Finding the matrix of the operator in a given basis

Question

I have the matrix of the operator in the the basis $e_1,e_2,e_3$ :

$A=\begin{bmatrix}2&0&-6\\2&6&-2\\2&0&-6\end{bmatrix}\ $and I have to find the matrix of the operator in the basis $f_1,f_2,f_3$, where $f_1 = e_1+e_2+e_3, f_2 = e_1-e_2+e_3 ,f_3 = e_1+2e_2+2e_3$.

To do that do I just have to do $A \begin{bmatrix}1\\1\\1\end{bmatrix}\,A \begin{bmatrix}1\\-1\\1\end{bmatrix}\ ,A\begin{bmatrix}1\\2\\2\end{bmatrix}\ $ where $(1,1,1),(1,-1,1),(1,2,2)$ are the coefficients in $f_1,f_2,f_3$ and put the results vertical in a new matrix

Answer 1

You can begin with that, but you then have to express the images of $f_1, f_2$ ad $f_3$ in the new basis, not in the initial basis.

So it should be as simple to find the inverse of the change of basis matrix: $$P\begin{bmatrix} 1&1&1\\1&-1&2\\1&1&2 \end{bmatrix}$$ and apply the standard formula $A'=P^{-1}AP$.

Answer 2

You are almost there. So far you found $Af_1$,$ Af_2$ , and $Af_3.$ $$A \begin{bmatrix}1\\1\\1\end{bmatrix}\,A \begin{bmatrix}1\\-1\\1\end{bmatrix}\ ,A\begin{bmatrix}1\\2\\2\end{bmatrix}$$ These are still expressed in the standard basis.

Now you have to find the new coordinates of each of these vectors in your new basis { $f_1$, $f_1$,$f_1$}.

That is you have to solve a system $$Af_1 =\alpha f_1 + \beta f_2 +\gamma f_3 $$ and that $ \alpha , \beta, \gamma $ will be the entries of your first column in the desired matrix.

Similarly you find the other two columns.