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First published on Saturday, Jul 6, 2024 and last modified on Thursday, Apr 10, 2025
Mathedu SAS
With the eigenvalues and eigenvectors determination, we are often able to diagonalise the matrices, that allow us to manipulate them easily.
You will discover how in that document.
Assume that \( h_{\lambda}\) is the homothety of factor \( \lambda\) .
Then any non zero vector \( \overrightarrow{u}\in\mathbb{P}^*\) is so that \( h_{\lambda}(\overrightarrow{u})=\lambda\overrightarrow{u}\) .
We say that:
\( \lambda\) is an eigenvalue of \( h_{\lambda}\) ,
and all the non zero vectors of the plane are eigenvectors of \( h_{\lambda}\) for the eigenvalue \( \lambda\) .
Assume that \( d_{(\lambda,\mu)}\) is the linear mapping with matrix \( D_{(\lambda,\mu)}=\begin{bmatrix} \lambda&0\\0&\mu \end{bmatrix}\) in the canonical basis.
Then we may observe that, by definition of the matrix of a linear mapping in the canonical basis \( B_{0}=(\overrightarrow{i},\overrightarrow{j})\) :
the non zero vector \( \overrightarrow{i}\) is so that \( d_{(\lambda,\mu)}(\overrightarrow{i})=\lambda\overrightarrow{i}\) ,
and the non zero vector \( \overrightarrow{j}\) is so that \( d_{(\lambda,\mu)}(\overrightarrow{j})=\mu\overrightarrow{j}\)
We say that:
\( \lambda\) and \( \mu\) are eigenvalues of \( d_{(\lambda,\mu)}\) ,
the vector \( \overrightarrow{i}\) is an eigenvector of \( d_{(\lambda,\mu)}\) for the eigenvalue \( \lambda\) .
and the vector \( \overrightarrow{j}\) is an eigenvector of \( d_{(\lambda,\mu)}\) for the eigenvalue \( \mu\) .
Assume that \( f\) is a linear mapping with matrix \( D_{(\lambda,\mu)}=\begin{bmatrix} \lambda&0\\0&\mu \end{bmatrix}\) in some basis \( B=(\overrightarrow{I},\overrightarrow{J})\) .
Then we may observe that, by definition of the matrix of a linear mapping in the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) :
the non zero vector \( \overrightarrow{I}\) is so that \( f(\overrightarrow{I})=\lambda\overrightarrow{I}\) ,
and the non zero vector \( \overrightarrow{J}\) is so that \( f(\overrightarrow{J})=\mu\overrightarrow{J}\)
We say that:
\( \lambda\) and \( \mu\) are eigenvalues of \( f\) ,
the vector \( \overrightarrow{I}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda\) .
and the vector \( \overrightarrow{J}\) is an eigenvector of \( f\) for the eigenvalue \( \mu\) .
Assume that \( f:\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping.
Assume that the positive scalar \( \lambda\in\mathbb{R}_+^*\) and the non zero vector \( \overrightarrow{u}\in\mathbb{P}^*\) are so that \( f(\overrightarrow{u})=\lambda\overrightarrow{u}\) .
Then \( \overrightarrow{u}\) and \( f(\overrightarrow{u})\) are positively aligned and we say that:
\( \lambda\) is a positive eigenvalue of \( f\) ,
and the vector \( \overrightarrow{u}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda\) .
Assume that \( f:\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping.
Assume that the negative scalar \( \lambda\in\mathbb{R}_-^*\) and the non zero vector \( \overrightarrow{u}\in\mathbb{P}^*\) are so that \( f(\overrightarrow{u})=\lambda\overrightarrow{u}\) .
Then \( \overrightarrow{u}\) and \( f(\overrightarrow{u})\) are negatively aligned and we say that:
\( \lambda\) is a negative eigenvalue of \( f\) ,
and the vector \( \overrightarrow{u}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda\) .
Assume that \( f:\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping.
Assume that the non zero vector \( \overrightarrow{u}\in\mathbb{P}^*\) is so that \( \overrightarrow{u}\in\ker(f)\) , the kernel of \( f\) , i.e. \( f(\overrightarrow{u})=\overrightarrow{0}\) .
Then \( f(\overrightarrow{u})=0\times \overrightarrow{u}\) and we say that:
\( 0\) is a negative eigenvalue of \( f\) ,
and the vector \( \overrightarrow{u}\) is an eigenvector of \( f\) for the eigenvalue \( 0\) .
Definition 1
Assume that \( f:\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping.
Then the scalar \( \lambda\in\mathbb{R}\) is said to be an eigenvalue of \( f\) if and only if there exists a non zero vector \( \overrightarrow{u}\in\mathbb{P}^*\) such that \( f(\overrightarrow{u})=\lambda\overrightarrow{u}\) .
And if the scalar \( \lambda\in\mathbb{R}\) is an eigenvalue of \( f\) , then the non zero vector \( \overrightarrow{u}\in\mathbb{P}^*\) is said to be an eigenvector of \( f\) for the eigenvalue \( \lambda\) if and only if \( f(\overrightarrow{u})=\lambda\overrightarrow{u}\) .
It is important that the definition of eigenvalues and eigenvectors is limited to non zero vectors, because for any linear mapping \( f:\mathbb{P}\rightarrow\mathbb{P}\) and any real number \( \mu\in\mathbb{R}\) , \( f(\overrightarrow{0})=\mu\overrightarrow{0}=\overrightarrow{0}\) and the definition would be meaningless.
Assume that \( f:\mathbb{P}\rightarrow\mathbb{P}\) is the linear mapping having \( A=\begin{bmatrix} 3&4\\4&-3 \end{bmatrix}\) as matrix in the canonical basis.
We shall prove that:
\( 5\) and \( -5\) are eigenvalues of \( f\) ,
\( \overrightarrow{u}=\begin{bmatrix} 2\\1 \end{bmatrix}\) is an eigenvector of \( f\) for the eigenvalue \( 5\) ,
and \( \overrightarrow{v}=\begin{bmatrix} 1\\{-2} \end{bmatrix}\) is an eigenvector of \( f\) for the eigenvalue \( -5\) .
Indeed, \( 5\) is an eigenvalue of \( f\) , with \( \overrightarrow{u}=\begin{bmatrix} 2\\1 \end{bmatrix}\) as eigenvector, because:
And \( -5\) is an eigenvalue of \( f\) , with \( \overrightarrow{u}=\begin{bmatrix} 1\2 \end{bmatrix}\) as eigenvector, because:
Assume that \( h_{\lambda}\) is the homothety of factor \( \lambda\) .
Then any non zero vector \( \overrightarrow{u}\in\mathbb{P}^*\) is so that \( h_{\lambda}(\overrightarrow{u})=\lambda\overrightarrow{u}\) .
We say that:
\( \lambda\) is an eigenvalue of \( h_{\lambda}\) ,
and all the non zero vectors of the plane are eigenvectors of \( h_{\lambda}\) for the eigenvalue \( \lambda\) .
Moreover, \( \lambda\) is the unique eigenvalue of \( h_{\lambda}\) .
This is because, if for some scalar \( \mu\in\mathbb{R}\) and for some non zero vector \( \overrightarrow{v}\in\mathbb{P}^*\) , we have \( h_{\lambda}(\overrightarrow{v})=\mu\overrightarrow{v}\) , as we have also \( h_{\lambda}(\overrightarrow{v})=\lambda\overrightarrow{v}\) , it follows that \( (\lambda-\mu)\overrightarrow{v}=\overrightarrow{0}\) with \( \overrightarrow{v}\ne\overrightarrow{0}\) , so that \( \lambda-\mu=0\) , and thus \( \lambda=\mu\) .
Assume that \( p\) is the projection on the line \( (D_{1})\) of equation \( ax+by=0\) along the line \( (D_{2})\) of equation \( cx+dy=0\) , where \( (a,b,c,d)\in\mathbb{R}^{4}\) are real numbers so that \( ad-bc\ne 0\) .
As \( ad-bc\ne 0\) , the lines \( (D_{1})\) and \( (D_{2})\) are secant in the origin, because the linear system
has a unique solution, and that solution is \( (0,0)\) that is a solution of the system.
We observe that:
The non zero vector \( \overrightarrow{u}_1=\begin{bmatrix} b\\{-a} \end{bmatrix}\) is on the line \( (D_{1})\) , so that \( p(\overrightarrow{u}_1)=\overrightarrow{u}_1\) .
Consequently, \( 1\) is an eigenvalue of \( p\) with eigenvector \( \overrightarrow{u}_1=\begin{bmatrix} b\\{-a} \end{bmatrix}\) .
The non zero vector \( \overrightarrow{u}_2=\begin{bmatrix} d\\{-c} \end{bmatrix}\) is on the line \( (D_{2})\) , so that \( p(\overrightarrow{u}_2)=\overrightarrow{0}\) .
Consequently, \( 0\) is an eigenvalue of \( p\) with eigenvector \( \overrightarrow{u}_2=\begin{bmatrix} d\\{-c} \end{bmatrix}\) .
Assume that \( \sigma\) is the symmetry on the line \( (D)\) of equation \( ax+by=0\) .
We observe that:
The non zero vector \( \overrightarrow{u}_1=\begin{bmatrix} b\\{-a} \end{bmatrix}\) is on the line \( (D)\) , so that \( \sigma(\overrightarrow{u}_1)=\overrightarrow{u}_1\) .
Consequently, \( 1\) is an eigenvalue of \( \sigma\) with eigenvector \( \overrightarrow{u}_1=\begin{bmatrix} b\\{-a} \end{bmatrix}\) .
The non zero vector \( \overrightarrow{u}_2=\begin{bmatrix} a\\b \end{bmatrix}\) is orthogonal to the vector \( \overrightarrow{u}_1\) and thus to the line \( (D)\) , so that \( \sigma(\overrightarrow{u}_2)=-\overrightarrow{u}_2\) .
Consequently, \( -1\) is an eigenvalue of \( \sigma\) with eigenvector \( \overrightarrow{u}_2=\begin{bmatrix} a\\b \end{bmatrix}\) .
The systems of linear equations having \( 0\) ’s as right hand side are said to be homogeneous, and they have special properties that may be used to find eigenvalues of a linear mapping.
Definition 2
Assume that \( (a,b,c,d)\in\mathbb{R}^4\) are real numbers.
Then the homogeneous system with coefficients \( (a,b,c,d)\) is the system with right hand sides \( 0\) ’s:
As \( (x,y)=(0,0)\) is always a solution of an homogeneous system, any homogeneous system has at least one solution.
Theorem 1
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 points of the line passing through the origin and directed by the vector of coordinates \( (-b,a)\) , unless \( a=b=0\) , in which case the solutions are the points of the line passing through the origin and directed by the vector of coordinates \( (-d,c)\) .
And if \( a=b=c=d=0\) the solutions are all the elements of \( \mathbb{R}^2\) .
Proof
Assume that \( (a,b,c,d)\in\mathbb{R}^4\) are real numbers and consider the homogeneous system:
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 equivalent 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 \( ax+by=0\) , that is the equation of the straight line passing the the origin and directed by the vector of coordinates \( (-b,a)\) .
The solutions are thus the points of the line passing through the origin and directed by the vector of coordinates \( (-b,a)\) .
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 \( ax+by=0\) , that is the equation of the straight line passing the the origin and directed by the vector of coordinates \( (-b,a)\) .
The solutions are thus the points of the line passing through the origin and directed by the vector of coordinates \( (-b,a)\) .
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 \( cx+dy=0\) , that is the equation of the straight line passing the the origin and directed by the vector of coordinates \( (-d,c)\) .
The solutions are thus the points of the line passing through the origin and directed by the vector of coordinates \( (-d,c)\) .
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 \( cx+dy=0\) , that is the equation of the straight line passing the the origin and directed by the vector of coordinates \( (-d,c)\) .
The solutions are thus the points of the line passing through the origin and directed by the vector of coordinates \( (-d,c)\) .
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.
The following theorem is a straightforward application of the lecture 9 about the matrix view of a linear sytem.
Theorem 2
Assume that \( (a,b,c,d)\in\mathbb{R}^4\) are real numbers and consider the homogeneous system:
Then the matrix equation equivalent to the homogeneous system is
\( AX=O_{21}\) , with \( A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) and \( O_{21}=\begin{bmatrix}0\\0\end{bmatrix}\) , the null column vector.
The following theorem is a simple transcription of the theorem 2 for the matrix equivalent equation.
Theorem 3
Assume that \( (a,b,c,d)\in\mathbb{R}^4\) are real numbers and consider the matrix \( A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) 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 \( (x,y)\in\mathbb{R}^2\) .
They are the roots of the characteristic polynomial, that is a quadratic polynomial easy to calculate .
Assume that \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping of matrix \( A\) in the canonical basis.
Assume that \( \lambda\in\mathbb{R}\) is a scalar.
Then \( \lambda\) is an eigenvalue of \( f\) if and only if the matrix equation \( AX=\lambda X\) has some non zero solution.
This is equivalent to the fact that the ‘homogeneous’ matrix equation \( (\lambda I-A) X=O_{21}\) has some non zero solution.
Consequently, \( \lambda\) is an eigenvalue of \( f\) if and only if \( \det(\lambda I-A)=0\) .
Definition 3
Assume that \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping of matrix \( A\) in the canonical basis.
We define the characteristic polynomial of the matrix \( A\) as:
Theorem 4
The eigenvalues of the linear mapping \( f\) are the roots of the characteristic polynomial \( \chi(\lambda)\) of its matrix \( A\) in the canonical basis.
Theorem 5
Assume that \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping of matrix \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) in the canonical basis.
Then the characteristic polynomial of the matrix \( A\) is the quadratic polynomial:
Proof
Assume that \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping of matrix \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) in the canonical basis.
We have then:
\( \lambda I-A=\begin{bmatrix} \lambda-a&-b\\{-c}&\lambda-d \end{bmatrix}\) .
so that
\( \chi(\lambda)=(\lambda-a)(\lambda-d)-(-b)(-c)=\lambda^2-(a+d)\lambda+ad-bc\) .
Because a quadratic polynomial has \( 0\) , \( 1\) or \( 2\) real roots, we have the following corollary.
Corollary 1
Any linear mapping in the plane has \( 0\) , \( 1\) or \( 2\) eigenvalues.
Assume that \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping of matrix \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) in the canonical basis, and consider the characteristic polynomial of the matrix \( A\) :
Consider the discriminant of the quadratic polynomial \( \chi(\lambda)\) :
Then the eigenvalues of\( f\) , i.e. the root(s) of \( \chi(\lambda)\) are the following.
If \( \Delta>0\) , then \( f\) has 2 distinct eigenvalues:
\( \lambda_1=\frac{a+d-\sqrt{\Delta}}{2}\) and \( \lambda_2=\frac{a+d+\sqrt{\Delta}}{2}\) .
If \( \Delta=0\) , then \( f\) has 1 eigenvalue: \( \lambda_0=\frac{a+d}{2}\)
If \( \Delta<0\) , then \( f\) has no eigenvalue.
Let’s calculate the eigenvalues of the linear mapping \( f\) with matrix \( A=\begin{bmatrix} 3&4\\4&-3 \end{bmatrix}\) in the canonical basis.
The characteristic polynomial of \( A\) is the determinant of the matrix \( \lambda I-A=\begin{bmatrix} \lambda-3&-4\\{-4}&\lambda+3 \end{bmatrix}\) .
It is thus
\( \chi(\lambda)=(\lambda-3)(\lambda+3)-(-4)\times (-4)=\lambda^2+(3+(-3))\lambda-9-16\) ,
so that \( \chi(\lambda)=\lambda^2-25=(\lambda-5)(\lambda+5)\) .
The roots of \( \chi(\lambda)\) are thus \( 5\) and \( -5\) .
Consequently, \( f\) has two eigenvalues, \( 5\) and \( -5\) .
We find again the two eigenvalues of the example of the paragraph 3.4.1, and we discover that they are the only eigenvalues of \( f\) .
Let’s calculate the eigenvalues of the rotation \( \rho_{\frac{\pi}{2}}\) of angle \( \frac{\pi}{2}\) , with matrix \( R_{\frac{\pi}{2}}=\begin{bmatrix} 0&-1\\1&0 \end{bmatrix}\) in the canonical basis.
The characteristic polynomial of \( R_{\frac{\pi}{2}}\) is the determinant of the matrix \( \lambda I-R_{\frac{\pi}{2}}=\begin{bmatrix} \lambda&1\\{-1}&\lambda \end{bmatrix}\) .
It is thus \( \chi(\lambda)=\lambda\times\lambda-1\times (-1)=\lambda^2+1\) .
\( \chi(\lambda)\) has no real root.
Consequently, the rotation \( \rho_{\frac{\pi}{2}}\) has no real root.
In fact, it may be proved that no rotation of any non zero angle modulo \( \pi\) has any eigenvalue, at least in \( \mathbb{R}\) .
However, such a rotation has two complex eingenvalues that are conjugate to eachother.
Let’s calculate the eigenvalues of the projection \( p\) on the line \( (D_{1})\) of equation \( ax+by=0\) , along the line \( (D_{2})\) of equation \( cx+dy=0\) , where \( (a,b,c,d)\in\mathbb{R}^{4}\) are realnumbers such that \( ad-bc\ne 0\) .
Its matrix in the canonical basis is \( \Pi=\begin{bmatrix} -\frac{bc}{ad-bc}&-\frac{bd}{ad-bc}\\ \\ \frac{ac}{ad-bc}&\frac{ad}{ad-bc}\end{bmatrix}\) , that we denote \( \Pi=\begin{bmatrix} A&B\\C&D \end{bmatrix}\) .
The characteristic polynomial of \( \Pi\) is equal to
\( \chi(\lambda)=\lambda^2-(A+D)\lambda+AD-BC=\lambda^2-\frac{-bc+ad}{ad-bc}\lambda+\frac{-bcad+acbd}{(ad-bc)^2}\)
so that \( \chi(\lambda)=\lambda^2-\lambda=\lambda(\lambda-1)\) .
The roots of \( \chi(\lambda)\) are thus \( 0\) and \( 1\) .
Consequently, \( p\) has two eigenvalues, \( 0\) and \( 1\) .
We find again the two eigenvalues of the example of the paragraph 3.4.3, and we discover that they are the only eigenvalues of \( p\) .
Let’s calculate the eigenvalues of the symmetry \( \sigma\) along the line \( (D)\) of equation \( ax+by=0\) , where \( (a,b)\in\mathbb{R}^{2}\) are real numbers that are not zero together, so that \( a^{2}+b^{2}\ne 0\) .
Its matrix in the canonical basis is \( S=\begin{bmatrix} \frac{a^2-b^2}{a^2+b^2}&\frac{2ab}{a^2+b^2}\\ \\ \frac{2ab}{a^2+b^2}&\frac{b^2-a^2}{a^2+b^2} \end{bmatrix}\) , that we denote \( S=\begin{bmatrix} A&B\\C&D \end{bmatrix}\) .
The characteristic polynomial of \( S\) is equal to
\( \chi(\lambda)=\lambda^2-(A+D)\lambda+AD-BC=\lambda^2-\frac{a^2-b^2+b^2-a^2}{a^2+b^2}\lambda +\frac{(a^2-b^2)(b^2-a^2)-4a^2b^2}{(a^2+b^2)^2}\) ,
so that \( \chi(\lambda)=\lambda^2-\frac{(a^2-b^2)^2+4a^2b^2}{(a^2+b^2)^2} =\lambda^2-\frac{(a^2+b^2)^2}{(a^2+b^2)^2} =\lambda^2-1=(\lambda-1)(\lambda+1)\) .
The roots of \( \chi(\lambda)\) are thus \( 1\) and \( -1\) .
Consequently, \( \sigma\) has two eigenvalues, \( 1\) and \( -1\) .
We find again the two eigenvalues of the example of the paragraph 3.4.4, and we discover that they are the only eigenvalues of \( \sigma\) .
The eigenvectors are strongly relates to the diagonalisation of matrices, and it is important because the diagonal matrices are very easy to manipulate.
To emphasize the advantages of the diagonal matrices, let’s discover the powers of a matrix.
Definition 4
Assume that \( A\) is a \( 2\times 2\) matrix with real elements.
Then the powers \( A^{n}\) of the matrix \( A\) are defined the following way.
\( A^{1}=A\)
\( A^{2}=A\times A\)
\( A^{3}=A\times A\times A\)
…
\( A^{n}=A\times A\times…\times A\)
…
That definition leads straightly to the following definition by recursion.
Theorem 6
Assume that \( A\) is a \( 2\times 2\) matrix with real elements.
Then the powers \( A^{n}\) of the matrix \( A\) may defined by recursion the following way.
Initialisation \( A^{1}=A\) .
Recursion For any \( n\ge 1\) , \( A^{n+1}=A^{n} A\) .
Theorem 7
Assume that \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) is a \( 2\times 2\) diagonal matrix with real elements and that \( n\in\mathbb{N}^*\) is a positive integer.
Then the following assertions hold.
\( D^n=\begin{bmatrix}\lambda_1^n&0\\0&\lambda_2^n\end{bmatrix}\) .
\( A\) is invertible if and only if \( \lambda_1\ne 0\) and \( \lambda_2\ne 0\) .
If \( \lambda_1\ne 0\) and \( \lambda_2\ne 0\) , then the inverse of \( D\) is equal to \( D^{-1}=\begin{bmatrix}1/\lambda_1&0\\0&1/\lambda_2\end{bmatrix}\) .
Lemma 1
Assume that \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) and \( D'=\begin{bmatrix}\mu_1&0\\0&\mu_2\end{bmatrix}\) are \( 2\times 2\) diagonal matrices with real elements. Then the following assertions hold.
\( DD'=\begin{bmatrix}\lambda_1\mu_1&0\\0&\lambda_2\mu_2\end{bmatrix}\) .
\( \det(D)=\lambda_1\lambda_2\)
Proof (of the lemma 1)
Assume that \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) and \( D'=\begin{bmatrix}\mu_1&0\\0&\mu_2\end{bmatrix}\) are \( 2\times 2\) diagonal matrices with real elements. Then the following calculations may be performed.
\( DD'=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix} \begin{bmatrix}\mu_1&0\\0&\mu_2\end{bmatrix} = \begin{bmatrix}\lambda_1\mu_1+0\times 0& \lambda_1\times 0+0\times\mu_2\\ 0\times\mu_1+\lambda_2\times 0& 0\times 0+\lambda_2\mu_2\end{bmatrix} =\begin{bmatrix}\lambda_1\mu_1&0\\0&\lambda_2\mu_2\end{bmatrix}\) .
\( \det(D)=\lambda_1\lambda_2-0\times 0=\lambda_1\lambda_2\) .
Proof (of the theorem 7)
Assume that \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) is a \( 2\times 2\) diagonal matrix with real elements and that \( n\in\mathbb{N}^*\) is a positive integer.
Let’s prove that \( D^n=\begin{bmatrix}\lambda_1^n&0\\0&\lambda_2^n\end{bmatrix}\) by recursion on \( n\ge 1\) .
Initialisation For \( n=1\) , we have \( D^1=D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix} =\begin{bmatrix}\lambda_1^1&0\\0&\lambda_2^1\end{bmatrix}\) .
Recursion Assume that, for some \( n\ge 1\) , \( D^{n}=\begin{bmatrix}\lambda_1^n&0\\0&\lambda_2^n\end{bmatrix}\) .
Then, because of the theorem 6, the item (I) of the lemma 1, and by hypothesi of recusion, we have:
\( D^{n+1}=D^nD =\begin{bmatrix}\lambda_1^n&0\\0&\lambda_2^n\end{bmatrix} \begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix} =\begin{bmatrix}\lambda_1\lambda_1^n&0\\0&\lambda_2^n\lambda_2\end{bmatrix} =\begin{bmatrix}\lambda_1^{n+1}&0\\0&\lambda_2^{n+1}\end{bmatrix}\) .
That ends the proof by recursion.
Because of the item (II) of the lemma 1, the determinant \( \lambda_1\lambda_2\) of \( D\) is non zero if and only if \( \lambda_1\ne 0\) and \( \lambda_2\ne 0\) .
So that \( D\) is invertible if and only if \( \lambda_1\ne 0\) and \( \lambda_2\ne 0\) .
Assume that \( \lambda_1\ne 0\) and \( \lambda_2\ne 0\) .
Then \( A\) is invertible and its inverse is equal to/
\( D^{-1} =\frac{1}{\lambda_1\lambda_2} \begin{bmatrix}\lambda_2&-0\\{-0}&\lambda_1\end{bmatrix} =\begin{bmatrix}1/\lambda_1&0\\0&1/\lambda_2\end{bmatrix}\)
Definition 5
Assume that \( A\) is a \( 2\times 2\) matrix with real elements, and consider the linear mapping \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) having the matrix \( A\) in the canonical basis.
Then \( A\) is said to be diagonalisable if and only if there exists a basis
\( B=(\overrightarrow{I},\overrightarrow{J})\) in which the matrix of \( f\) is a diagonal \( 2\times 2\) matrix with real elements \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) .
Theorem 8
Assume that \( A\) is a \( 2\times 2\) matrix with real elements, and consider the linear mapping \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) having the matrix \( A\) in the canonical basis.
Then the matrix \( A\) is diagonalisable if and only if there exists an invertible matrix \( P\) and a diagonal matrix \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) such that \( A=PDP^{-1}\) .
Moreover, if \( A\) is diagonalisable, then the matrix \( P\) such that \( A=PDP^{-1}\) is the transition matrix from the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) in which the matrix of \( f\) is the diagonal matrix \( D\) , to the canonical basis.
Proof
Assume that \( A\) is a \( 2\times 2\) matrix with real elements, and consider the linear mapping \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) having the matrix \( A\) in the canonical basis.
Assume that \( A\) is diagonalisable.
Consider a basis \( B=(\overrightarrow{I},\overrightarrow{J})\) in which the matrix of \( f\) is the diagonal \( 2\times 2\) matrix with real elements \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) .
Consider the transition matrix \( P\) from the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) to the canonical basis.
Then \( D=P^{-1}AP\) , so that \( A=PDP^{-1}\) .
Assume now that there exists an invertible matrix \( P\) and a diagonal matrix \( D\) such that \( A=PDP^{-1}\) , so that \( D=P^{-1}AP\) .
Consider the basis \( B\) made of the vectors with column vectors of coordinates the column of \( P\) .
Then \( P\) is the transition matrix from the basis \( B\) to the canonical basis, so that the matrix of \( f\) in the basis \( B\) is \( P^{-1}AP=D\) , that is a diagonal matrix.
Consequently, \( A\) is diagonalisable.
Assume that \( \lambda\in\mathbb{R}\) is a real number,
and consider the homothety \( h_{\lambda}\)
of factor \( \lambda\) .
Then in any basis of the plane, \( h_{\lambda}\) has the matrix \( H_{\lambda}=\begin{bmatrix} \lambda&0\\0&\lambda \end{bmatrix}\) .
As it is a scalar matrix and thus a diagonal matrix, the homothety
\( h_{\lambda}\) is diagonalisable.
Consider the symmetry \( \sigma\) along the line \( (D)\) .
Consider a basis \( B=(\overrightarrow{I},\overrightarrow{J})\) made of:
a non zero vector \( \overrightarrow{I}\) of the line \( (D)\) ,
and a non zero vector \( \overrightarrow{J}\) orthogonal to \( (D)\) .
Then the following asserions hold.
\( \sigma(\overrightarrow{I})=\overrightarrow{I}\) ,
and \( \sigma(\overrightarrow{J})=-\overrightarrow{J}\)
So that the matrix of \( f\) in the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) is the diagonal matrix \( S_0=\begin{bmatrix} 1&0\\0&-1 \end{bmatrix}\) .
Consequently, the symmetry \( \sigma\) is diagonalisable.
Consider the projection \( p\) on the line \( (D_{1})\) along the other line \( (D_{2})\) .
Consider a basis \( B=(\overrightarrow{I},\overrightarrow{J})\) made of:
a non zero vector \( \overrightarrow{I}\) of the line \( (D_{1})\) ,
and a non zero vector \( \overrightarrow{J}\) of the line \( (D_{2})\) .
Then the following asserions hold.
\( \sigma(\overrightarrow{I})=\overrightarrow{I}\) ,
and \( \sigma(\overrightarrow{J})=\overrightarrow{0}\)
So that the matrix of \( f\) in the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) is the diagonal matrix \( \Pi_0=\begin{bmatrix} 1&0\\0&0 \end{bmatrix}\) .
Consequently, the projection \( p\) is diagonalisable.
Theorem 9
Assume that \( A\) is a \( 2\times 2\) diagonalisable matrix with real elements.
Consider the linear mapping \( f\) having the matrix \( A\) in the canonical basis.
Consider a basis \( B=(\overrightarrow{I},\overrightarrow{J})\) in which the matrix of \( f\) is the diagonal matrix \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) .
Then the following assertions hold.
\( \lambda_1\) and \( \lambda_2\) are eigenvalues of \( f\) .
\( \overrightarrow{I}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda_1\) .
And \( \overrightarrow{J}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda_2\) .
Proof
Assume that \( A\) is a \( 2\times 2\) diagonalisable matrix with real elements.
Consider the linear mapping \( f\) having the matrix \( A\) in the canonical basis.
Consider a basis \( B=(\overrightarrow{I},\overrightarrow{J})\) in which the matrix of \( f\) is the diagonal matrix \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) .
Then, by definition of a matrix of a linear mapping in some basis, the following assertions hold.
\( f(\overrightarrow{I})=\lambda_1\overrightarrow{I}\) , so that:
\( \lambda_1\) is an eigenvalue of \( f\) ,
and \( \overrightarrow{I}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda_1\) .
And \( f(\overrightarrow{J})=\lambda_2\overrightarrow{J}\) , so that:
\( \lambda_2\) is an eigenvalue of \( f\) ,
and \( \overrightarrow{J}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda_2\) .
Assume that \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping of matrix \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) in the canonical basis that is not a scalar matrix.
Note that if \( A\) is a scalar matrix, then it is also a diagonal matrix, and it is thus already diagonalised.
Assume that \( f\) has an eigenvalue \( \lambda\) , and let’s calculate the column vector of coordinates \( X=\begin{bmatrix}x\\y\end{bmatrix}\) of some eigenvector of \( f\) in the canonical basis.
\( X\) is a non zero solution of the matrix equation \( (\lambda I-A)X=O_{21}\) , that has an infinity of solutions.
Consquently, \( (x,y)\) is a non zero solution of the homogeneous system:
Because of the theorem 1, a non zero solution of that system is the following.
\( x=b\) and \( y=\lambda-a\) , unless \( b=0\) and \( \lambda=a\) ,
in which case \( \lambda\ne d\) because \( A\) is not a scalar matrix, so that a non zero solution of the system is \( x=\lambda-d\) and \( y=c\) .
Consequently, the column vector of coordinates of an eigenvector of \( f\) for the eigenvalue \( \lambda\) is the following.
If it’s not the null vector, the vector with column vector of coordinates \( X=\begin{bmatrix}b\\\lambda-a\end{bmatrix}\) .
Or else the vector with column vector of coordinates \( X'=\begin{bmatrix}\lambda-d\\c\end{bmatrix}\) , that is then non zero.
Theorem 10
Assume that \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping.
Assume that \( f\) has two distinct eigenvalues \( \lambda_1\) and \( \lambda_2\) .
Then any eigenvectors \( \overrightarrow{I}\) and \( \overrightarrow{J}\) of \( f\) for the eigenvalues \( \lambda_1\) and \( \lambda_2\) respectively constitute a basis \( B=(\overrightarrow{I},\overrightarrow{J})\) in which the matrix of \( f\) is the diagonal matrix \( D=\begin{bmatrix} \lambda_1&0\\0&\lambda_2 \end{bmatrix}\) .
Consequently, the matrix of any linear mapping with two distinct eigenvalues is diagonalisable.
Proof
Assume that \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping.
Assume that \( f\) has two distinct eigenvalues \( \lambda_1\) and \( \lambda_2\) .
Assume that \( \overrightarrow{I}\) and \( \overrightarrow{J}\) are eigenvectors of \( f\) for the eigenvalues \( \lambda_1\) and \( \lambda_2\) respectively.
\( \overrightarrow{I}\) and \( \overrightarrow{J}\) are non zero vectors because they are eigenvectors.
Let’s prove by contradiction that \( \overrightarrow{I}\) and \( \overrightarrow{J}\) are non aligned.
Assume that \( \overrightarrow{I}\) and \( \overrightarrow{J}\) are aligned.
Consider the non zero scalar \( k\in\mathbb{R}^{*}\) such that \( \overrightarrow{J}=k\overrightarrow{I}\) .
As \( \overrightarrow{I}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda_1\) , we have \( f(\overrightarrow{I})=\lambda_{1}\overrightarrow{I}\) .
As \( \overrightarrow{J}=k\overrightarrow{I}\) , we have:
\( f(\overrightarrow{J})=kf(\overrightarrow{I})=k(\lambda_{1}\overrightarrow{I}) =\lambda_{1}(k\overrightarrow{I})=\lambda_{1}\overrightarrow{J}\) .
But \( \overrightarrow{J}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda_2\) , we have \( f(\overrightarrow{J})=\lambda_{2}\overrightarrow{J}\) .
Consequently, \( (\lambda_{1}-\lambda_{2})\overrightarrow{J}=\overrightarrow{0}\) , with \( \overrightarrow{J}\ne \overrightarrow{0}\) , so that \( \lambda_{1}-\lambda_{2}=0\) and \( \lambda_{1}=\lambda_{2}\) .
This is contradictory with the fact that \( \lambda_{1}\) and \( \lambda_{2}\) are distinct eigenvalues of \( f\) .
Consequently, \( \overrightarrow{I}\) and \( \overrightarrow{J}\) are non aligned non zero vectors, so that they constitute a basis.
But because of the theorem 8, the following corollary may be deduce from the theorem 10.
Corollary 2
Assume that \( A\) is a \( 2\times 2\) matrix with real elements, and consider the linear mapping \( f:\;\mathbb{P}\rightarrow\mathbb{P}\) having the matrix \( A\) in the canonical basis.
Assume that \( f\) has two distinct eigenvalues \( \lambda_1\) and \( \lambda_2\) , with respective eigenvectors \( \overrightarrow{I}\) and \( \overrightarrow{J}\) .
Then \( A=PDP^{-1}\) , with the following matrices \( P\) and \( D\) .
\( D\) is the diagonal matrix \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) .
And \( P\) is the matrix with columns the column vectors of coordinates of \( \overrightarrow{I}\) and \( \overrightarrow{J}\) .
Theorem 11
If the matrix \( A\) of some linear mapping \( f\) is diagonalisable, then one of the two following assertions hold.
Either\( f\) has two distinct eigenvalues.
Or \( f\) is a homothety, and thus \( A\) is a scalar matrix.
Proof
Assume that \( A\) is a \( 2\times 2\) diagonalisable matrix with real elements.
Consider the linear mapping \( f\) having the matrix \( A\) in the canonical basis.
Consider the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) in which the matrix of \( f\) is the diagonal matrix \( D=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\) .
Then, because of the theorem 9, the following assertions hold.
\( \lambda_1\) and \( \lambda_2\) are eigenvalues of \( f\) .
\( \overrightarrow{I}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda_1\) .
And \( \overrightarrow{J}\) is an eigenvector of \( f\) for the eigenvalue \( \lambda_2\) .
If \( \lambda_{1}=\lambda_{2}\) , then \( D\) is a scalar matrix, that is the matrix in any basis of the homothety of factor \( \lambda_{1}\) , so that \( f\) is an homothety.
And if \( \lambda_{1}\ne \lambda_{2}\) , then \( f\) has two distinct eigenvalues.
We have discovered how to diagonalize matrices with the help of the eigenvalues and eigenvectors of the linear mappings.
We have learned how to calculate the eigenvalues as roots of the so called characteristic polynomial, and how to deduce the correponding eigenvectors.
In the euclidean plane, the characteristic polynomial is quadratic, so that its roots may be calculated algebraically.
It is however not the case in greater dimensions, but this is for the next course.
And you will see in the last document of that publication that most rotations don’t have any real eigenvalues, so that they are not diagonalisable in \( \mathbb{R}\) .
They are however usefully diagonalisable in the field \( \mathbb{C}\) of complex numbers.
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