Eigenvalues ​​and eigenvectors of matrices and endomorphism

 

matrices-matrix-eigenvalues-eigenvectors

In this article, we learn to calculate the eigenvalues and eigenvectors of matrices and endomorphism. In fact, we will introduce simple techniques to calculate the eigenvalues ​​and the associated eigenspace (characteristic space).

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 by a combinations of the others lines (also if we replace any column by 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*}

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 $c\in \mathbb{C}^\ast$.

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