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

The characteristic equation of matrix A is

$$| A - \lambda I| = 0 \\ (-1)^3 \lambda^3 + (-1)^2 s_{1} \lambda^2 + (-1) s_{2} \lambda + |A| = 0.....(1)$$

where $$s_{1} = trace(A) = (2 + 3 + 4) = 9$$

$$\therefore s_{1} = 9 \\ s_{2} = \begin{vmatrix} 3 & 2 \\ 3 & 4 \end{vmatrix} + \begin{vmatrix} 2 & 1 \\ 3 & 4 \end{vmatrix} + \begin{vmatrix} 2 & 1 \\ 2 & 3 \end{vmatrix} \\ = 6 + 5 + 4 = 15 \\ \therefore s_{2} = 15 \\ |A| = 12 - 2 - 3 \\ \therefore |A| = 7$$

Equation (1) becomes,

$$-\lambda^3 + 9\lambda^2 - 15\lambda + 7 = 0$$

Hence eigen values of A are

$$\lambda = 7, 1, 1$$

To find eigen vectors corresponding to these eigen values, we use the system of homogeneous equations $(A - \lambda I) = 0$

Using Cramers' Rule

When $\lambda = 7$, corresponding system of homogeneous equations is

$$\begin{bmatrix} -5 & 1 & 1 \\\ 2 & -4 & 2 \\ 3 & 3 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

The linear equations are:

$$-5x + y + z = 0 \\ 2x - 4y + 2z = 0 \\ 3x + 3y - 3z = 0$$

Considering any two lienar equations and solving them by Cramer's rule, we get

$$-5x + y + z = 0 \\ 2x - 4y + 2z = 0$$

$$\frac{x}{\begin{vmatrix} 1 & 1 \\ -4 & 2 \end{vmatrix}} = \frac{-y} {\begin{vmatrix} -5 & 1 \\ 2 & 2 \end{vmatrix}} = \frac{z} {\begin{vmatrix} -5 & 1 \\ 2 & -4 \end{vmatrix}} \\ \frac{x}{6} = -\frac{y}{-12} = \frac{z}{18} \\ \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$$

Hence, the eigen vector corresponding to the eigen value $\lambda$ = 12 is

$$X_{1} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$

When $\lambda = 1$, corresponding system of homogenous equations is

$$\begin{bmatrix} 1 & 1 & 1 \\\ 2 & 2 & 2 \\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

The linear equations are:

$$x + y + z = 0 \\ 2x + 2y + 2z = 0 \\ 3x + 3y + 3z = 0$$

Only one equations is independent: x + y + z = 0

i.e. the rank of the coefficient matrix is 1. Therefore system has 3 - 1 = 2 linearly independent solutions.

Let, $y = k_{1}$ and $z = k_{2}$ $\hspace{1cm} \rightarrow x = -k_{1} - k_{2}$

$$\begin{bmatrix} -k_{1} - k_{2} \\ k_{1} \\ k_{2} \end{bmatrix} = k_{1} \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + k_{2} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$$

Hence setting $k_{1} = k_{2} = 1$ there are two independent eigen vectors corresponding to $\lambda = 1$

$$X_{2} = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} \hspace{0.75cm} X_{3} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$$

Hence, the eigen values are $\lambda = 12, -3, -6$ and the corresponding eigen vectors are $$X_{1} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \hspace{0.5cm} X_{2} = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} \hspace{0.5cm} X_{3} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$$