Answer to Question #292571 in Linear Algebra for Zain

Question

Question. Given the matrix A = 3 1 1

2 4 2

1 1 3

a. Solve A for its eigen values and given vectors

b. Constant Similar matrix for A if possible

Answer 1

a.

$$A=\begin{bmatrix} 3 & 1 & 1 \\ 2 & 4 & 2 \\ 1 & 1 & 3 \end{bmatrix}$$

$$A-\lambda I=\begin{bmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}$$

$$\det A-\lambda I=\begin{vmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{vmatrix}$$

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

$$-2(3-\lambda)=36-24\lambda+4\lambda^2-9\lambda+6\lambda^2-\lambda^3$$

$$+4-4+\lambda-12+4\lambda=-\lambda^3+10\lambda^2-28\lambda+24=0$$

$$-\lambda^2(\lambda-2)+8\lambda(\lambda-2)-12(\lambda-2)=0$$

$$-(\lambda-2)^2(\lambda-6)=0$$

$\lambda_1=6, \lambda_2=2, \lambda_3=2$

These are the eigenvalues.

b. Find the eigenvectors.

$\lambda=6$

$$\begin{bmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}=\begin{bmatrix} -3 & 1 & 1 \\ 2 & -2 & 2 \\ 1 & 1 & -3 \end{bmatrix}$$

$R_1=R_1/(-3)$

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

$R_2=R_2-2R_1$

$$\begin{bmatrix} 1 & -1/3 & -1/3 \\ 0 & -4/3 & 8/3 \\ 1 & 1 & -3 \end{bmatrix}$$

$R_3=R_3-R_1$

$$\begin{bmatrix} 1 & -1/3 & -1/3 \\ 0 & -4/3 & 8/3 \\ 0 & 4/3 & -8/3 \end{bmatrix}$$

$R_2=-3R_2/4$

$$\begin{bmatrix} 1 & -1/3 & -1/3 \\ 0 & 1 & -2 \\ 0 & 4/3 & -8/3 \end{bmatrix}$$

$R_1=R_1+R_2/3$

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 4/3 & -8/3 \end{bmatrix}$$

$R_3=R_3-4R_2/3$

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

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $x_3=t,$ then $x_1=t, x_2=2t.$

The eigenvector is $\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$

$\lambda=2$

$$\begin{bmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}=\begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 1 & 1 & 1 \end{bmatrix}$$

$R_2=R_2-2R_1$

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

$R_3=R_3-R_1$

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

$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $x_2=t,x_3=s,$ then $x_1=-t-s.$

The eigenvectors are $\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}.$

$$P=\begin{bmatrix} 1 & -1 & -1 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}, D=\begin{bmatrix} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$

$$PDP^{-1}=A$$

The matrices $A=\begin{bmatrix} 3 & 1 & 1 \\ 2 & 4 & 2 \\ 1 & 1 & 3 \end{bmatrix}$ and $D=\begin{bmatrix} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$ are similar.