Find bases B3 and B2 for $\Bbb R^3$ and $\Bbb R^2$ given a linear transformation and its matrix A linear transformation T is defined by
T: $\Bbb R^3$ $\rightarrow$ $\Bbb R^2$ $\Rightarrow$ T$\begin{pmatrix}x\\y\\z\end{pmatrix}$ = $\begin{pmatrix}2x+y\\y+2z\end{pmatrix}$
Find bases $\mathscr B_3'$ and $\mathscr B_2'$ for $\Bbb R^3$ and $\Bbb R^2$ respectively such that $Mat_{\mathscr B_3' , \mathscr B_2'}$ (T) = $\begin{pmatrix} 1&0&-1\\0&1&2 \end{pmatrix}$
So far I have attempted to answer this by producing an arbitrary basis for $\Bbb R^3$, and then using the definition of the transformation matrix, express it as a linear combination of the basis vectors that will make up the basis for $\Bbb R^2$. If my arbitrary basis for $\Bbb R^3$ is {$\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}$,$\begin{pmatrix}a_2\\b_2\\c_2\end{pmatrix}$,$\begin{pmatrix}a_3\\b_3\\c_3\end{pmatrix}$}
then I can simplify this to $(\alpha_2+2\alpha_3) \begin{pmatrix}2a_2+b_2\\b_2+2c_2\end{pmatrix}$ + $(\alpha_1-\alpha_3) \begin{pmatrix}2a_1+b_1\\b_1+2c_1\end{pmatrix}$
I'm not sure what else to try or how to proceed if this is the correct approach
 A: I do not think your approach is correct.
We can simplify the answer by taking the ordered bases
$$\begin{align}
\mathscr B_3' & = \left\{
\begin{pmatrix}1\\0\\0\end{pmatrix},
\begin{pmatrix}0\\1\\0\end{pmatrix},
\begin{pmatrix}0\\0\\1\end{pmatrix}\right\},\\\\
\mathscr B_2' & = \left\{
\begin{pmatrix}\beta_{11}\\ \beta_{12} \end{pmatrix},
\begin{pmatrix}\beta_{21}\\ \beta_{22} \end{pmatrix}\right\}\end{align}$$
and solving for $\mathscr B_2'.$ The simplification is that with our choice for $\mathscr B_3',$ coordinate matrices in $\Bbb R^3$ have the same entries as vectors in $\Bbb R^3,$ so we do not have to worry about converting from one to the other. Then with $[ \cdot ]_{\mathscr B_2'}$ denoting the coordinate matrix with respect to ordered basis $\mathscr B_2',$ we have
$$\begin{align}
\left[ T \begin{pmatrix} x\\ y\\ z\end{pmatrix} \right]_{\mathscr B_2'} & = \mbox{Mat}_{\mathscr B_3', \mathscr B_2'}(T) \begin{pmatrix} x\\ y\\ z\end{pmatrix}\\\\
\left[ \begin{pmatrix} 2x + y\\ y + 2z\end{pmatrix} \right]_{\mathscr B_2'} & = \begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & 2\end{pmatrix} \begin{pmatrix} x\\ y\\ z\end{pmatrix}\\\\
& = \begin{pmatrix} x - z\\ y + 2z\end{pmatrix}\\\\
\begin{pmatrix} 2x + y\\ y + 2z\end{pmatrix} & =
 (x -  z) \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
 (y + 2z) \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix}\end{align}$$
To solve for the four $\beta$s, equate the coefficients of $x, y,$ and $z$ and solve to get
$$\mathscr B_2' = \left\{
\begin{pmatrix} 2\\ 0\end{pmatrix},
\begin{pmatrix} 1\\ 1\end{pmatrix}\right\}.$$
As amd stated in his answer to your previous question, other bases will satisfy the hypotheses of your question.
A: I do not think your approach is correct.
We can simplify the answer by taking the ordered bases
$$\begin{align}
\mathscr B_3' & = \left\{
\varepsilon_1 = \begin{pmatrix}1\\0\\0\end{pmatrix},
\varepsilon_2 = \begin{pmatrix}0\\1\\0\end{pmatrix},
\varepsilon_3 = \begin{pmatrix}0\\0\\1\end{pmatrix}\right\},\\\\
\mathscr B_2' & = \left\{
\begin{pmatrix}\beta_{11}\\ \beta_{12} \end{pmatrix},
\begin{pmatrix}\beta_{21}\\ \beta_{22} \end{pmatrix}\right\}\end{align}$$
and solving for $\mathscr B_2'.$ The simplification is that with our choice for $\mathscr B_3',$ coordinate matrices in $\Bbb R^3$ have the same entries as vectors in $\Bbb R^3,$ so we do not have to worry about converting from one to the other.
Now the $i$th column of Mat$_{\mathscr B_3', \mathscr B_2'}(T)$ is $[T\varepsilon_i]_{\mathscr B_2'}$ where $[ \cdot ]_{\mathscr B_2'}$ denotes the coordinate matrix with respect to ordered basis $\mathscr B_2'.$ That gives us the following three systems of equations:
$$\begin{align}
\begin{pmatrix} 1\\ 0\end{pmatrix} & = \left[ T \begin{pmatrix} 1\\ 0\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\
& = \left[ \begin{pmatrix} 2\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\
1 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
0 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & =
  \begin{pmatrix} 2         \\ 0         \end{pmatrix},\\\\
\end{align}$$

$$\begin{align}
\begin{pmatrix} 0\\ 1\end{pmatrix} & = \left[ T \begin{pmatrix} 0\\ 1\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\
& = \left[ \begin{pmatrix} 1\\ 1\end{pmatrix} \right]_{\mathscr B_2'}\\\\
0 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
1 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & =
  \begin{pmatrix} 1         \\ 1         \end{pmatrix},\\\\
\end{align}$$

$$\begin{align}
\begin{pmatrix} -1\\ 2\end{pmatrix} & = \left[ T \begin{pmatrix} 0\\ 0\\ 1\end{pmatrix} \right]_{\mathscr B_2'}\\\\
& = \left[ \begin{pmatrix} 0\\ 2\end{pmatrix} \right]_{\mathscr B_2'}\\\\
-1 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
 2 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & =
   \begin{pmatrix} 0         \\ 2         \end{pmatrix}.\end{align}$$
Solving for the four $\beta$s, we get
$$\mathscr B_2' = \left\{
\begin{pmatrix} 2\\ 0\end{pmatrix},
\begin{pmatrix} 1\\ 1\end{pmatrix}\right\}.$$
As amd stated in his answer to your previous question, other bases will satisfy the hypotheses of your question.
