Answer to Question #282371 in Linear Algebra for JOY

Question

Find eigen values and associated eigen vectors of the matrix

A=[ 1 −3 3

−3 −5 3

6 −6 4 ] in the field R. Also find an invertible matrix p such that p−1 Ap is diagonal.

Answer 1

$$A-\lambda I=\begin{bmatrix} 1-\lambda & -3 & 3 \\ -3 & -5-\lambda & 3 \\ 6 & -6 & 4-\lambda \end{bmatrix}$$

The characteristic equation

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

$$(1-\lambda)(-5-\lambda)(4-\lambda)-54+54$$

$$-18(-5-\lambda)+18(1-\lambda)-9(4-\lambda)=0$$

$$-\lambda^3+30\lambda+52=0$$

$$-\lambda^2(\lambda+2)+2\lambda(\lambda+2)+26(\lambda+2)=0$$

$$\lambda_1=-2$$

$$\lambda^2-2\lambda-26=0$$

$$(\lambda-1)^2=27$$

$$\lambda_2=1-3\sqrt{3}, \lambda_3=1+3\sqrt{3}$$

$$\lambda_1=-2,\lambda_2=1-3\sqrt{3}, \lambda_3=1+3\sqrt{3}$$

These are eigenvalues.

$\lambda=-2$

$$\begin{bmatrix} 1-\lambda & -3 & 3 \\ -3 & -5-\lambda & 3 \\ 6 & -6 & 4-\lambda \end{bmatrix}=\begin{bmatrix} 3 & -3 & 3 \\ -3 & -3 & 3 \\ 6 & -6 & 6 \end{bmatrix}$$

$R_1=R_1/3$

$$\begin{bmatrix} 1 & -1 & 1 \\ -3 & -3 & 3 \\ 6 & -6 & 6 \end{bmatrix}$$

$R_2=R_2+3R_1$

$$\begin{bmatrix} 1 & -1 & 1 \\ 0 & -6 & 6 \\ 6 & -6 & 6 \end{bmatrix}$$

$R_3=R_3-6R_1$

$$\begin{bmatrix} 1 & -1 & 1 \\ 0 & -6 & 6 \\ 0 & 0 & 0 \end{bmatrix}$$

$R_2=R_2/(-6)$

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

$R_1=R_1+R_2$

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

If we take $v_3=t,$ then $v_1=0, v_2=t, v_3=t.$

Thus the eigenvector is

$$\bold v=\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$$

$\lambda=1-3\sqrt{3}$

$$\begin{bmatrix} 1-\lambda & -3 & 3 \\ -3 & -5-\lambda & 3 \\ 6 & -6 & 4-\lambda \end{bmatrix}$$

$$=\begin{bmatrix} 3\sqrt{3} & -3 & 3 \\ -3 & -6+3\sqrt{3} & 3 \\ 6 & -6 & 3+3\sqrt{3} \end{bmatrix}$$

$R_1=\sqrt{3}R_1/9$

$$\begin{bmatrix} 1 & -\sqrt{3}/3 & \sqrt{3}/3 \\ -3 & -6+3\sqrt{3} & 3 \\ 6 & -6 & 3+3\sqrt{3} \end{bmatrix}$$

$R_2=R_2+3R_1$

$$\begin{bmatrix} 1 & -\sqrt{3}/3 & \sqrt{3}/3 \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \\ 6 & -6 & 3+3\sqrt{3} \end{bmatrix}$$

$R_3=R_3-6R_1$

$$\begin{bmatrix} 1 & -\sqrt{3}/3 & \sqrt{3}/3 \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \end{bmatrix}$$

$R_2=R_2/(-6+2\sqrt{3})$

$$\begin{bmatrix} 1 & -\sqrt{3}/3 & \sqrt{3}/3 \\ 0 & 1 & -1-\sqrt{3}/2 \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \end{bmatrix}$$

$R_1=R_1+R_2/\sqrt{3}$

$$\begin{bmatrix} 1 & 0 & -1/2 \\ 0 & 1 & -1-\sqrt{3}/2 \\ 0 & -6+2\sqrt{3} & 3+\sqrt{3} \end{bmatrix}$$

$R_3=R_3+(6-2\sqrt{3})R_2$

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

If we take $u_3=t,$ then $u_1=t/2, u_2=(2+\sqrt{3})t/2, u_3=t.$

Thus the eigenvector is

$$\bold u=\begin{bmatrix} 1/2 \\ (2+\sqrt{3})/2 \\ 1 \end{bmatrix}$$

$\lambda=1+3\sqrt{3}$

$$\begin{bmatrix} 1-\lambda & -3 & 3 \\ -3 & -5-\lambda & 3 \\ 6 & -6 & 4-\lambda \end{bmatrix}$$

$$=\begin{bmatrix} -3\sqrt{3} & -3 & 3 \\ -3 & -6-3\sqrt{3} & 3 \\ 6 & -6 & 3-3\sqrt{3} \end{bmatrix}$$

$R_1=-\sqrt{3}R_1/9$

$$\begin{bmatrix} 1 & \sqrt{3}/3 & -\sqrt{3}/3 \\ -3 & -6-3\sqrt{3} & 3 \\ 6 & -6 & 3-3\sqrt{3} \end{bmatrix}$$

$R_2=R_2+3R_1$

$$\begin{bmatrix} 1 & \sqrt{3}/3 & -\sqrt{3}/3 \\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \\ 6 & -6 & 3-3\sqrt{3} \end{bmatrix}$$

$R_3=R_3-6R_1$

$$\begin{bmatrix} 1 & \sqrt{3}/3 & -\sqrt{3}/3 \\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \end{bmatrix}$$

$R_2=R_2/(-6-2\sqrt{3})$

$$\begin{bmatrix} 1 & \sqrt{3}/3 & -\sqrt{3}/3 \\ 0 & 1 & -1+\sqrt{3} /2\\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \end{bmatrix}$$

$R_1=R_1-R_2/\sqrt{3}$

$$\begin{bmatrix} 1 & 0 & -1/2 \\ 0 & 1 & -1+\sqrt{3} /2\\ 0 & -6-2\sqrt{3} & 3-\sqrt{3} \end{bmatrix}$$

$R_3=R_3+(6+2\sqrt{3})R_2$

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

If we take $w_3=t,$ then $w_1=t/2, w_2=(2-\sqrt{3})t/2, w_3=t.$

Thus the eigenvector is

$$\bold w=\begin{bmatrix} 1/2 \\ (2-\sqrt{3})/2 \\ 1 \end{bmatrix}$$

Form the matrix $P,$ whose column $i$ is eigenvector no. $i:$

$$P=\begin{bmatrix} 0 & 1/2 & 1/2 \\ 1 & 1+\sqrt{3}/2 & 1-\sqrt{3} /2\\ 1 & 1 & 1 \end{bmatrix}$$

Form the matrix $D$ whose element at row $i,$ column $i$ is eigenvalue no. $i:$

$$D=\begin{bmatrix} -2 & 0 & 0 \\ 0 & 1-3\sqrt{3} & 0\\ 0 & 0 & 1+3\sqrt{3} \end{bmatrix}$$

The matrices $P$ and $D$  are such that the initial matrix

$$A=\begin{bmatrix} 1 & -3 & 3 \\ -3 & -5 & 3 \\ 6 & -6 & 4 \end{bmatrix}=PDP^{-1}$$