Eigenvalues and Eigenvectors of Matrices. Methods for Finding Them.

Literature: Collection of Problems in Mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

Let the number $\lambda$ and the vector $x \in L, x \neq 0$ be such that $$Ax=\lambda x.\qquad\qquad\qquad\qquad\qquad(1)$$ Then the number $\lambda$ is called an eigenvalue of the linear operator $A$, and the vector $x$ is called an eigenvector of this operator, corresponding to the eigenvalue $\lambda$.

In a finite-dimensional space $L_n$, the vector equality (1) is equivalent to the matrix equality $$(A-\lambda E)X=0,\,\,\,\, X\neq 0.\qquad\qquad\quad\quad (2)$$

From this, it follows that the number $\lambda$ is an eigenvalue of the operator $A$ if and only if the determinant $\det(A - \lambda E) = 0$, i.e., $\lambda$ is a root of the polynomial $p(\lambda) = \det(A - \lambda E)$, called the characteristic polynomial of the operator $A$. The coordinate column $X$ of any eigenvector corresponding to the eigenvalue $\lambda$ is a nontrivial solution of the homogeneous system (2).

Examples.

Find the eigenvalues and eigenvectors of linear operators given by their matrices.

4.134. $A = \begin{pmatrix} 2 & -1 & 2 \ 5 & -3 & 3 \ -1 & 0 & -2 \end{pmatrix}.$

Solution.

Let's find the eigenvectors of the given linear operator. The number $\lambda$ is an eigenvalue of the operator $A$ if and only if $\det(A - \lambda E) = 0$. Let's write the characteristic equation:

$$A-\lambda E=\begin{pmatrix}2&-1&2\\5&-3&3\\-1&0&-2\end{pmatrix}-\lambda\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=$$ $$=\begin{pmatrix}2-\lambda&-1&2\\5&-3-\lambda&3\\-1&0&-2-\lambda\end{pmatrix}.$$

$$det(A-\lambda E)=\begin{vmatrix}2-\lambda&-1&2\\5&-3-\lambda&3\\-1&0&-2-\lambda\end{vmatrix}=$$ $$=(2-\lambda)(-3-\lambda)(-2-\lambda)+3+2(-3-\lambda)+5(-2-\lambda)=$$ $$=-\lambda^3-3\lambda^2+4\lambda+12+3-6-2\lambda-10-5\lambda=-\lambda^3-3\lambda^2-3\lambda-1=0.$$

Let's solve the found equation to find the eigenvalues.

$$\lambda^3+3\lambda^2+3\lambda+1=(\lambda^3+1)+3\lambda(\lambda+1)=$$ $$=(\lambda+1)(\lambda^2-\lambda+1)+3\lambda(\lambda+1)=(\lambda+1)(\lambda^2-\lambda+1+3\lambda)=$$ $$=(\lambda+1)(\lambda^2+2\lambda+1)=(\lambda+1)^3=0\Rightarrow \lambda=-1.$$

The eigenvector for the eigenvalue $\lambda = -1$ can be found from the system $$(A-\lambda E)X=0, X\neq 0, \Rightarrow (A+E)X=0, X\neq 0$$

$$(A+E)X=\begin{pmatrix}2+1&-1&2\\5&-3+1&3\\-1&0&-2+1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=$$ $$=\begin{pmatrix}3x_1-x_2+2x_3\\5x_1-2x_2+3x_3\\-x_1-x_3\end{pmatrix}=0.$$

Let's solve the homogeneous system of equations:

$$\left\{\begin{array}{lcl}3x_1-x_2+2x_3=0\\ 5x_1-2x_2+3x_3=0\\-x_1-x_3=0\end{array}\right.$$

Let's compute the rank of the coefficient matrix $A = \begin{pmatrix}3 & -1 & 2 \ 5 & -2 & 3 \ -1 & 0 & -1\end{pmatrix}$ using the method of bordered minors:

We fix a non-zero second-order minor $M_2 = \begin{vmatrix}3 & -1 \ 5 & -2\end{vmatrix} = -6 + 5 = -1 \neq 0.$

Next, we consider the bordered third-order minor: $\begin{vmatrix}3 & -1 & 2 \ 5 & -2 & 3 \ -1 & 0 & -1\end{vmatrix} = 6 + 3 - 4 - 5 = 0;$

Thus, the rank of the matrix $A$ is two.

Choosing the basic minor $M = \begin{vmatrix}3 & -1 \ 5 & -2\end{vmatrix} = -1 \neq 0.$ Then, assuming $x_3 = c,$ we obtain:

$$\left\{\begin{array}{lcl}3x_1-x_2+2с=0\\ 5x_1-2x_2+3с=0\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}3x_1-x_2=-2c\\5x_1-2x_2=-3c\end{array}\right.$$

Using Cramer's rule, we find $x_1$ and $x_2:$

$\Delta=\begin{vmatrix}3&-1\\5&-2\end{vmatrix}=-6+5=-1;$

$\Delta_1=\begin{vmatrix}-2c&-1\\-3c&-2\end{vmatrix}=4c-3c=c;$

$\Delta_2=\begin{vmatrix}3&-2c\\5&-3c\end{vmatrix}=-9c+10c=c;$

$x_1=\frac{\Delta_1}{\Delta}=\frac{c}{-1}=-c;$ $x_2=\frac{\Delta_2}{\Delta}=\frac{c}{-1}=-c.$

Thus, the general solution of the system is $X(c) = \begin{pmatrix} -c \ -c \ c \end{pmatrix}.$

From the general solution, we find the fundamental system of solutions: $E = X(1) = \begin{pmatrix} -1 \ -1 \ 1 \end{pmatrix}.$

Using the fundamental system of solutions, the general solution can be written as $X(c) = cE.$

Answer: $\lambda = -1;$ $X = c \begin{pmatrix} -1 \ -1 \ 1 \end{pmatrix}, c \neq 0.$

4.143. $A=\begin{pmatrix}0&-1&0\\1&1&-2\\1&-1&0\end{pmatrix}.$

Solution.

Let's find the eigenvectors of the given linear operator. The number $\lambda$ is an eigenvalue of the operator $A$ if and only if $\det(A - \lambda E) = 0$. Let's write down the characteristic equation:

$$A-\lambda E=\begin{pmatrix}0&-1&0\\1&1&-2\\1&-1&0\end{pmatrix}-\lambda\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=$$ $$=\begin{pmatrix}-\lambda&-1&0\\1&1-\lambda&-2\\1&-1&-\lambda\end{pmatrix}.$$

$$det(A-\lambda E)=\begin{vmatrix}-\lambda&-1&0\\1&1-\lambda&-2\\1&-1&-\lambda\end{vmatrix}=$$ $$=-\lambda(1-\lambda)(-\lambda)+2-\lambda+2\lambda=$$ $$=-\lambda^3+\lambda^2+\lambda+2=0.$$

Let's solve the found equation to find the eigenvalues.

$$-\lambda^3+\lambda^2+\lambda+2=(\lambda-2)(-\lambda^2-\lambda-1)=0\Rightarrow \lambda=2.$$

We will find the eigenvector for the eigenvalue $\lambda = 2$ from the system$$(A-\lambda E)X=0, X\neq 0, \Rightarrow (A-2E)X=0, X\neq 0$$

$$(A-2E)X=\begin{pmatrix}-2&-1&0\\1&-1&-2\\1&-1&-2\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=$$ $$=\begin{pmatrix}-2x_1-x_2\\x_1-x_2-2x_3\\x_1-x_2-2x_3\end{pmatrix}=0.$$

Let's solve the homogeneous system of equations:

$$\left\{\begin{array}{lcl}-2x_1-x_2=0\\ x_1-x_2-2x_3=0\\x_1-x_2-2x_3=0\end{array}\right.$$

Let's compute the rank of the coefficient matrix $A = \begin{pmatrix}-2 & -1 & 0 \ 1 & -1 & -2 \ 1 & -1 & -2\end{pmatrix}$ using the method of bordered minors:

We fix a non-zero second-order minor $M_2 = \begin{vmatrix}-2 & -1 \ 1 & -1\end{vmatrix} = 2 + 1 = 3 \neq 0.$

Consider the bordered third-order minor: $\begin{vmatrix}-2 & -1 & 0 \ 1 & -1 & -2 \ 1 & -1 & -2\end{vmatrix} = 0;$

Thus, the rank of matrix $A$ is two.

Choose as the basic minor $M = \begin{vmatrix}-2 & -1 \ 1 & -1\end{vmatrix} = 3 \neq 0.$ Then, assuming $x_3 = c,$ we obtain: $$\left\{\begin{array}{lcl}-2x_1-x_2=0\\ x_1-x_2-2с=0\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}-2x_1-x_2=0\\x_1-x_2=2c\end{array}\right.$$

Using Cramer's rule, we find $x_1$ and $x_2:$

$\Delta=\begin{vmatrix}-2&-1\\1&-1\end{vmatrix}=2+1=3;$

$\Delta_1=\begin{vmatrix}0&-1\\2c&-1\end{vmatrix}=2c;$

$\Delta_2=\begin{vmatrix}-2&0\\1&2c\end{vmatrix}=-4c;$

$x_1=\frac{\Delta_1}{\Delta}=\frac{2c}{3};$ $x_2=\frac{\Delta_2}{\Delta}=\frac{-4c}{3}.$

Thus, the general solution of the system is $X(c) = \begin{pmatrix}\frac{2c}{3} \ -\frac{4c}{3} \ c \end{pmatrix}.$

From the general solution, we find the fundamental system of solutions: $E = X(1) = \begin{pmatrix}\frac{2}{3} \ -\frac{4}{3} \ 1 \end{pmatrix}.$

Using the fundamental system of solutions, the general solution can be written in the form $X(c) = cE.$ By redefining the constant, $\alpha = 3c,$ we obtain the eigenvector $X = \alpha \begin{pmatrix}2 \ -4 \ 3\end{pmatrix}, \alpha \neq 0.$

Answer: $\lambda = 2;$ $X = \alpha \begin{pmatrix}2 \ -4 \ 3\end{pmatrix}, \alpha \neq 0.$

Homework.

Find the eigenvalues and eigenvectors of linear operators given by their matrices.

4.135. $A = \begin{pmatrix}0 & 1 & 0 \\ -4 & 4 & 0 \\ -2 & 1 & 2\end{pmatrix}.$

Answer: $\lambda = 2;$ $X = c_1 \begin{pmatrix}1 \ 2 \ 0\end{pmatrix} + c_2 \begin{pmatrix}0 \ 0 \ 1\end{pmatrix},$ where $c_1$ and $c_2$ are not both zero.

4.142. $A = \begin{pmatrix}1 & -3 & 1 \\ 3 & -3 & -1 \\ 3 & -5 & 1\end{pmatrix}.$

Answer:$\lambda_1=-1,$ $X(\lambda_1)=c\begin{pmatrix}1\\1\\1\end{pmatrix};$ $\lambda_2=2,$ $X(\lambda_2)=c\begin{pmatrix}4\\1\\7\end{pmatrix};$ $\lambda_3=-2,$ $X(\lambda_3)=c\begin{pmatrix}2\\3\\3\end{pmatrix}, c\neq 0.$

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