91Ó°ÊÓ

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

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$\mathit{|}\mathit{A}\mathit{-}\mathit{λ}\mathit{l}\mathit{|}\mathbf{=}\mathbf{0}$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${\mathit{λ}}_{\mathbf{i}}$is given by$\mathit{|}\mathit{A}\mathit{-}{\mathit{λ}}_i\mathit{l}\mathit{|}{\mathit{X}}_i\mathbf{=}\mathbf{0}$where Ois null matrix.

The characteristic matrix of matrix A is$\left(A-\mathrm{λ}l\right)$

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

Solve further

$$\begin{gathered} \left|A-\mathrm{λ}\mathrm{l}\right|=0 \\ \left|\begin{array}{ccc}1-\mathrm{λ}& 1& -1\\ 1& 1-\mathrm{λ}& 1\\ -1& 1& -1-\mathrm{λ}\end{array}\right|=0 \\ \left(1-\mathrm{λ}\right)\left[\left(1-\mathrm{λ}\right)\left(-1-\mathrm{λ}\right)-1\right]-1\left[1\left(-1-\mathrm{λ}\right)\right]+1\left[1+1\left(1-\mathrm{λ}\right)\right]=0 \end{gathered}$$

The characteristic equation of the matrix A is$\left(1-\mathrm{λ}\right)\left[-l+{\mathrm{λ}}^2-1\right]-1\mathrm{l}-\mathrm{λ}]-1\left|2-\mathrm{λ}\right|=0$

$$\begin{gathered} \left(1-\mathrm{λ}\right)\left[-1+{\mathrm{λ}}^2-1\right]+2\left(\mathrm{λ}-1\right)=0 \\ \left(1-\mathrm{λ}\right)\left[{\mathrm{λ}}^2-2\right]-2\left(1-\mathrm{λ}\right)=0 \\ \left(1-\mathrm{λ}\right)\left[{\mathrm{λ}}^2-4\right]=0 \\ \left(\mathrm{λ}+2\right)\left(\mathrm{λ}-2\right)\left(1-\mathrm{λ}\right)=0 \end{gathered}$$

Hence, the eigen values of matrix A are $\mathrm{λ}=-2,\mathrm{λ}=1$, and $\mathrm{λ}=2$.

$x=1⇒y=1$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$\mathrm{λ}=-2$then$\left(A-\mathrm{λ}l\right)X_1=0$

$$\begin{gathered} \left[A-\left(-2\right)X_3\right]X_1=[A+2C_3\mathrm{l}{X}_1 \\ \left[\left(\begin{array}{ccc}1& 1& -1\\ 1& 1& 1\\ -1& 1& -1\end{array}\right)+2\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right]\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}3& 1& -1\\ 1& 3& 1\\ -1& 1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \end{gathered}$$

Then,

$$\begin{gathered} \left(\begin{array}{c}3x+y-z\\ x+3y+z\\ -x+y+z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \begin{array}{c}3x+y-z=0,\\ x+3y+z=0,\\ -x+y+z=0\end{array} \end{gathered}$$

Solving these equations, we get; $z=2x$and $x+y=0$. Further setting $x=1$, we get$z=2$and $y=-1$.

Thus, the eigenvector corresponding to eigen valuelocalid="1658831212317" $\mathrm{λ}=-2$is $X_1=\left(\begin{array}{c}1\\ -1\\ 2\end{array}\right)$.

Let$X_2=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$be the eigenvector corresponding to eigen$\mathrm{λ}=1$ value then $\left(A-\mathrm{λ}l\right)X_2=0$

$$\begin{gathered} \left[A-\left(2\right)J_3\right]X_1=\mathrm{l}A-l_3]X_1 \\ \left[\left(\begin{array}{ccc}1& 1& -1\\ 1& 1& 1\\ -1& 1& -1\end{array}\right)-\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right]\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}0& 1& -1\\ 1& 0& 1\\ -1& 1& -2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \end{gathered}$$

Then,

$$\begin{gathered} \left(\begin{array}{c}0x+y-z\\ x+0y+z\\ -x+y-2z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \begin{array}{c}y-z=0,\\ x+z=0,\\ -x+y-2z=0.\end{array} \end{gathered}$$

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

Hence, the eigenvector corresponding to eigen value $\mathrm{λ}=1$is $X_2=\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)$.

Let $X_3=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$be the eigenvector corresponding to eigenvalue $\mathrm{λ}=2$thenrole="math" localid="1658832438679" $\left(A-\mathrm{λ}i\right)X_3=0$

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

Then,

$$\begin{gathered} \left(\begin{array}{c}-x+y-z\\ x-y+z\\ -x+y-3z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ -x+y+z=0, \\ x-y+z=0, \\ -x+y-3z=0. \end{gathered}$$

The solution is of these equations are $y=x+3z,y=x+z$, $z=0$therefore, . Now we set Hence, the eigenvector corresponding to eigenvalue$\mathrm{λ}=2$ is $X_3=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)$.