MHB E2.3 Express T_b^b as the product of three matrices

[karush]

https://www.physicsforums.com/attachments/8962
ok this is my overleaf homework page but did not do (c) and (d)
this class is over but trying to do some stuff I missed.
I am only auditing so I may sit in again next year...;)
also if you see typos much grateful

I don't see a lot of replies on these DE questions so maybe there isn't an army of eager help?

 

[HallsofIvy]

Look at what linear transformation $T\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x- y \\ y- z\\ 2x+ 3y- 3z\end{bmatrix}$ does to each basis vector in $\alpha$:
$T\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}1- 0 \\ 0- 0\\ 2+ 0- 0\end{bmatrix}= \begin{bmatrix}1 \\ 0\\ 2\end{bmatrix}$.

$T\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}0- 1 \\ 1- 0\\ 0+ 3- 0\end{bmatrix}= \begin{bmatrix}-1 \\ 1\\ 3\end{bmatrix}$.​

$T\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}= \begin{bmatrix}0- 0 \\ 0- 1\\ 0+ 0- 3\end{bmatrix}= \begin{bmatrix}0 \\ -1\\ -3\end{bmatrix}$.

Those vectors will be the columns of the matrix representing T in this basis.​

 

[karush]

So $[T]_\beta^\beta$ would be the product of
$$\begin{bmatrix}1 \\ 0\\ 2\end{bmatrix}
\cdot \begin{bmatrix}-1 \\ 1\\ 3\end{bmatrix}
\cdot \begin{bmatrix}0 \\ -1\\ -3\end{bmatrix}$$
?