What is eigenvalue decomposition?

For eigenvalue $7$: $(\begin{bmatrix} 1 & 4\\ 9 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}.(+7)).v_{1} = 0$

$(\begin{bmatrix} 1 & 4\\ 9 & 1 \end{bmatrix} - \begin{bmatrix} 7 & 0\\ 0 & 7 \end{bmatrix}).v_{1} = 0$

$(\begin{bmatrix} -6 & 4\\ 9 & -6 \end{bmatrix}).(\begin{bmatrix} x_{1}\\ y_{1} \end{bmatrix}) = 0$

$-6x_{1}+4y_{1} = 0 \\ 9x_{1}-6y_{1} = 0 \\$

This system of linear equation can have infinitely many solutions. One of the solutions is:

$x_{1} = 2\\ y_{1} = 3$

Hence, $v_{1} = \begin{bmatrix} 2\\ 3 \end{bmatrix}$

Normalizing Eigenvectors: $v_{1n} = \begin{bmatrix} \frac{2}{\sqrt{13}}\\ \frac{3}{\sqrt{13}} \end{bmatrix}$

For eigenvalue $-5$: $(\begin{bmatrix} 1 & 4\\ 9 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}.(-5)).v_{2} = 0$

$(\begin{bmatrix} 1 & 4\\ 9 & 1 \end{bmatrix} - \begin{bmatrix} -5 & 0\\ 0 & -5 \end{bmatrix}).v_{2} = 0$

$(\begin{bmatrix} 6 & 4\\ 9 & 6 \end{bmatrix}).(\begin{bmatrix} x_{2}\\ y_{2} \end{bmatrix}) = 0$

$6x_{2}+4y_{2} = 0 \\ 9x_{2}+6y_{2} = 0 \\$

This system of linear equation can have infinitely many solutions. One of the solutions is:

$x_{2} = 2\\ y_{2} = -3$

Hence, $v_{2} = \begin{bmatrix} 2\\ -3 \end{bmatrix}$

Normalizing Eigenvectors: $v_{2n} = \begin{bmatrix} \frac{2}{\sqrt{13}}\\ \frac{-3}{\sqrt{13}} \end{bmatrix}$