The characteristic equation is $\begin{bmatrix} 6 - \lambda & -2 & 2 \\ -2 & 3 - \lambda & -1 \\ 2 & -1 & 3 - \lambda \end{bmatrix} = 0$

Simplifying we get

$(6 - \lambda) [(3 - \lambda)^2 - 1] + 2[-6 + 2\lambda + 2] + 2[2 - 6 + 2\lambda] = 0 \\ \therefore [6 - \lambda] [\lambda^2 - 6\lambda + 9 - 1]-8 + 4\lambda - 8 + 4\lambda = 0 \\ \therefore 6\lambda^2 - 36\lambda + 48 - \lambda^3 + 6\lambda^2 - 8\lambda - 16 + 8\lambda = 0 \\ \therefore -\lambda^3 + 12\lambda^2 - 36\lambda + 32 = 0 \\ \lambda = 8,2,2$

For $\lambda = 2, (A - \lambda_{1}I)X = 0$ given

$\begin{bmatrix} 4 & -2 & 2 \\ -2 & 1 & -1 \\ 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{2} \times 2, R_{3} \times 2 \\ \begin{bmatrix} 4 & -2 & 2 \\ -4 & 2 & -2 \\ 4 & -2 & 2 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ \therefore R_{2} + R_{1}, R_{3} - R_{1} \\ \begin{bmatrix} 4 & -2 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Rank is 1 and variable is 3

$\therefore$ 3 - 1 = 2 independent solution

$\therefore 4x_{1} - 2x_{2} + 2x_{3} = 0$

Putting $x_{1} = -1$ and $x_{2} = 0 \\ X_{3} = 2$

Again putting $x_{1} = 1$ and $x_{2} = 2 \\ X_{3} = 0$

$\therefore$ For eigen value 2 we get two eigen vectors they are

$X_{1} = [-1, 0, 2], X_{2} = [1,2,0]$

2) For $\lambda = 8, [A - \lambda I]X = 0 \\ \begin{bmatrix} -2 & -2 & 2 \\ -2 & -5 & -1 \\ 2 & -1 & -5 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{3} + R_{1}, R_{2} - R_{1} \\ \begin{bmatrix} -2 & -2 & 2 \\ 0 & -3 & -3 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Rank = 2, Variables = 3

3 - 2 = 1 independent solution

$\therefore -3 \times -2 - 3 \times 3 = 0$

Let $x_{2} = -1 \\ x_{3} = 1 \\ -2x_{1} - 2x_{2} + 2x_{3} = 0 \\ \therefore x_{1} = 2$

$\therefore$ For Eigen value 8

Eigen value i

[2, -1, 1]

Since $M^{-1}AM = D$ the given matrix is diagonalised to

matrix D = $\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 8 \end{bmatrix}$

By transforming matrix

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