Find the Eigen values & Eigen vectors of the following matrices

(i) A = $ \begin{bmatrix} 2&-1& 1 \\ 1& 2&-1 \\ 1&-1& 2 \end{bmatrix} $ (ii) A = $ \begin{bmatrix} 2&2&1 \\ 1&3&1 \\ 1&2&2 \end{bmatrix} $ (iii) A = $ \begin{bmatrix} 6&-2& 2 \\ -2& 3&-1 \\ 2&-1& 3 \end{bmatrix} $

Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

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

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

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

$ \lambda^3 - (6)\lambda^2 + (3+3+5)\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.

To find Eigen vectors of A

case(i): $\lambda$ = 1

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

By Crammer's rule,

$ \frac{x_1}{0} = -\frac{x_2}{2} = \frac{x_3}{-2} \\ \therefore \frac{x_1}{0} = -\frac{x_2}{1} = \frac{x_3}{1}$

$$ X_1 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} $$

case(ii): $\lambda$ = 2

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

By Crammer's rule,

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

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

case (iii): $\lambda$ = 3

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

$ \frac{x_1}{2} = \frac{-x_2}{0} = \frac{x_3}{2} \\ \therefore \frac{x_1}{1} = \frac{x_2}{0} = \frac{x_3}{1} $

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

Therefore, the three eigen vectors are:

$ X_1 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \hspace{0.5cm} X_2 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \hspace{0.5cm} X_3 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $

(ii) $ A = \begin{bmatrix} 2&2&1 \\ 1&3&1 \\ 1&2&2 \end{bmatrix} $

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

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

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

$ \lambda^3 - (7)\lambda^2 + (4+3+4)\lambda - 5 = 0 \\ \lambda^3 - 7 \lambda^2 + 11 \lambda - 5 = 0 \\ (\lambda - 1)(\lambda - 1)(\lambda - 5) = 0 $

Therefore, eigen values are $\lambda$ = 1,1,5.

For eigen vectors,

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

case(i) $\lambda$ = 5

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

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

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

case(ii) $\lambda$ = 1

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

$$ \begin{bmatrix} 1& 2 & 1 \\ 1& 2 &1 \\ 1&2& 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

$ x_1 + 2x_2 + x_3 = 0 $

[Note: Since rows are repeated, Crammer's rule is not applicable]

A is non-symmetric.

Put $ x_3 = 0; \hspace{0.5cm} x_1 + 2x_2 = 0 $

Put $ x_2 = 1; \hspace{0.5cm} x_1 = -1$

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

case(iii) $\lambda$ = 1

Put x$_2$ = 0

x$_1$ + x$_3$ = 0

x$_1$ = - x$_3$

Put x$_3$ = 1

x$_1$ = -1

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

(iii) A = $ \begin{bmatrix} 6&-2& 2 \\ -2& 3&-1 \\ 2&-1& 3 \end{bmatrix} $

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

$$ \begin{vmatrix} 6-\lambda&-2& 2 \\ -2& 3-\lambda &-1 \\ 2&-1& 3-\lambda \end{vmatrix} = 0 $$

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

$ \lambda^3 - (12)\lambda^2 + (8+14+14)\lambda - 32 = 0 \\ \lambda^3 - 12 \lambda^2 + 36 \lambda - 32 = 0 \\ (\lambda - 2)(\lambda - 2)(\lambda - 8) = 0 $

Therefore, eigen values are $\lambda$ = 2,2,8.

For eigen vectors,

$$ \begin{bmatrix} A - \lambda I \end{bmatrix} \begin{bmatrix} X \end{bmatrix} = 0 $$

case(i) $\lambda$ = 8

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

By Crammer's rule,

$ \frac{x_1}{24} = \frac{-x_2}{12} = \frac{x_3}{12} \\ \therefore \frac{x_1}{2} = \frac{x_2}{-1} = \frac{x_3}{1} $

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

Case(ii) $\lambda$ = 2

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

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

$ 2x_1 - x_2 + x+3 = 0 $

Put x$_3$ = 0

x$_2$ =2

2x$_1$ = 2

x$_1$ = 1

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

case(iii) $\lambda$ = 2

Note: If A is symmetric then the eigen vectors are orthogonal to each others.

Here, A is symmetric

$ X_1X_3^T = 0 \\ X_2X_3^T = 0 $

Let, $ X_3 = \begin{bmatrix} l \\ m \\ n \end{bmatrix} $ be the third eigen vector

$ \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} \begin{bmatrix} l & m & n \end{bmatrix} = 0 \hspace{1cm} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \begin{bmatrix} l & m & n \end{bmatrix} = 0 $

$ 2l - m + n = 0 \hspace{1cm} l + 2m + 0n = 0 $

By Creammer's rule

$ \frac{l}{-2} = \frac{m}{1} = \frac{n}{5} $

$$ X_3 = \begin{bmatrix} -2 \\ 1 \\ 5 \end{bmatrix} $$