1. Given $A = \begin{bmatrix} -9 & 4 & 4 \\ -8 & 3 & 4 \\ -16 & 8 & 7 \end{bmatrix}$

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 $$s_{1} = trace(A) = (-9 + 3 + 7) = 1 \\ \therefore s_{1} = 1 \\ s_{2} = \begin{vmatrix} 3 & 4 \\ 8 & 7 \end{vmatrix} + \begin{vmatrix} - 9 & 4 \\ -16 & 7 \end{vmatrix} + \begin{vmatrix} -9 & 4 \ -8 & 3 \end{vmatrix} \\ = - 11 + 1 + 5 = -5 \\ \therefore s_{2} = -5 \\ |A| = -9(-11) - 4(8) + 4(-16) = 3 \\ \therefore |A| = 3$$

Equation (1) becomes,

Hence

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

$\therefore$ 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

$$\begin{bmatrix} -8 & 4 & 4 \\ -8 & 4 & 4 \\ -16 & 8 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

The linear equations are:

$$-8x + 4y + 4z = 0 \\ -8x + 4y + 4z = 0 \\ -16x + 8y + 8z = 0$$

Only one equation is independent: -2x + 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 = \frac{1}{2} [k_{1} + k_{2}] $

$$\begin{bmatrix} \frac{1}{2} [k_{1} + k_{2}] \\ k_{1} \\ k_{2} \end{bmatrix} = k_{1} \begin{bmatrix} \frac{1}{2} \\ 1 \\ 0 \end{bmatrix} + k_{2} \begin{bmatrix} \frac{1}{2} \\ 0 \\ 1 \end{bmatrix}$$

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

$$X_{1} = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \hspace{0.25cm} X_{2} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}$$

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

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

$$\begin{bmatrix} -12 & 4 & 4 \\ -8 & 0 & 4 \\ -16 & 8 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{2} \rightarrow R_{2} - R_{1}, R_{3} \rightarrow R_{3} - R_{1} \\ \begin{bmatrix} -12 & 4 & 4 \\ 4 & -4 & 0 \\ -4 \\ 4 \\ 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{3} \rightarrow R_{3} + R_{2} \\ \begin{bmatrix} -12 & 4 & 4 \\ 4 & -4 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{1} \rightarrow - \frac{1}{4}R_{1}, R_{2} \rightarrow \frac{1}{4} R_{2} \\ \begin{bmatrix} 3 & - 1 & -1 \\ 1 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

The linear equations are:

$$3x - y - z = 0 \\ x - y = 0 \\ x = y$$

Put x = y = k then z = 2k

For k = 1 we get $X_{3} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$

Although eigen values of A are not distinct the geomatric mulitplicity 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 transforming matrix $M = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 0 & 2 & 2 \end{bmatrix}$