Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

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

$$ \begin{vmatrix} -9-\lambda&4& 4 \\ -8& 3-\lambda &4 \\ -16&8& 7-\lambda \end{vmatrix} = 0 $$

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

$ \lambda^3 - (-9+10)\lambda^2 + (-11+1+5)\lambda - 3 = 0 \\ \lambda^3 - \lambda^2 - 5 \lambda - 3 = 0 \\ \lambda = 3,-1,-1 $

For eigen vector,

$$ \begin{bmatrix} A - \lambda I \end{bmatrix} \begin{bmatrix} X \end{bmatrix} = 0 \\ \begin{bmatrix} -9-\lambda&4& 4 \\ -8& 3-\lambda &4 \\ -16&8& 7-\lambda \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

Case(i) $\lambda$ = 3

$$ \begin{bmatrix} -12 & 4 & 4 \\ -8 & 0 & 4 \\ -16 & 8 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

Using Crammer's rule,

$ \frac{x_1}{-32} = \frac{-x_2}{32} = \frac{x_3}{-64} \\ \therefore \frac{x_1}{1} = \frac{x_2}{1} = \frac{x_3}{2} $

$$ X_1 = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} $$

Case(ii) $\lambda$ = -1

$$ \begin{bmatrix} -9+1 & 4 & 4 \\ -8 & 3+1 & 4 \\ -16 & 8 & 7+1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ \begin{bmatrix} -8 & 4 & 4 \\ -8 & 4 & 4 \\ -16 & 8 & 8 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ \begin{bmatrix} -2 & 1 & 1 \\ -2 & 1 & 1 \\ -2 & 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

$ -2x_1 +x_2 + x_3 = 0 $

Put x$_3$ = 0

-2x$_1$ + x$_2$ = 0

2x$_1$ = x$_2$

Put x$_2$ = 2

2x$_1$ = 2

x$_1$ = 1

$$ X_2 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} $$

A is non-symmetric

Case(iii) $\lambda$ = -1

$ -2x_1 + x_2 + x_3 = 0 $

Put x$_2$ = 0

-2x$_1$ + x$_3$ = 0

-2x$_1$ = -x$_3$

2x$_1$ = x$_3$

Put x$_3$ = 2

x$_1$ = 1

$$ X_3 = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} $$

Therefore, the transforming matrix is given by:

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

$ M^{-1} = \frac{1}{|M|}[Adj M] \\ = - \frac{1}{2} \begin{bmatrix} 4 & -2 & -2 \\ -2 & 0 & 1 \\ -4 & 2 & 1 \end{bmatrix} $

Consider, $ M^{-1}AM $

$ = - \frac{1}{2} \begin{bmatrix} 4 & -2 & -2 \\ -2 & 0 & 1 \\ -4 & 2 & 1 \end{bmatrix} \begin{bmatrix} -9&4&4 \\ -8&3&4 \\ -16&8&7 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 2 & 0 & 2 \end{bmatrix} \\ = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \end{bmatrix} = D $

Where D is a diagonal matrix whose diagonal entries are eigen values of A.

Therefore A is diagonalisable.