# Basis which makes TWO linear transformation diagonalised at once

Find a basis $\gamma$ with respect to which both of the following lienar transformations on $\mathbb{R^3}$ become diagionalised (the matrices below are the matrices with respect to the standard basis):

$$S=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

and $$T=\begin{bmatrix} 2 & -1 & 2 \\ -1 & 2 & 2 \\ 2 & 2 & -1 \\ \end{bmatrix}$$

I found a basis of eigenvectors for S and another one for T but they are not the same, how should I proceed?

• Hint: Does $ST = TS$? – Eric Towers May 5 '16 at 13:10
• Hint: $T$ has distinct eigenvalues, so any eigenvector of $T$ is also an eigenvector of $S$ – Omnomnomnom May 5 '16 at 13:40
• @OOmnomnomnom: Maxima says $T$ has only $2$ eigenvalues, $-3$ and $3$. – Bernard May 5 '16 at 13:46
• 3 is a repeated eigenvalue – Lemma May 5 '16 at 14:10
• Could you further you explanation please, I am missing pieces of the jigsaw puzzle – Kam May 6 '16 at 12:46

## 1 Answer

Hint: Try the transformation matrix (or it's inverse depending on what you call the transformation matrix) $$\begin{bmatrix} 1 & 1 & -2\\ 1 & -1 &-2\\ -2& 0 &-2\\ \end{bmatrix}$$