Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 8M

Year: Dec 2015

A = $\begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix}$ $A^T = \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix}$

Now we compute

$A^T A = \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 4 & 8 \end{bmatrix}$

Now characteristic equation by this polynomial is

$\begin{vmatrix} 2 - \lambda & 4 \\ 4 & 8 - \lambda \end{vmatrix} = 0 \\ \therefore (2 - \lambda) (8 - \lambda) - 16 = 0 \\ \therefore 16 - 8\lambda - 2\lambda + \lambda^2 - 16 = 0 \\ \therefore \lambda^2 - 10\lambda = 0$

For $\lambda = 10 \hspace{0.5cm} [A - \lambda I]X = 0$ we get

$\begin{bmatrix} -8 & 4 \\ 8 & -4 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ R_{2} + R_{1} \\ \begin{bmatrix} -8 & 4 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Rank is 1

$\therefore$ Let $x_{1} = 1 \hspace{1cm} \therefore x_{2} = 2$

$\therefore$ The unit eigen vector is $\sqrt{1^2 + 2^2} \\ V_{1} = V_{1} = \begin{bmatrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{bmatrix}$