Given $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 0 & 0 & 1 \end{bmatrix}$

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 $$s_{1} = trace(A) = (2 + 2 + 1) = 5 \\ s_{1} = 5 \\ s_{2} = \begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} + \begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} + \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} \\ = 2 + 2 + 3 = 7 \\ \therefore s_{2} = 7 \\ |A| = 2(2) - 1(1) + 1(0) = 3 \\ \therefore |A| = 3$$

Equation (1) becomes,

Hence

$$\phi(\lambda) = - \lambda^3 + 5 \lambda^2 - 7 \lambda + 3 = 0 \\ \phi(\lambda) = \lambda^3 - 5 \lambda^2 + 7 \lambda - 3 = 0 \\ \therefore \phi(\lambda) = (\lambda - 3)(\lambda^2 - 2\lambda + 1) = 0$$

$\therefore$ Characteristic roots are $\lambda$ = 1, 1, 3

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

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

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

The linear equations are:

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

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

i.e. the rank of the coefficient matrix is 1. Therefore system has 3 - 1 = 2 linearly indepndent 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 $$X_{1} = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} \hspace{1cm} X_{2} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$$

$\therefore$ Geometric multiplicity of $\lambda$ = 1 is 2 which is equal to algebraic multiplicity.

When $\lambda$ = 3, corresponding system of homogeneous equations is

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

The linear equations are:

$$2z = 0, z = 0 \\ -x + y = 0 \\ x = y$$

Put X = y = k

For k = 1 we get $X_{3} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} $

Although eigen values of A are not distinct the geometric multiplicity of $\lambda$ = 3 is one which is equal to algebraic multiplicity.

$\therefore$ A is diagonisable.

Diagonal form: $\because M^{-1}AM = D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}$

Where M is tranforming matrix $M = \begin{bmatrix} -1 & -1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$