We're given $L: \mathbb{R}^3 \rightarrow \mathbb{R}^3$, which is a linear transformation defined by: $$ L \left( \begin{bmatrix}x_{1}\\x_{2}\\x_{3} \end{bmatrix} \right) = \begin{bmatrix}4x_{3}\\3x_{1}+5x_{2}-2x_{3}\\x_{1}+x_{2}+4x_{3} \end{bmatrix}$$
We're also given a basis $$B = \left(\begin{bmatrix}1\\1\\1 \end{bmatrix} , \begin{bmatrix}1\\1\\0 \end{bmatrix}, \begin{bmatrix}1\\0\\0 \end{bmatrix} \right) $$
The point of the exercise is to find a matrix $M$ associated to $L$ with respect to $B$, such that $[L(x)]_{B} = M[x]_{B}$ for all $x\in \mathbb{R}^3$.
Here's my approach. I transformed the vectors in $B$ to get the columns of the matrix $M$ giving me: $$M = \begin{bmatrix}4 & 0 & 0\\ 6&8&3\\6&2&1 \end{bmatrix} $$
The solution sheed says that $M =\begin{bmatrix}6&2&1\\ 0&6&2\\-2&-8&-3 \end{bmatrix}$.
I have no idea how to find the the correct solution nor why my approach is wrong.
Best Answer
The key idea is that the matrix:
$$B=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}$$
transforms a vector in the base B to a vector in the standard basis and its inverse transform a vector in the standard basis to a vector in the base B:
$$v_S=B v_B \iff v_B=B^{-1}v_S$$
Thus, since the linear transformation in the standard basis is expressed by:
$$L=\begin{bmatrix}0&0&4\\3&5&-2\\1&1&4\end{bmatrix} \quad w_S=Lv_S$$
we have
$$w_S=Lv_S \implies Bw_B=LBv_B\implies w_B=B^{-1}LBv_B$$
and therefore the matrix
$$B^{-1}LB=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}^{-1}.\begin{bmatrix}0&0&4\\3&5&-2\\1&1&4\end{bmatrix}.\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}$$
represents the linear transformation with respect to the new basis.