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.

$\left(\begin{array}{ccc}-3& 2& 2\\ 2& 1& 3\\ 2& 3& 1\end{array}\right)$

Short answer

The Eigen values of given matrix are $\lambda =-2,\lambda =-4$, and $\lambda =5$, and corresponding eigenvectors are $X_1=\left(\left.\begin{array}{c}0\\ -i\\ 1\end{array}\right),X_2=\left(\begin{array}{c}4\\ 1\\ 1\end{array}\right.\right)$ and$X_3=\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)$

Step-by-step solution

Given information

The given matrix is $A=\left(\begin{array}{ccc}-3& 2& 2\\ 2& 1& 3\\ 2& 3& 1\end{array}\right)$.

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. The eigenvector corresponding to Eigen value${\mathit{\lambda }}_{\mathbf{i}}$is given by$\left(A-{\lambda }_{i}l\right){\mathit{X}}_{\mathbf{i}}\mathbf{=}\mathit{O}$where$\mathit{O}$is null matrix.

Find the Eigen values of given function

The characteristic matrix of matrix is

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

The characteristic equation of the matrix Ais

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

Solve Further

$$\begin{gathered} -\left(3+\lambda \right)\left[\left(1-\lambda \right)\left(1-\lambda \right)-9\right]-2\left[2\left(1-\lambda \right)-6\right]+2\left[6-2\left(1-\lambda \right)\right]=0 \\ -\left(3+\lambda \right)\left[{\lambda }^2-2\lambda -8\right]+2\left[2\left(\lambda +2\right)\right]+2\left[2\left(\lambda +2\right)\right]=0 \\ -\left(3+\lambda \right)\left[\left(\lambda -4\right)\left(\lambda +2\right)\right]+8\left(\lambda +2\right)=0 \\ \left(\lambda +2\right)\left[-\left(\lambda -4\right)\left(3+\lambda \right)+8\right]=0 \end{gathered}$$

Then

$$\begin{gathered} \left(\lambda +2\right)\left[{\lambda }^2-\lambda -20\right]=0 \\ \left(\lambda +2\right)\left(\lambda +4\right)\left(\lambda -5\right)=0 \\ \lambda =-2,\lambda =-4,\lambda =5 \end{gathered}$$

Hence, the eigen values of matrix A are $\lambda =-2,\lambda =-4$, and $\lambda =5$.

Find the Eigen vectors of given function

Now, compute eigenvectors corresponding to these eigen values as follows:

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

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

If we set $x=0$then $y+z=0$, further setting $z=1$and$y=-1$.

Thus, the eigenvector corresponding to eigenvaluelocalid="1658819919486" $\lambda =-2$is localid="1658814549665">X1=0-11.

Letlocalid="1658819926026" $X_2=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$be the eigenvector corresponding to eigenvaluelocalid="1658819932116" $\lambda =-4$then

localid="1658819941982" $\left(A-\lambda l\right)X_2=0$

$$\begin{gathered} \left[A-\left(-4\right)l_3\right]\left|X_1=\left[A+4l_3\right]\right|X_1 \\ \left[\left(\begin{array}{ccc}-3& 2& 2\\ 2& 1& 3\\ 2& 3& 1\end{array}\right)+4\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right]\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}-1& 2& 2\\ 2& 5& 3\\ 2& 3& 5\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \\ \left(\begin{array}{c}-x+2y+2z\\ 2x+5y+3z\\ 2x+3y+5z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

If we set$y=1$and $z=1$, then $x=-4$.

Hence, the eigenvector corresponding to eigen valuelocalid="1658820132506" $\lambda =-4$is localid="1658820068146" $X_2=\left(\begin{array}{c}-4\\ 1\\ 1\end{array}\right)$.

Letlocalid="1658814833575">X3=xyzbe the eigenvector corresponding to eigenvaluelocalid="1658819979586" $\lambda =5$then$\left(A-\lambda l\right)X_3=0$

$$\begin{gathered} \left[A-\left(5\right)l_3\right]\left|X_1=\left[A5l_3\right]\right|X_1 \\ \left[\left(\begin{array}{ccc}-3& 2& 2\\ 2& 1& 3\\ 2& 3& 1\end{array}\right)-5\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right]\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}-8& 2& 2\\ 2& -4& 3\\ 2& 3& -4\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \\ \left(\begin{array}{c}-8x+2y+2z\\ 2x-4y+3z\\ 2x+3y-4z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

If we set$x=1$and,22 then $z=2$

Hence, the eigenvector corresponding to eigenvaluelocalid="1658819997260" $\lambda =5$is localid="1658819987305" $X_3=\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)$.