# Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.

$\boldsymbol{ \left [ \begin{array}{ c c c } 2 & 5 & 5 \\ 5 & 2 & 5 \\ 5 & 5 & 2 \end{array} \right ] \ ; \ \lambda \ = \ 12 }$

The aim of this question is to understand the diagonalization process of a given matrix at given eigenvalues.

To solve this question, we first evaluate the expression $\boldsymbol{ A \ – \ \lambda I }$. Then we solve the system $\boldsymbol{ ( A \ – \ \lambda I ) \vec{x}\ = 0 }$ to find the eigen vectors.

Given that:

$A \ = \ \left [ \begin{array}{ c c c } 2 & 5 & 5 \\ 5 & 2 & 5 \\ 5 & 5 & 2 \end{array} \right ]$

And:

$\lambda \ = \text{ Eigen Values }$

For $\lambda \ = \ 12$:

$A \ – \ \lambda I \ = \ \left [ \begin{array}{ c c c } 2 & 5 & 5 \\ 5 & 2 & 5 \\ 5 & 5 & 2 \end{array} \right ] \ – \ 12 \ \left [ \begin{array}{ c c c } 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right ]$

$A \ – \ \lambda I \ = \ \left [ \begin{array}{ c c c } 2 \ – \ 12 & 5 & 5 \\ 5 & 2 \ – \ 12 & 5 \\ 5 & 5 & 2 \ – \ 12 \end{array} \right ]$

$A \ – \ \lambda I \ = \ \left [ \begin{array}{ c c c } -10 & 5 & 5 \\ 5 & -10 & 5 \\ 5 & 5 & -10 \end{array} \right ]$

Converting to row echelon form through row operations:

$\begin{array}{ c } R_2 = 2R_2 + R_1 \\ \longrightarrow \\ R_3 = 2R_3+R_1 \end{array} \left [ \begin{array}{ c c c } -10 & 5 & 5 \\ 0 & -15 & 15 \\ 0 & 15 & -15 \end{array} \right ]$

$\begin{array}{ c } R_1 = R_1 + \frac{ R_2 }{ 3 } \\ \longrightarrow \\ R_3 = R_2 + R_3 \end{array} \left [ \begin{array}{ c c c } -10 & 0 & 10 \\ 0 & -15 & 15 \\ 0 & 0 & 0 \end{array} \right ]$

$\begin{array}{ c } R_1 = \frac{ -R_1 }{ 10 } \\ \longrightarrow \\ R_2 = \frac{ -R_2 }{ 3 } \end{array} \left [ \begin{array}{ c c c } 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array} \right ]$

So:

$A \ – \ \lambda I \ = \ \left [ \begin{array}{ c c c } 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array} \right ]$

To find the eigenvectors:

$( A \ – \ \lambda I ) \vec{x}\ = 0$

Substituting Values:

$\left [ \begin{array}{ c c c } 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array} \right ] \ \left [ \begin{array}{ c } x_1 \\ x_2 \\ x_3 \end{array} \right ] \ = \ 0$

Solving this simple system yields:

$\vec{x} \ = \ \left [ \begin{array}{ c } 1 \\ 1 \\ 1 \end{array} \right ]$

## Numerical Result

$A \ – \ \lambda I \ = \ \left [ \begin{array}{ c c c } 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array} \right ]$

$\vec{x} \ = \ \left [ \begin{array}{ c } 1 \\ 1 \\ 1 \end{array} \right ]$

## Example

Diagonalize the same matrix given in the above question for $lambda \ = \ -3$:

For $\lambda \ = \ -3$:

$A \ – \ \lambda I \ = \ \left [ \begin{array}{ c c c } 5 & 5 & 5 \\ 5 & 5 & 5 \\ 5 & 5 & 5 \end{array} \right ]$

Converting to row echelon form through row operations:

$\begin{array}{ c } R_2 = R_2 – R_1 \\ \longrightarrow \\ R_3 = R_3 – R_1 \end{array} \left [ \begin{array}{ c c c } 5 & 5 & 5 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right ]$

$\begin{array}{ c } R_1 = \frac{ R_1 }{ 5 } \\ \longrightarrow \end{array} \left [ \begin{array}{ c c c } 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right ]$

So:

$A \ – \ \lambda I \ = \ \left [ \begin{array}{ c c c } 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right ]$