How would I find the standard matrix of this linear transformation? Let $T$ be a linear transformation on the plane with 
$T\begin{pmatrix}-5\\ -4\end{pmatrix}=\begin{pmatrix}-5\\ -5\end{pmatrix}$
and 
$T\begin{pmatrix}-4\\ -3\end{pmatrix}=\begin{pmatrix}-1\\ 2\end{pmatrix}$
Find a standard matrix for $T(x)$.
Since the answer should be a $2$x$2$ matrix, I thought that I would solve for $T$ in the first equation and then solve for $T$ in the second equation. Then combine the two results into a standard matrix. But I am not sure how I would solve for $T$ as the inverse of a $2$x$1$ matrix is not possible. 
So I am assuming there must be a different approach.
If so, any help or guidance will be highly appreciated!
 A: Express the the standard basis in terms of $\begin{pmatrix}-5\\ -4\end{pmatrix}$ and $\begin{pmatrix}-4\\ -3\end{pmatrix}$ ,i.e.,   find $x_1$ and $x_2$ such that $\begin{pmatrix} 1\\ 0\end{pmatrix} = x_1\begin{pmatrix}-5\\ -5\end{pmatrix} + x_2\begin{pmatrix}-4\\ 3\end{pmatrix} $. Do the same for the basis $\begin{pmatrix} 0\\ 1\end{pmatrix} $.  Now, compute $T(\begin{pmatrix} 0\\ 1\end{pmatrix})$ and $T(\begin{pmatrix} 1\\ 0\end{pmatrix})$ using the fact that $T$ is a linear transformation. From the values of $T(\begin{pmatrix} 0\\ 1\end{pmatrix})$ and $T(\begin{pmatrix} 1\\ 0\end{pmatrix})$, you can construct your matrix.
A: There is a rather general approach to it using changes of bases.


*

*The image of $T$ is spanned by the vectors $u_1 = \left( \begin{matrix} -5 \\ -5\end{matrix}  \right),\, u_2 = \left( \begin{matrix} -1 \\ 2\end{matrix} \right)$.

*We deal with 2 bases $B=\{b_1,b_2\}$ and $E=\{e_1,e_2\}$ of $\mathbb{R}^2$: $$b_1 = \left( \begin{matrix} -5 \\ -4\end{matrix}  \right),\, b_2 = \left( \begin{matrix} -4 \\ -3\end{matrix} \right) \mbox{ and } e_1 = \left( \begin{matrix} 1 \\ 0\end{matrix}  \right),\, e_2 = \left( \begin{matrix} 0 \\ 1\end{matrix} \right)$$

*The matrix $T_{B,E}$ for $T$ with respect to basis $B$ in the domain and $E$ in the codomain is already given by $Tb_1 = u_1$ and $Tb_2 = u_2$:$$T_{B,E}: \mathbb{R}^2, B \longrightarrow \mathbb{R}^2, E:\, T_{B,E} = (u_1 \, u_2) =\left( \begin{matrix} -5 & -1 \\ -5 & 2\end{matrix} \right)$$

*You are looking for $T_{E,E}$. So, we need to transform vectors given in $E$ into vectors in $B$. The matrix $I_{B,E}$ of the identity mapping $I$ is already given: $$I_{B,E}: \mathbb{R}^2, B \longrightarrow \mathbb{R}^2, E:\, I_{B,E} = (b_1 \, b_2) =\left( \begin{matrix} -5 & -4 \\ -4 & -3\end{matrix} \right)$$

*Now, you get $T_{E,E}$ immediately:
$$T_{E,E} = T_{B,E}I_{E,B}=T_{B,E}(I_{B,E})^{-1} = \left( \begin{matrix} -5 & -1 \\ -5 & 2\end{matrix} \right)\left( \begin{matrix} -5 & -4 \\ -4 & -3\end{matrix} \right)^{-1}=\left( \begin{matrix} -11 & 15 \\ -23 & 30\end{matrix} \right)$$
As you can see:
$$\left( \begin{matrix} -11 & 15 \\ -23 & 30\end{matrix} \right)\left( \begin{matrix} -5 \\ -4\end{matrix}  \right)= \left( \begin{matrix} -5 \\ -5\end{matrix}  \right) \mbox{ and }\left( \begin{matrix} -11 & 15 \\ -23 & 30\end{matrix} \right)\left( \begin{matrix} -4 \\ -3\end{matrix} \right) =  \left( \begin{matrix} -1 \\ 2\end{matrix} \right)$$

