<s>[INST] Transformation matrix with respect to two bases 
Let $V=\mathbb{Q^{2\times3}}$ , $W=\mathbb{Q^{2\times2}}$be vector spaces over $\mathbb{Q}$ and consider the linear map $\phi:V\rightarrow W$ given by: $$\phi(M)=MA, \ A=\begin {pmatrix} 1&-3 \\2 &-2 \\3&-1\end{pmatrix}. $$
Given two bases $B=\{B_1,B_2,B_3,B_4,B_5,B_6\}$ of $V$ and $C=\{C_1,C_2,C_3,C_4\}$ of $W$: $$B_1=\begin {pmatrix} 1&0&0 \\0&0&0\end{pmatrix}, B_2=\begin {pmatrix} 0&1&0 \\0&0&0\end{pmatrix}, B_3=\begin {pmatrix} 0&0&1 \\0&0&0\end{pmatrix}, \\B_4=\begin {pmatrix} 0&0&0 \\1&0&0\end{pmatrix}, B_5=\begin {pmatrix} 0&0&0 \\0&1&0\end{pmatrix}, B_6=\begin {pmatrix} 0&0&0 \\0&0&1\end{pmatrix}. \\C_1=\begin {pmatrix} 1&0 \\0&0\end{pmatrix}, C_2=\begin {pmatrix} 0&0 \\0&1\end{pmatrix},C_3=\begin {pmatrix} 0&1/2 \\1/2&0\end{pmatrix},C_4=\begin {pmatrix} 0&1/2 \\-1/2&0\end{pmatrix}.  $$Determine the transformation matrix of $\phi$ with respect to $B$ and $C$.

I know that fundamentally the columns of the the transformation matrix are the images of the  vectors of $B$ under $\phi$, written on the basis $C$ but I'm not sure of how to apply it in this question. I first just calculated the image of each $B_i$ under $\phi$ as $B_i'$.
$$B_1'=\begin {pmatrix} 1&-3 \\0&0\end{pmatrix}, B_2'=\begin {pmatrix} 2&-2 \\0&0\end{pmatrix} ,B_3'=\begin {pmatrix} 3&-1 \\0&0\end{pmatrix}, \\B_4'=\begin {pmatrix} 0&0 \\1&-3\end{pmatrix},B_5'=\begin {pmatrix} 0&0 \\2&-2\end{pmatrix}, B_6'=\begin {pmatrix} 0&0 \\3&-1\end{pmatrix}$$
Then to express the matrices in terms of other matrices I had: $$B_1'=C_1-3C_3-3C_4, B_2'=2C_1-2C_3-2C_4, B_3'=3C_1-C_3-C_4 \\B_4'=-3C_2+C_3-C_4, B_5'=-2C_2+2C_3-2C_4,B_6'=-C_2+3C_3-3C_4 $$ Turning the remains into a 4x6 Matrix I ended up with: $$\begin {pmatrix} 1&2&3&0&0&0\\0&0&0&-3&-2&-1\\-3&-2&-1&1&2&3\\-3&-2&-1&-1&-2&-3 \end {pmatrix}$$ However I'm not sure of my answer which is why I wanted to ask if someone could please verify it for me.
 [/INST] Since $\phi: V\to W$ with $\dim_{{\bf Q}} V=2\times 3$ and $\dim_{{\bf Q}} W=2\times 2$. Then, the matrix of representation for the linear transformation respect to ordered basis $B$ and $C$ satisfy $[\phi]_{B}^{C}\in M_{4\times 6}({\bf Q})$.
By definition the matrix $[\phi]_{B}^{C}$ has in its columns the coordinates $[\phi(B_k)]_{C}$ for all $k=1,\ldots,6$. Then,

*

*$\phi(B_1)=\begin{bmatrix}1&0&0\\0&0&0 \end{bmatrix}\begin{bmatrix} 1&-3 \\2 &-2 \\3&-1\end{bmatrix}=\begin{bmatrix}1&-3\\0&0 \end{bmatrix}$, then $[\phi(B_1)]_{B}^{C}=\begin{bmatrix} 1\\0\\-3\\-3\end{bmatrix}$, since $\phi(B_1)=1C_1+0C_2+(-3)C_3+(-3)C_4$.

Continuing with that process, I get,
$$[\phi]_{B}^{C}=\begin{bmatrix} 1&2&3&0&0&0\\0&0&0&-3&-2&-1\\-3&-2&-1&1&2&3\\-3&-2&-1&-1&-2&-3 \end{bmatrix}$$
as you said.
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