Am I correct about this standard matrix and other definitions? I have this linear transformation $T:\mathbb R^2\to\mathbb R^3$ such that $T\left[\begin{matrix}1\\0\end{matrix}\right]=\left[\begin{matrix}2\\3\\1\end{matrix}\right]$ and $T\left[\begin{matrix}1\\5\end{matrix}\right]=\left[\begin{matrix}-3\\3\\21\end{matrix}\right]$ and I need to find the standard matrix of $T$ and I need to determine if $T$ is onto and is one to one. However I'm a little unsure of my reasoning for each one.
So, if I'm correct here, the standard matrix $A_T=(T(\vec e_1)\quad T(\vec e_2)...T(\vec e_n))$ and if we said $\vec e_1=\left[\begin{matrix}1\\0\end{matrix}\right]$ and $\vec e_2=\left[\begin{matrix}1\\5\end{matrix}\right]$, then the standard matrix $A_T=\left[\begin{matrix}1&1\\0&5\end{matrix}\right]$. Is this right? Something seems strange about it. And then since the matrix $A_T$ is linearly independent and the columns span $\mathbb R^2$, then the matrix is one to one and onto.
This is just what I have. I have a feeling it's not correct though.
 A: If $$T\left(\begin{matrix}1\\0\end{matrix}\right)=\left(\begin{matrix}2\\3\\1\end{matrix}\right)$$ and $$T\left(\begin{matrix}1\\5\end{matrix}\right)=\left(\begin{matrix}-3\\3\\21\end{matrix}\right)$$ so by setting $\epsilon_1=\left(\begin{matrix}1\\0\end{matrix}\right)$ and $\epsilon_2=\left(\begin{matrix}0\\1\end{matrix}\right)$ you have $\left(\begin{matrix}1\\5\end{matrix}\right)=\left(\begin{matrix}1\\0\end{matrix}\right)+5\left(\begin{matrix}0\\1\end{matrix}\right)$ and then $T(\epsilon_1)=\left(\begin{matrix}2\\3\\1\end{matrix}\right)$ and $$T(\epsilon_2)=T\left(\begin{matrix}1\\5\end{matrix}\right)-5T\left(\begin{matrix}0\\1\end{matrix}\right)=\left(\begin{matrix}-3\\3\\21\end{matrix}\right)-5\left(\begin{matrix}2\\3\\1\end{matrix}\right)=\left(\begin{matrix}-13\\-12\\17\end{matrix}\right)$$
A: I presume by standard matrix, you mean with respect to the standard basis $e_1, e_2$?
Note that $e_2 = \begin{bmatrix}0\\1\end{bmatrix} = \frac{1}{5}\left( \begin{bmatrix}1\\5\end{bmatrix} - \begin{bmatrix}1\\0\end{bmatrix} \right) $, so you can compute $T e_2 = \frac{1}{5}\left( T\begin{bmatrix}1\\5\end{bmatrix} - T\begin{bmatrix}1\\0\end{bmatrix} \right) $.
Then the standard matrix is $\begin{bmatrix}Te_1 & Te_2\end{bmatrix}$.
To see if $T$ is one to one, you need to check if $\ker T = \{0 \}$. One way is to check if $T e_1, T e_2$ are linearly independent. This also tells you the dimension of the range space, which will let you determine if $T$ is onto or not.
A: You started out fine, but then you went wrong by redefining the standard basis vectors! $e_2$ is not the vector $\binom{5}{1}$: it is the vector $\binom{0}{1}$. If you really need to find the standard matrix, then you'll have to think about how one might go about using the information you have to figure out $T e_2$ -- e.g. by solving for $e_2$ in terms of $e_1$ and $\binom{1}{5}$.
Or, you might instead try to organize the information you're given in a fashion that's easy to do matrix arithmetic with:
$$ T \binom{1\ \ 1}{0\ \ 5} = \left(\begin{matrix}2 & -3 \\ 3 & 3 \\ 1 & 21 \end{matrix}\right)$$
