I have an endomorphism T(x) = Ax where
$A = \begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{pmatrix}$ in the canonical basis. I want the transformation to go from $B'$ to $B'$. How do I do that?
This is what I've tried to do so far:
I tell myself that I want the vector columns of A to be expressed in B' = {(1,0,0),(0,1,0),(1,1,1)). Thus, I know that, for example, for the first column of $A$, we have
$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} * [x_1]_B' = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$
And now, I tell myself that we need to left multiply both sides by $A^{-1}$, and we get
$[x_1]_B' = A^{-1}\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$.
I do this calculation for the three columns, and I would get a linear transformation taking a vector in basis $B'$ to a vector in basis $B'$. Is this right? Thank you.
The matrix
$$M = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$
represents the change of basis form the new basis to the canonical then the matrix A' in the new basis is
$$A'=M^{-1}AM$$
Indeed since in the canonical
and
we have that
$$y=Ax\implies My'=AMx'\implies y'=M^{-1}AMx'$$