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