We learn to you how to find the eigenvalues of a matrix as well as the associated eigenvectors. This notion is very useful because it makes it possible to simplify the matrices only by changing the base of the space. We can prove in some cases that some matrices are similar to diagonal matrices, so matrix power calculations become simple in such cases. We assume that you are familiarized with the matrix operations.

## The answer to how to find the eigenvalues of a matrix.

Exercise: Determine the eigenvalues and the associated characteristic spaces of the following matrix: \begin{align*} A=\begin{pmatrix} 2&-1&1\\ -1&2&-1\\ -1&1&0\end{pmatrix}\end{align*}.

Solution: If we denote by $\sigma(A)$ the set of all eigenvalues of $A$ ”called also the spectrum of $A$”, then by definition we have \begin{align*} \sigma(A)&=\{\lambda \in\mathbb{C}: \lambda I_3-A\;\text{is not injective}\}\cr &= \{\lambda\in\mathbb{C}:\det( \lambda I_3-A)=0\}. \end{align*} For any $\lambda\in\mathbb{C}$ we have \begin{align*} \det( \lambda I_3-A)= \begin{vmatrix} \lambda-2&1&-1\\1&\lambda-2&1\\1&-1&\lambda\end{vmatrix}. \end{align*} We know that the determinant is unchanged if we replace any line of the matrix with a combination of the other lines “also if we replace any column with a linear combination of the others columns”.

We denote by $L_i$ for $i=1,2,3,$ the lines of any matrix of order $3$. By replacing the line $L_1$ by $L_1+L_3$, we obtain \begin{align*} \det( \lambda I_3-A)&= \begin{vmatrix} \lambda-1&0&\lambda-1\\1&\lambda-2&1\\1&-1&\lambda\end{vmatrix}\cr &= (\lambda-1)\begin{vmatrix} 1&0&1\\1&\lambda-2&1\\1&-1&\lambda\end{vmatrix} \end{align*} In the lest determinant we replace the line $L_2$ by $L_2-L_3,$ we obtain \begin{align*} \det( \lambda I_3-A)&= (\lambda-1)\begin{vmatrix} 1&0&1\\0&\lambda-1&1-\lambda\\1&-1&\lambda\end{vmatrix}\cr &=(\lambda-1)^2\begin{vmatrix} 1&0&1\\0&1&-1\\1&-1&\lambda\end{vmatrix} \end{align*} We replace the line $L_3$ by $L_3+L_2-L_1$, we obtain \begin{align*} \det( \lambda I_3-A) &=(\lambda-1)^2\begin{vmatrix} 1&0&1\\0&1&-1\\0&0&\lambda-2\end{vmatrix}\cr &= (\lambda-1)^2(\lambda-2). \end{align*} Then the matrix $A$ process two eigenvalues “$\lambda_1=1$ is a double eigenvalue and $\lambda_2=2$ is a simple eigenvalue”.

Let us denote by $E_1$ the characteristic space associated to eigenvalue $\lambda_1=1$ and $E_2$ the characteristic space associated to eigenvalue $\lambda_2=2$. By definition we have \begin{align*} E_1=\ker(I_3-A),\quad E_2=\ker(2I_3-A). \end{align*} Then \begin{align*} X=\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)\in E_1&\;\Longleftrightarrow\; A X=X\cr &\;\Longleftrightarrow\;\begin{cases} x-y+z=0\\ -x+y_z=0\\ -x+y-z=0\end{cases} \cr &\;\Longleftrightarrow\; x-y+z=0 \cr &\;\Longleftrightarrow\; X=\left(\begin{smallmatrix}y-z\\y\\z\end{smallmatrix}\right) \cr &\;\Longleftrightarrow\; X=\left(\begin{smallmatrix}y\\y\\0\end{smallmatrix}\right)+\left(\begin{smallmatrix}-z\\0\\z\end{smallmatrix}\right)\cr &\;\Longleftrightarrow\; X=y\left(\begin{smallmatrix}1\\1\\0\end{smallmatrix}\right)+z\left(\begin{smallmatrix}-1\\0\\1\end{smallmatrix}\right)\cr &;\Longleftrightarrow\; X\in {\rm span}\left\{\left(\begin{smallmatrix}1\\1\\0\end{smallmatrix}\right),\left(\begin{smallmatrix}-1\\0\\1\end{smallmatrix}\right)\right\}. \end{align*} This shows that \begin{align*} E_1={\rm span}\left\{\left(\begin{smallmatrix}1\\1\\0\end{smallmatrix}\right),\left(\begin{smallmatrix}-1\\0\\1\end{smallmatrix}\right)\right\}. \end{align*} Now we calculate $E_2$. Let $X=\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)$. Then \begin{align*} X\in E^2&\;\Longleftrightarrow\; AX=2X\cr &\;\Longleftrightarrow\;\begin{cases} -y+z=0\\ -x-z=0\\-x+y-2z=0\end{cases} \cr &\;\Longleftrightarrow\; y=z,\quad x=-z \cr &\;\Longleftrightarrow\; X=\left(\begin{smallmatrix}-z\\z\\z\end{smallmatrix}\right)\cr &\;\Longleftrightarrow\; X=z\left(\begin{smallmatrix}-1\\1\\1\end{smallmatrix}\right)\cr &\;\Longleftrightarrow\; X\in {\rm span}\left\{\left(\begin{smallmatrix}-1\\1\\1\end{smallmatrix}\right)\right\}. \end{align*} Hence \begin{align*} E_2={\rm span}\left\{\left(\begin{smallmatrix}-1\\1\\1\end{smallmatrix}\right)\right\}. \end{align*}

## Eigenvalues of endomorphism

Exercise: Determine eigenvalues and eigenvectors of the endomorphism \begin{align*}f:\mathbb{C}&[X]\longrightarrow \mathbb{C}[X]\cr & P\longmapsto f(P)=P-(X-1)P’.\end{align*}

Solution: Let $\lambda\in\mathbb{C}$ be an eigenvalue of $f$. Then there exists a polynomial $P\in \mathbb{C}$ non-null such that $f(P)=\lambda P$. This means that \begin{align*} (1-\lambda)P-(X-1)P’=0. \end{align*}If $\lambda=1$ then we have $P’=0$. So $P$ is a non-null constant.

Assume that $\lambda\neq 1$. Define the polynomial function $u(x)=P(x)$ for any $x\in (1,\infty)$. This implies that the function $u$ is the solution of the following differential equation \begin{align*} (x-1)u’-(1-\lambda)u. \end{align*} As $P\neq 0$ then $u$ is non null. Then \begin{align*} \frac{u’}{u}=\frac{1-\lambda}{x-1}. \end{align*} This is equivalent to \begin{align*} u(x)=c(x-1)^{1-\lambda},\qquad c\in \mathbb{C}^\ast. \end{align*} For $u$ to be a polynomial function it is necessary that $1-\lambda\in \mathbb{N}$. This means that $\lambda=1-n$ with $n\in\mathbb{N}^\ast$, because $\lambda\neq 1$. We then have $u(x)=c (x-1)^n$ for any $x\in (1,+\infty)$ and $n\in\mathbb{N}^\ast$. Then \begin{align*} P=c(X-1)^n. \end{align*} Conversely, if $n\in\mathbb{N}$ and $P=c(X-1)^n$ with $c\in \mathbb{C}^\ast,$ then $f(P)=(1-n)P$. Thus the eigenvalues of $f$ are $\{1-n:n\in\mathbb{N}\}$ and the associated eigenvectors are $P=c(X-1)^n$ with $c\in \mathbb{C}^\ast$.