in the canonical basis.
Then the matrix \( A\) is a unitary matrix if and only if \( f\) preserves the norms of the vectors: \( \forall\overrightarrow{u}\in\mathbb{P},\left\|f(\overrightarrow{u})\right\|=\left\|\overrightarrow{u}\right\|\) .
An example of unitary matrix is the identity matrix \( I\) , because:
- \( I\) is the matrix of the identity mapping \( \text{Id}_{\mathbb{P}}\) ,
- and \( \forall\overrightarrow{u}\in\mathbb{P},\left\|\text{Id}_{\mathbb{P}}(\overrightarrow{u})\right\|=\left\|\overrightarrow{u}\right\|\) .
5.2 The Product of two Unitary Matrices is a Unitary Matrix
Theorem 11
Assume that \( A\) and \( B\) are unitary matrices.
Then the matrix product \( AB\) is a unitary matrix.
Proof
Assume that \( A\) and \( B\) are unitary matrices.
Assume that \( A\) is the matrix of the linear mapping \( f:\mathbb{P}\rightarrow\mathbb{P}\) in the canonical basis.
Assume that \( B\) is the matrix of the linear mapping \( g:\mathbb{P}\rightarrow\mathbb{P}\) in the canonical basis.
Then \( AB\) is the matrix of the linear mapping \( f\circ g\) in the canonical basis.
Moreover, for any vector \( \overrightarrow{u}\in\mathbb{P}\) , we have
\( \left\|(f\circ g)(\overrightarrow{u})\right\|=\left\|f(g(\overrightarrow{u}))\right\| =\left\|g(\overrightarrow{u})\right\|=\left\|\overrightarrow{u}\right\|\) .
Consequently, the matrix product \( AB\) is a unitary matrix.
5.3 Preservation of the Dot Product
Theorem 12
A linear mapping that preserves the norm of vectors also preserves the dot product of vectors.
This is because the dot product of two vectors may be written using norms of vectors.
Indeed, for any vectors \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^2\) , we have
\( \left\| \overrightarrow{u+v} \right\|^2= \left\| \overrightarrow{u} \right\|^2 +\left\| \overrightarrow{v} \right\|^2 +2\overrightarrow{u}\cdot\overrightarrow{v}\) , so that
\( \overrightarrow{u}\cdot\overrightarrow{v}=\frac{1}{2} \left( \left\| \overrightarrow{u+v} \right\|^2-\left\| \overrightarrow{u} \right\|^2 -\left\| \overrightarrow{v} \right\|^2 \right)\) .
5.4 Equivalent Conditions for a Matrix to be Unitary
Theorem 13
Assume that \( U\) is a \( 2\times 2\) matrix with real elements.
Then the following assertions are equivalent.
- \( U\) is a unitary matrix.
- The vectors with column vectors of coordinates the columns of \( U\) form an orthonormal basis.
- \( U\) is the transition matrix from an orthonormal basis to an orthonormal basis.
- \( U^{T}U=I\) , the identity matrix.
- \( U\) is invertible and \( U^{-1}=U^{T}\) .
Proof
Assume that \( U\) is a \( 2\times 2\) matrix with real elements.
Assume that \( U\) is the matrix of the linear mapping \( u:\mathbb{P}\rightarrow\mathbb{P}\) in the canonical basis.
Then we prove the following equivalences.
(I) \( \Leftrightarrow\) (II)
Assume that \( U\) is a unitary matrix.
Then the columns of \( U\) are the column vectors of coordinates of the images \( u(\overrightarrow{i})\) and \( u(\overrightarrow{j})\) of the canonical basis.
Moreover, \( u\) preserves the norm and the dot product.
Consequently, \( B=(u(\overrightarrow{i}),u(\overrightarrow{j}))\) is an orthonormal basis as is the canonical basis.
So that the vectors with column vectors of coordinates the columns of \( U\) form an orthonormal basis.
Assume now that the vectors with column vectors of coordinates the columns of \( U\) form an orthonormal basis.
Denote \( U=\begin{bmatrix} u_{11}&u_{12}\\u_{21}&u_{22} \end{bmatrix}\) , and consider the vectors \( \overrightarrow{I}\) and \( \overrightarrow{J}\) with column vectors of coordinates the columns of \( U\) .
Then \( U^{T}U=\begin{bmatrix} u_{11}&u_{21}\\u_{12}&u_{22} \end{bmatrix}\begin{bmatrix} u_{11}&u_{12}\\u_{21}&u_{22} \end{bmatrix} =\begin{bmatrix} u_{11}^{2}+u_{21}^{2}&u_{11}u_{12}+u_{21}u_{22}\\ u_{12}u_{11}+u_{22}u_{21}&u_{12}^{2}+u_{22}^{2} \end{bmatrix}\)
\( =\begin{bmatrix} \left\|\overrightarrow{I}\right\|^{2}& \overrightarrow{I}\cdot\overrightarrow{J}\\ \overrightarrow{J}\cdot\overrightarrow{I}& \left\|\overrightarrow{J}\right\|^{2}\end{bmatrix} =\begin{bmatrix} 1&0\\0&1 \end{bmatrix}=I\) .
So that, because of theorem 10, for any vector \( \overrightarrow{v}\in\mathbb{P}\) with column vectors of coordinates \( X\) in the canonical basis, \( \left\| u(\overrightarrow{v})\right\|^{2}=X^{T}U^{T}UX=X^{T}IX=X^{T}X=\left\| \overrightarrow{v}\right\|^{2}\) .
Consequently, \( u\) preserves the norms of vectors and \( U\) is a unitary matrix.
(II) \( \Leftrightarrow\) (III)
Assume that the vectors with column vectors of coordinates the columns of \( U\) form an orthonormal basis.
Consider the basis \( B\) made of the vectors having the columns of \( U\) as column vectors of coordinates.
Then \( B\) is an orthonormal basis and \( U\) is the transition matrix from \( B\) to the canonical basis, that is orthonormal too.
Consequently, \( U\) is the transition matrix from an orthonormal basis to an orthonormal basis.
Assume now that \( U\) is the transition matrix from an orthonormal basis to an orthonormal basis.
Then \( U\) is either the matrix of a rotation or the matrix of a symmetry along a line.
But you will see in the section 5.5 that these matrices are unitary, so that, because (I) \( \Leftrightarrow\) (II), the vectors with column vectors of coordinates the columns of \( U\) form an orthonormal basis.
(IV) \( \Leftrightarrow\) (V)
That equivalence is clear, by definition of the inverse of a matrix.
(I) \( \Leftrightarrow\) (IV)
Consider the vectors \( \overrightarrow{I}\) and \( \overrightarrow{J}\) with column vectors of coordinates the columns of \( U\) .
We have seen that \( U^{T}U =\begin{bmatrix} \left\|\overrightarrow{I}\right\|^{2}& \overrightarrow{I}\cdot\overrightarrow{J}\\ \overrightarrow{J}\cdot\overrightarrow{I}& \left\|\overrightarrow{J}\right\|^{2}\end{bmatrix}\) .
So that \( U^{T}U=I\) if and only if the vectors with column vectors of coordinates the columns of \( U\) form an orthonormal basis.
5.5 Classification of the Unitary Matrices (1/3)
5.5.1 The Matrices of Rotations and Symmetries Along Lines are Unitary
Theorem 14
Assume that \( \theta\in\mathbb{R}\) is a real number, and consider the matrices:
\( R_{\theta}=\begin{bmatrix} \cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta) \end{bmatrix}\) of the rotation of angle \( \theta\) in the canonical basis.
\( \Sigma_{\theta}=\begin{bmatrix} -\cos(2\theta)&\sin(2\theta)\\\sin(2\theta)&\cos(2\theta)\end{bmatrix}\) of the symmetry along the line \( (D_{\theta})\) of angle \( \theta\) with the \( x\) axis, in the canonical basis.
Then \( R_{\theta}\) and \( \Sigma_{\theta}\) are unitary matrices.
Proof
We use the criterium (V) of the theorem 13.
Assume that \( \theta\in\mathbb{R}\) is a real number.
\( R_{\theta}=\begin{bmatrix} \cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta) \end{bmatrix}\) .
Then \( R_{\theta}^T=\begin{bmatrix} \cos(\theta)&\sin(\theta)\\{-\sin(\theta)}&\cos(\theta) \end{bmatrix} =\begin{bmatrix} \cos(-\theta)&-\sin(-\theta)\\\sin(-\theta)&\cos(-\theta) \end{bmatrix}\) .
It is the matrix in the canonical basis of the rotation of angle \( -\theta\) , that is the inverse of the matrix in the canonical basis of the rotation of angle \( \theta\) .
Consequently, \( R_{\theta}^{T}=R_{\theta}^{-1}\) .
\( \Sigma_{\theta}=\begin{bmatrix} -\cos(2\theta)&\sin(2\theta)\\\sin(2\theta)&\cos(2\theta)\end{bmatrix}\) .
Then \( \Sigma_{\theta}\) is symmetric, so that \( \Sigma_{\theta}^{T}=\Sigma_{\theta}\) .
Moreover, a symmetry along a straight line is the inverse of itself.
Consequently, \( \Sigma_{\theta}^{-1}=\Sigma_{\theta}^{T}\) .
5.5.2 The Unitary Matrices are the Matrices of Rotations and of Symmetries Along Lines
Theorem 15
Assume that is a unitary matrix.
Then, one of the following assertions holds.
Either \( U\) is the matrix of a rotation, and then:
\( \det(U)=1\) ,
and \( U\) is the transition matrix from the canonical basis to a direct orthonormal basis.
Or \( U\) is the matrix of a symmetry along a line, and then:
\( \det(U)=-1\) ,
and \( U\) is the transition matrix from the canonical basis to a reverse orthonormal basis.
Proof
Assume that \( U\) is a unitary matrix and consider the linear mapping \( f\) having \( U\) as matrix in the canonical basis.
We have to prove that \( f\) is either a rotation or a symmetry along a line.
Consider the images by \( f\) of the vectors of the canonical basis, \( \overrightarrow{I}=f(\overrightarrow{i})\) and \( \overrightarrow{J}=f(\overrightarrow{j})\) .
As \( U\) is a unitary matrix, then \( f\) preserves the norms and dot products of vectors.
Consequently, as the canonical basis is an orhtonormal basis, the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) is an orthonormal basis too.
Let’s denote \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{I})}\) .
If the orthonormal basis \( B=(\overrightarrow{I},\overrightarrow{J})\) is direct, then \( \widehat{(\overrightarrow{I},\overrightarrow{J})}=\frac{\pi}{2}\) , so that \( \widehat{(\overrightarrow{i},\overrightarrow{J})}=\theta+\frac{\pi}{2}\) .
The matrix \( U\) is then equal to:
\( U=\begin{bmatrix} \cos(\theta)&\cos(\theta+\frac{\pi}{2})\\ \sin(\theta)&\sin(\theta+\frac{\pi}{2}) \end{bmatrix} =\begin{bmatrix} \cos(\theta)&-\sin(\theta)\\ \sin(\theta)&\cos(\theta) \end{bmatrix}\) .
Consequently, \( f\) is the rotation of angle \( \theta\) .
\( U\) is also the transition matrix from the canonical basis to the direct orthonormal basis
\( B=(\overrightarrow{I},\overrightarrow{J})\) .
Moreover, \( \det(U)=\cos^{2}(\theta)+\sin^{2}(\theta)=1\) .
And if the orthonormal basis \( B=(\overrightarrow{I},\overrightarrow{J})\) is reverse, then \( \widehat{(\overrightarrow{I},\overrightarrow{J})}=-\frac{\pi}{2}\) , so that \( \widehat{(\overrightarrow{i},\overrightarrow{J})}=\theta-\frac{\pi}{2}\) .
The matrix \( U\) is then equal to:
\( U=\begin{bmatrix} \cos(\theta)&\cos(\theta-\frac{\pi}{2})\\ \sin(\theta)&\sin(\theta-\frac{\pi}{2}) \end{bmatrix} =\begin{bmatrix} \cos(\theta)&\sin(\theta)\\ \sin(\theta)&-\cos(\theta) \end{bmatrix}\) .
Consequently, \( U\) is the transition matrix from the canonical basis to the reverse orthonormal basis \( B=(\overrightarrow{I},\overrightarrow{J})\) .
Moreover, \( \det(U)=-\cos^{2}(\theta)-\sin^{2}(\theta)=-1\) .
And \( f\) is a symmetry along the line (D) making an angle \( \alpha\) to be found with the \( x\) axis.
The angle \( \alpha\) must be such that:
\( U=\Sigma_{\alpha}=\begin{bmatrix} -\cos(2\alpha)&\sin(2\alpha)\\\sin(2\alpha)&\cos(2\alpha)\end{bmatrix} =\begin{bmatrix} \cos(\pi-2\alpha)&\sin(\pi-2\alpha)\\\sin(\pi-2\alpha)&-\cos(\pi-2\alpha)\end{bmatrix}\) .
It is thus sufficient to take \( \alpha\) so that \( \pi-2\alpha=\theta\) , that is \( \alpha=\frac{\theta}{2}+\frac{\pi}{2}\) .
5.6 Inversion of a Unitary Matrix
Theorem 16
Assume that \( U\) is a unitary matrix.
Then \( U\) is invertible and its inverse \( U^{-1}\) is a unitary matrix.
The proof relies on the theorem 15 and on the following facts.
- The matrix of the rotation of angle \( \theta\) is invertible, and its inverse is the rotation of angle \( -\theta\) .
- The matrix of a symmetry along a line is invertible, and its inverse is itself.
5.7 The Multiplicative Group of Unitary Matrices
Because of the previous result, we may set the following result.
The set of all the unitary matrices is, with the matrix multiplication, a group, because:
- The product of two unitary matrices is a unitary matrix.
- The matrix multiplication is associative.
- The identity matrix is a unitary matrix.
- Any unitary matrix is invertible, and its inverse is a unitary matrix.
6 Conclusion
The symmetric, skew symmetric and unit matrices have important nice properties, as you could see in that text.
Moreover, it may be proved that a symmetric matrix is diagonalisable in an orthtonormal basis.
That is that there exists an orthonormal basis in which the matrix of the correspondant linear mapping is a diagonal matrix.
You will learn more about diagonalisation of matrices in the next section about eigenvalues and iegenvectors.
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