Question

Find the Eigen values and eigenvectors of the following matrices. Do some problems by hand to be sure you understand what the process means. Then check your results by computer.

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

Short answer

The eigen values of given matrix are $\lambda =-2,\lambda =2\mathrm{and}\lambda =3$and corresponding eigenvectors are $X_1=\left(\begin{array}{c}1\\ 0\\ -2\end{array}\right),X_2=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\mathrm{and}{X}_3=\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)$

Step-by-step solution

Given information

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

Definition of Eigen values and Eigen vectors

Eigen values are the special set of scalar values that is associated with the set of linear equations most probably in the matrix equations

An Eigen vector or characteristic vector of a linear transformation is a nonzero vector that changes at most by a scalar factor when that linear transformation is applied to it.

The roots of characteristic equation$\left|A-\lambda l\right|\mathbf{=}\mathbf{0}$of matrix A are known as the Eigen values of matrix A(I the unit matrix of same order as of A. The eigenvector corresponding to Eigen value${\mathit{\lambda }}_{\mathbf{1}}$is given by $\left(A-\lambda ,l\right){\mathit{X}}_{\mathbf{1}}\mathbf{=}\mathit{O}$ where Ois null matrix.

Find the Eigen values of given function

The characteristic matrix of matrix A is $\left(A-\lambda l\right)$

role="math" localid="1658830844704" $$\begin{gathered} A-\lambda l=\left(\begin{array}{ccc}2& 0& 2\\ 0& 2& 0\\ 2& 0& -1\end{array}\right)-\lambda \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right) \\ =\left(\begin{array}{ccc}2-\lambda & 0& 2\\ 0& 2-\lambda & 0\\ 2& 0& -1-\lambda \end{array}\right) \end{gathered}$$

The characteristic equation of the matrix A is

$\left|A-\lambda l\right|=0$

$$\begin{gathered} \left(\begin{array}{ccc}2-\lambda & 0& 2\\ 0& 2-\lambda & 0\\ 2& 0& -1-\lambda \end{array}\right)=0 \\ -(2-\lambda )\left[\left(\lambda -2\right)(\lambda +1)-4\right]=0 \\ (\lambda -2)\left[{\lambda }^2-\lambda -6\right]=0 \\ (\lambda -2)(\lambda +2)(\lambda -3)=0 \\ \lambda =-2,\lambda =2,\lambda =3 \end{gathered}$$

Hence, the eigenvalues values of matrix A are $\lambda =-2,\lambda =2\mathrm{and}\lambda =3$.

Find the Eigen vectors of given function

$\lambda =3is{X}_3=\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)$Now, compute eigenvectors corresponding to these eigen values as follows:

Eigenvector corresponding to $\lambda =-2$.

Let $X_1=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$be the eigenvector corresponding to eigen value $\lambda =-2$then

$\left(A=\lambda l\right)X_1=0$

$\left|A-(-2)l_3\right|X_1=\left|A+2l_3\right|X_1$

$$\begin{gathered} \left[\left(\begin{array}{ccc}2& 0& 2\\ 0& 2& 0\\ 2& 0& -1\end{array}\right)+2\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right]\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}4& 0& 2\\ 0& 4& 0\\ 2& 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \\ \left(\begin{array}{c}4x+0y+2z\\ 0x+4y+0z\\ 2x+0y+z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ 4x+0y+2z=0, \\ 0x+4y+0z=0,2x+0y+z \end{gathered}$$

Hence, the eigenvector corresponding to eigenvalue $\lambda =-2\mathrm{is}{X}_1=\left(\begin{array}{c}1\\ 0\\ -2\end{array}\right)$.

Let $X_2=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$be the eigenvector corresponding to eigenvalue $\lambda =2$then $\left(A-\lambda l\right)X_2=0$

$$\begin{gathered} [A-\left(2\right)l_3\overline{)X_1=\left|A-2l_3\right|X_1=O} \\ \left[\left(\begin{array}{ccc}2& 0& 2\\ 0& 2& 0\\ 2& 0& -1\end{array}\right)-2\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right]\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \\ =\left(\begin{array}{ccc}0& 0& 2\\ 0& 0& 0\\ 2& 0& -3\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \\ \left(\begin{array}{c}0x+0y+2z\\ 0x+0y+0z\\ 2x+0y-3z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ 0\mathrm{x}+0\mathrm{y}+2\mathrm{z}+\mathrm{oy}-3\mathrm{z}=0 \end{gathered}$$

Setting $z=0\Rightarrow x=0,y$ is nonzero we sety = 1.

Hence, the eigenvector corresponding to eigenvalue $\lambda =2\mathrm{is}{X}_2=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$.

$[A-(3N_3]X_1=[A-3J_3\overline{)X_1}$

$$\begin{gathered} \left[\left(\begin{array}{ccc}2& 0& 2\\ 0& 2& 0\\ 2& 0& -1\end{array}\right)-3\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right]\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}-1& 0& 2\\ 0& -1& 0\\ 2& 0& -4\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \\ \left(\begin{array}{c}-x+0y+2z\\ 0x-y+0z\\ 2x+0y-4z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ -x+0y+2z=0 \\ 0x-y+0z=0 \end{gathered}$$

Setting z = 1 , we have x = 2, and we have y = 0.

Hence, the eigenvector corresponding to eigenvalue $\lambda =3is{X}_3=\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)$.