Eigenvalues and Eigenvectors of Matrices. Methods for Finding Them.

Let the number $\lambda$ and the vector $x\in L,x\ne 0$ be such that $$Ax=\lambda x.(1)$$ 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 $$(A-\lambda E)X=0,X\ne 0.(2)$$

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.

1. $A=\left(\begin{array}{ccccccc}2& -1& 25& -3& 3-1& 0& -2\end{array}\right).$

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:

$$A-\lambda E=\left(\begin{array}{ccc}2& -1& 2\\ 5& -3& 3\\ -1& 0& -2\end{array}\right)-\lambda \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=$$ $$=\left(\begin{array}{ccc}2-\lambda & -1& 2\\ 5& -3-\lambda & 3\\ -1& 0& -2-\lambda \end{array}\right).$$

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

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 $$(A-\lambda E)X=0,X\ne 0,\Rightarrow (A+E)X=0,X\ne 0$$

$$(A+E)X=\left(\begin{array}{ccc}2+1& -1& 2\\ 5& -3+1& 3\\ -1& 0& -2+1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=$$ $$=\left(\begin{array}{c}3x_1-x_2+2x_3\\ 5x_1-2x_2+3x_3\\ -x_1-x_3\end{array}\right)=0.$$

Let's solve the homogeneous system of equations:

$$\{\begin{array}{l}3x_1-x_2+2x_3=0\\ 5x_1-2x_2+3x_3=0\\ -x_1-x_3=0\end{array}$$

Let's compute the rank of the coefficient matrix $A=\left(\begin{array}{ccccccc}3& -1& 25& -2& 3-1& 0& -1\end{array}\right)$ using the method of bordered minors:

We fix a non-zero second-order minor $M_2=\left|\begin{array}{ccc}3& -15& -2\end{array}\right|=-6+5=-1\ne 0.$

Next, we consider the bordered third-order minor: $\left|\begin{array}{ccccccc}3& -1& 25& -2& 3-1& 0& -1\end{array}\right|=6+3-4-5=0;$

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

Choosing the basic minor $M=\left|\begin{array}{ccc}3& -15& -2\end{array}\right|=-1\ne 0.$ Then, assuming $x_3=c,$ we obtain:

$$\{\begin{array}{l}3x_1-x_2+2с=0\\ 5x_1-2x_2+3с=0\end{array}\Rightarrow \{\begin{array}{l}3x_1-x_2=-2c\\ 5x_1-2x_2=-3c\end{array}$$

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

$\mathrm{\Delta }=\left|\begin{array}{cc}3& -1\\ 5& -2\end{array}\right|=-6+5=-1;$

${\mathrm{\Delta }}_1=\left|\begin{array}{cc}-2c& -1\\ -3c& -2\end{array}\right|=4c-3c=c;$

${\mathrm{\Delta }}_2=\left|\begin{array}{cc}3& -2c\\ 5& -3c\end{array}\right|=-9c+10c=c;$

$x_1=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{c}{-1}=-c;$ $x_2=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{c}{-1}=-c.$

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

From the general solution, we find the fundamental system of solutions: $E=X(1)=\left(\begin{array}{c}-1-11\end{array}\right).$

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

Answer: $\lambda =-1;$ $X=c\left(\begin{array}{c}-1-11\end{array}\right),c\ne 0.$

2. $A=\left(\begin{array}{ccc}0& -1& 0\\ 1& 1& -2\\ 1& -1& 0\end{array}\right).$

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:

$$A-\lambda E=\left(\begin{array}{ccc}0& -1& 0\\ 1& 1& -2\\ 1& -1& 0\end{array}\right)-\lambda \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=$$ $$=\left(\begin{array}{ccc}-\lambda & -1& 0\\ 1& 1-\lambda & -2\\ 1& -1& -\lambda \end{array}\right).$$

$$det(A-\lambda E)=\left|\begin{array}{ccc}-\lambda & -1& 0\\ 1& 1-\lambda & -2\\ 1& -1& -\lambda \end{array}\right|=$$ $$=-\lambda (1-\lambda )(-\lambda )+2-\lambda +2\lambda =$$ $$=-{\lambda }^3+{\lambda }^2+\lambda +2=0.$$

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$$(A-\lambda E)X=0,X\ne 0,\Rightarrow (A-2E)X=0,X\ne 0$$

$$(A-2E)X=\left(\begin{array}{ccc}-2& -1& 0\\ 1& -1& -2\\ 1& -1& -2\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=$$ $$=\left(\begin{array}{c}-2x_1-x_2\\ x_1-x_2-2x_3\\ x_1-x_2-2x_3\end{array}\right)=0.$$

Let's solve the homogeneous system of equations:

$$\{\begin{array}{l}-2x_1-x_2=0\\ x_1-x_2-2x_3=0\\ x_1-x_2-2x_3=0\end{array}$$

Let's compute the rank of the coefficient matrix $A=\left(\begin{array}{ccccccc}-2& -1& 01& -1& -21& -1& -2\end{array}\right)$ using the method of bordered minors:

We fix a non-zero second-order minor $M_2=\left|\begin{array}{ccc}-2& -11& -1\end{array}\right|=2+1=3\ne 0.$

Consider the bordered third-order minor: $\left|\begin{array}{ccccccc}-2& -1& 01& -1& -21& -1& -2\end{array}\right|=0;$

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

Choose as the basic minor $M=\left|\begin{array}{ccc}-2& -11& -1\end{array}\right|=3\ne 0.$ Then, assuming $x_3=c,$ we obtain: $$\{\begin{array}{l}-2x_1-x_2=0\\ x_1-x_2-2с=0\end{array}\Rightarrow \{\begin{array}{l}-2x_1-x_2=0\\ x_1-x_2=2c\end{array}$$

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

$\mathrm{\Delta }=\left|\begin{array}{cc}-2& -1\\ 1& -1\end{array}\right|=2+1=3;$

${\mathrm{\Delta }}_1=\left|\begin{array}{cc}0& -1\\ 2c& -1\end{array}\right|=2c;$

${\mathrm{\Delta }}_2=\left|\begin{array}{cc}-2& 0\\ 1& 2c\end{array}\right|=-4c;$

$x_1=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{2c}{3};$ $x_2=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{-4c}{3}.$

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

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

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 $X=\alpha \left(\begin{array}{c}2-43\end{array}\right),\alpha \ne 0.$

Answer: $\lambda =2;$ $X=\alpha \left(\begin{array}{c}2-43\end{array}\right),\alpha \ne 0.$

Homework.

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

3. $A=\left(\begin{array}{ccc}0& 1& 0\\ -4& 4& 0\\ -2& 1& 2\end{array}\right).$

Answer: $\lambda =2;$ $X=c_1\left(\begin{array}{c}120\end{array}\right)+c_2\left(\begin{array}{c}001\end{array}\right),$ where $c_1$ and $c_2$ are not both zero.

4. $A=\left(\begin{array}{ccc}1& -3& 1\\ 3& -3& -1\\ 3& -5& 1\end{array}\right).$

Answer:${\lambda }_1=-1,$ $X({\lambda }_1)=c\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right);$ ${\lambda }_2=2,$ $X({\lambda }_2)=c\left(\begin{array}{c}4\\ 1\\ 7\end{array}\right);$ ${\lambda }_3=-2,$ $X({\lambda }_3)=c\left(\begin{array}{c}2\\ 3\\ 3\end{array}\right),c\ne 0.$

Tags: eigenvalue, eigenvector, linear algebra, vector, vector algebra