Question

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Answer 1

Let

$$A=\begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{bmatrix}$$

$$A-\lambda I=\begin{bmatrix} 1-\lambda & 1 & 2 \\ 1 & 2-\lambda & 1 \\ 2 & 1 & 1-\lambda \end{bmatrix}$$

$$det(A-\lambda I)=\begin{vmatrix} 1-\lambda & 1 & 2 \\ 1 & 2-\lambda & 1 \\ 2 & 1 & 1-\lambda \end{vmatrix}=$$

$$= (1-\lambda)\begin{vmatrix} 2-\lambda & 1 \\ 1 & 1-\lambda \end{vmatrix}-(1)\begin{vmatrix} 1 & 1 \\ 2 & 1-\lambda \end{vmatrix}+(2)\begin{vmatrix} 1 & 2-\lambda \\ 2 & 1 \end{vmatrix}=$$

$$=(1-\lambda)(2-2\lambda-\lambda+\lambda^2-1)-(1-\lambda-2)+$$

$$+2(1-4+2\lambda)=1-3\lambda+\lambda^2-\lambda+3\lambda^2-\lambda^3+5\lambda-5=$$

$$=-\lambda^3+4\lambda^2+\lambda-4$$

This is a characteristic polynomial.

Solve $det(A-\lambda I)=0$

$$-\lambda^3+4\lambda^2+\lambda-4=0$$ $$-\lambda^2(\lambda-4)+(\lambda-4)=0$$ $$(\lambda-4)(1-\lambda)(1+\lambda)=0$$

The roots are: $\lambda_1=4,\lambda_2=1,\lambda_3=-1.$

These are the eigenvalues.

Find the eigenvectors.

$\lambda=4$

$$\begin{bmatrix} 1-\lambda & 1 & 2 \\ 1 & 2-\lambda & 1 \\ 2 & 1 & 1-\lambda \end{bmatrix}=\begin{bmatrix} -3 & 1 & 2 \\ 1 & -2 & 1 \\ 2 & 1 & -3 \end{bmatrix}$$

Perform row operations to obtain the rref of the matrix:

$R_2=R_2+(1/3)R_1$

$$\begin{bmatrix} -3 & 1 & 2 \\ 0 & -5/3 & 5/3 \\ 2 & 1 & -3 \end{bmatrix}$$

$R_3=R_3+(2/3)R_1$

$$\begin{bmatrix} -3 & 1 & 2 \\ 0 & -5/3 & 5/3 \\ 0 & 5/3 & -5/3 \end{bmatrix}$$

$R_3=R_3+R_2$

$$\begin{bmatrix} -3 & 1 & 2 \\ 0 & -5/3 & 5/3 \\ 0 & 0 & 0 \end{bmatrix}$$

$R_2=(-3/5)R_2$

$$\begin{bmatrix} -3 & 1 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$

$R_1=R_1-R_2$

$$\begin{bmatrix} -3 & 0 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$

$R_1=(-1/3)R_1$

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$

Now, solve the matrix equation

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $v_3=t,$ then $v_1=t, v_2=t,v_3=t.$

Therefore

$$\mathbf {v}=\begin{bmatrix} t \\ t \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}t$$

$\lambda=1$

$$\begin{bmatrix} 1-\lambda & 1 & 2 \\ 1 & 2-\lambda & 1 \\ 2 & 1 & 1-\lambda \end{bmatrix}=\begin{bmatrix} 0 & 1 & 2 \\ 1 & 1 & 1 \\ 2 & 1 & 0 \end{bmatrix}$$

Perform row operations to obtain the rref of the matrix:

Swap rows 1 and 2

$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 2 & 1 & 0 \end{bmatrix}$$

$R_3=R_3-(2)R_1$

$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & -1 & -2 \end{bmatrix}$$

$R_3=R_3+R_2$

$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}$$

$R_1=R_1-R_2$

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}$$

Now, solve the matrix equation

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $u_3=t,$ then $u_1=t, u_2=-2t,u_3=t.$

Therefore

$$\mathbf {u}=\begin{bmatrix} t \\ -2t \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}t$$

$\lambda=-1$

$$\begin{bmatrix} 1-\lambda & 1 & 2 \\ 1 & 2-\lambda & 1 \\ 2 & 1 & 1-\lambda \end{bmatrix}=\begin{bmatrix} 2 & 1 & 2 \\ 1 & 3 & 1 \\ 2 & 1 & 2 \end{bmatrix}$$

Perform row operations to obtain the rref of the matrix:

$R_3=R_3-R_1$

$$\begin{bmatrix} 2 & 1 & 2 \\ 1 & 3 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

$R_2=R_2-(1/2)R_1$

$$\begin{bmatrix} 2 & 1 & 2 \\ 0 & 5/2 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

$R_2=(2/5)R_2$

$$\begin{bmatrix} 2 & 1 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

$R_1=R_1-R_2$

$$\begin{bmatrix} 2 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

$R_1=(1/2)R_1$

$$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

Now, solve the matrix equation

$$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $w_3=t,$ then $w_1=-t, w_2=0, w_3=t.$

Therefore

$$\mathbf {w}=\begin{bmatrix} -t \\ 0 \\ t \end{bmatrix}=\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}t$$

Eigen value: $4,$ eigenvector:$\begin{bmatrix} 1/\sqrt{3} \\ 1/\sqrt{3} \\ 1/\sqrt{3} \end{bmatrix}$

Eigen value: $1,$ eigenvector: $\begin{bmatrix} 1/\sqrt{6} \\ -2/\sqrt{6} \\ 1/\sqrt{6} \end{bmatrix}$

Figenvalue: $-1,$ eigenvector: $\begin{bmatrix} -1/\sqrt{2} \\ 0 \\ 1/\sqrt{2} \end{bmatrix}$