Find Eigen values & Eigen vectors of the matrix A$^2$ + 2I where

A = $ \begin{bmatrix} 8&-8&-2 \\ 4&-3&-2 \\ 3&-4& 1 \end{bmatrix} $

Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

A = $ \begin{bmatrix} 8&-8&-2 \\ 4&-3&-2 \\ 3&-4& 1 \end{bmatrix} $

For characteristic equation |A - $\lambda$ I| = 0

$$ \begin{vmatrix} 8-\lambda&-8& -2 \\ 4& -3-\lambda &-2 \\ 3&-4& 1-\lambda \end{vmatrix} = 0 $$

$\lambda^3$ - (Sum of diagonal elements)$\lambda^2$ + (Sum of Principal minors)$\lambda$ - |A| = 0

$ \lambda^3 - (8-3+1)\lambda^2 + (-11+14+8)\lambda - 6 = 0 \\ \lambda^3 - 6\lambda^2 + 11 \lambda - 6 = 0 \\ (\lambda - 1)(\lambda - 2)(\lambda - 3) = 0 $

Therefore, eigen values are $\lambda$ = 1,2,3.

Eigen Values of A: 1,2,3

Eigen values of A$^2$: 1,4,9

Eigen values of I: 1,1,1

Eigen values of A$^2$ + 2I: 1+2, 4+2, 9+2 = 3,6,11

Note:

For Eigen vector

Eigen vectors of A$^2$ + 2I is same as Eigen vector of A.

For eigen vectors of A,

$$ \begin{bmatrix} A - \lambda I \end{bmatrix} \begin{bmatrix} X \end{bmatrix} = 0 \\ \begin{bmatrix} 8-\lambda&-8& -2 \\ 4& -3-\lambda &-2 \\ 3&-4& 1-\lambda \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

case(i) $\lambda$ = 1

$$ \begin{bmatrix} 8-1 &-8& -2 \\ 4& -3-1&-2 \\ 3&-4& 1-1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

$$ \begin{bmatrix} 7 &-8& -2 \\ 4& -4&-2 \\ 3&-4& 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

By Crammer's rule,

$ \frac{x_1}{-8} = \frac{-x_2}{6} = \frac{x_3}{-4} \\ \therefore \frac{x_1}{4} = \frac{x_2}{3} = \frac{x_3}{2} $

$$ X_1 = \begin{bmatrix} 4 \\ 3 \\ 2 \end{bmatrix} $$

case(ii) $\lambda$ = 2

$$ \begin{bmatrix} 8-2 &-8& -2 \\ 4& -3-2&-2 \\ 3&-4& 1-2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

$$ \begin{bmatrix} 6&-8& -2 \\ 4& -5&-2 \\ 3&-4& -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

$ \frac{x_1}{-3} = \frac{-x_2}{2} = \frac{x_3}{-1} \\ \therefore \frac{x_1}{3} = \frac{x_2}{2} = \frac{x_3}{1} $

$$ X_2 = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} $$

Case(iii) $\lambda$ = 3

$$ \begin{bmatrix} 8-3 &-8& -2 \\ 4& -3-3&-2 \\ 3&-4& 1-3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

$$ \begin{bmatrix} 5&-8& -2 \\ 4& -6 &-2 \\ 3&-4& -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

$ \frac{x_1}{4} = \frac{-x_2}{-2} = \frac{x_3}{2} \\ \therefore \frac{x_1}{2} = \frac{x_2}{1} = \frac{x_3}{1} $

$$ X_3 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} $$

Therefore Eigen vectors of A$^2$ + 2I are:

$ X_1 = \begin{bmatrix} 4 \\ 3 \\ 2 \end{bmatrix} \hspace{1cm} X_2 = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} \hspace{1cm} X_3 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} $