TITLE: How to find operator matrix in a new bases?
QUESTION [0 upvotes]: Let's assume that a linear operator in V space in basis (e1, e2, e3, e4) has a matrix: 
$A = \begin{pmatrix}
0&1&2&3\\5&4&0&-1\\3&2&0&3\\6&1&-1&7
\end{pmatrix}$
What look matrix of this operator will have in bases $(e_2, e_1, e_3, e_4)$ and $(e_1, e_1+e_2, e_1+e_2+e_3, e_1+e_2+e_3+e_4)$ ?
I see that this equation should have a look like:
$\begin{pmatrix}
e_1\\e_2\\e_3\\e_4
\end{pmatrix}$ * $\begin{pmatrix}
0&1&2&3\\5&4&0&-1\\3&2&0&3\\6&1&-1&7
\end{pmatrix}$ = $\begin{pmatrix}e_2 \\ e_1 \\ e_3 \\ e_4\end{pmatrix}$
but what's next?

REPLY [2 votes]: (a) The matricial equation of the operator $T$ in $B$ is $$Y=AX,\quad X=\begin{pmatrix}x_1\\ \vdots\\{x_n}\end{pmatrix}\text { coordinates of }x\text{ in }B,\;Y=\begin{pmatrix}y_1\\ \vdots\\{y_n}\end{pmatrix}\text { coordinates of }f(x)\text{ in }B.$$ 
(b) The change of basis matrix from $B=(e_2, e_1, e_3, e_4)$ to $$B'=(e_1, e_1+e_2, e_1+e_2+e_3, e_1+e_2+e_3+e_4)$$ is (transposing coefficients)$$P=\begin{pmatrix}{1}&{1}&{1}&1\\{0}&{1}&{1}&1\\{0}&{0}&{1}&1\\{0}&{0}&{0}&1\end{pmatrix}$$ 
(c) According to a well known theorem, the matricial equation of the operator $T$ in $B'$ is  
$$Y'=\left(P^{-1}AP\right)X',\quad X'=\begin{pmatrix}x'_1\\ \vdots\\{x'_n}\end{pmatrix}\text { coordinates of }x\text{ in }B',\;Y'=\begin{pmatrix}y'_1\\ \vdots\\{y'_n}\end{pmatrix}\text { coordinates of }f(x)\text{ in }B'.$$
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