$A = \begin{bmatrix} 1&2&1 \\ 0&1&0 \\ 1&3&1 \\ \end{bmatrix} $
I computed the eigenvalues:
$λ_ 1 = 1$
$λ_ 2 = 0$
$λ_ 3 = 2$
The corresponding eigenvalues I computed are:
$X_{λ1} = \begin{bmatrix} x_1 \\ (-1/2) x_3 \\ x_3 \\ \end{bmatrix} $ Substitute $x_1$ as 1, $x_3$ as 2: $X_{λ1} = \begin{bmatrix} x1 \\ -1 \\ 2 \\ \end{bmatrix} $
$X_{λ2} = \begin{bmatrix} -x_3 \\ 0 \\ x_3 \\ \end{bmatrix} $ Substitute $x_3$ as 1: $X_{λ2} = \begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix} $
$X_{λ3} = \begin{bmatrix} x_3 \\ 0 \\ x_3 \\ \end{bmatrix} $ Substitute $x_3$ as 1: $X_{λ3} = \begin{bmatrix} 1 \\ 0 \\ 1 \\ \end{bmatrix} $
To show A is diagonalizable:
$S = \begin{bmatrix} 1&-1&1 \\ -1&0&0 \\ 2&1&1 \\ \end{bmatrix} $
$S^{-1}AS = D$
AS = SD
After matrix multiplication, I got
$ \begin{bmatrix} 1&0&2 \\ -1&0&0 \\ 0&0&2 \\ \end{bmatrix} $ = $ \begin{bmatrix} 1&0&2 \\ -1&0&0 \\ 2&0&2 \\ \end{bmatrix} $
I know I made a mistake somewhere because AS does not equal SD. I can't find my mistake. Any help would be appreciated.
Just to remove this question from the "Unanswered" queue, it's clear from the comments that Christina's problem has been solved.