Basis of one linear operator form the other linear operator. Suppose that I have two linear operators, $T_1$ and $T_2$ represented by matrices $3x3$.
Suppose that I have found the eigenbasis of $T_1$ equals to:
$$\{    \begin{pmatrix}
1\\ 
0\\
-1
\end{pmatrix} ,\begin{pmatrix}
1\\ 
-\sqrt{2}\\
1
\end{pmatrix},\begin{pmatrix}
1\\ 
\sqrt{2}\\
1
\end{pmatrix}\}$$
My question is how do I form $T_2$ in this basis,
Where $$T_2= \frac{1}{\sqrt{2}}\begin{pmatrix}
0 & i & 0 \\ 
i & 0 & i \\ 
0 & i & 0
\end{pmatrix}$$
Thank you!
 A: Let $B$ denote the matrix of the eigenbasis $\mathcal B=(b_1,b_2,b_3)$, i.e., $B:=\pmatrix{1&1&1\\0&-\sqrt2&\sqrt2\\-1&1&1}$, then the transformed matrix is
$$[T_2]_{\mathcal B}=B^{-1}T_2B\,.$$
This can be shown probably easiest in the form $B\,[T_2]_{\mathcal B}=T_2B$:
$$T_2B=T_2\,(b_1|b_2|b_3)=(T_2b_1|T_2b_2|T_2b_3) $$
and these columns are coordinated in basis $\mathcal B$: if $T_2b_i=\alpha_1b_1+\alpha_2b_2+\alpha_3b_3$, that means exactly
$$B\pmatrix{\alpha_1\\\alpha_2\\ \alpha_3}=T_2b_i\,. $$
If you calculated well, (applying to $T_1$, at least) you have to get the diagonal matrix with the corresponding eigenvalues, because of the definition of the matrix of a transformation w.r.t. a given basis (the $i$th column of the matrix will be $T_1b_i$ coordinated in $\mathcal B$). For another convenience of this:
$$B^{-1}T_1B=B^{-1}T_1\,(b_1|b_2|b_3)=B^{-1}(\lambda_1b_1|\lambda_2b_2|\lambda_3b_3)=\\
=B^{-1}\cdot B\pmatrix{\lambda_1&0&0\\0&\lambda_2 &0\\0&0&\lambda_3}=\pmatrix{\lambda_1&0&0\\0&\lambda_2 &0\\0&0&\lambda_3}\,.$$
