Ace test series

Question

Comments

Rank of the matrix = no. of independent eigen vectors

Answer 1

$I=\left[ \begin{array}{cc} 1 & 0 \\\\ 0 & 1 \end{array} \right]$

 

$\left[ \begin{array}{cc} 1 - \lambda & 0 \\\\ 0 & 1 - \lambda  \end{array} \right]$

 

$\left| \begin{array}{cc}1 - \lambda & 0 \\\\ 0 &1 - \lambda \end{array} \right|=\left(- \lambda + 1\right)^{2}$

The roots are

$\lambda_1 = 1$

$\lambda_2 = 1$

Next finding the eigen vectors

$\lambda_1 = 1$

$\left[ \begin{array}{cc} - \lambda + 1 & 0 \\\\ 0 & - \lambda + 1 \end{array} \right]=\left[ \begin{array}{cc} 0 & 0 \\\\ 0 & 0 \end{array} \right]$

$\left[ \begin{array}{cc} 0 & 0 \\\\ 0 & 0 \end{array} \right] \left[ \begin{array}{c} v_{1} \\\\ v_{2} \end{array} \right]=\left[ \begin{array}{c} 0 \\\\ 0 \end{array} \right]$

take $v_1 = t, v_2 = s$

$\mathbf{v}=\left[ \begin{array}{c} t \\\\ s \end{array} \right]=\left[ \begin{array}{c} 1 \\\\ 0 \end{array} \right] t+\left[ \begin{array}{c} 0 \\\\ 1 \end{array} \right] s$

So we got 2 linearly independent eigen vectors, $\left[ \begin{array}{c} 0 \\\\ 1 \end{array} \right]$,$\left[ \begin{array}{c} 1 \\\\ 0 \end{array} \right]$