Finding a basis for the kernel and range of $T$ 
Let $T\colon M_{23} \to M_{33}$ be the linear transformation defined by $T(A)=\begin{pmatrix} 2 & -1 \\ 1 & 2 \\ 3&1 \end{pmatrix}\,A$, for $A\in M_{23}$. Find a basis for the kernel and range of $T$. 

I don't know how to exactly approach this question. All I know is that the kernel of $T$ would be the nullspace. I row reduced $\begin{pmatrix} 2 & -1 \\ 1 & 2 \\ 3&1 \end{pmatrix}$ and got $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0&0 \end{pmatrix}$ but I don't know what to do from here.
 A: Let me give you a hint. If you write down a $2 \times 3$ matrix $A$ and compute $T(A)$ you can find what the range looks like and you can also compute the kernel from this. More explicitly, for an arbitrary $2 \times 3$ matrix
$$
A = \begin{bmatrix}
a & c & e\\
b & d & f\\
\end{bmatrix}
$$
you will get
\begin{align*}
T(A) = 
\begin{bmatrix}
2 &-1\\
1 & 2\\
3 & 1
\end{bmatrix}
\begin{bmatrix}
a & c & e\\
b & d & f\\
\end{bmatrix}
= 
\begin{bmatrix}
2a - b & 2c - d & 2e - f\\
a + 2b & c + 2d & e + 2f\\
3a + b & 3c + d & 3e + f\\
\end{bmatrix}
\end{align*}
Now, this tells you how a matrix in the range of $T$ looks like. 
Can you find a basis for the range using this? 
Also, for such a matrix $A$ to be in the kernel, you must have $T(A) = (0)_{3 \times 3}$. Then you can find conditions for $a, b, c, d, e, f$ so that you get the zero matrix and with that information you can find what a matrix in the kernel looks like and in that way find a basis for the kernel.
A: Note that for $A$ to be in the kernel, each column of $A$ must be orthogonal to each row of that $3\times2$ matrix. What can you find that's orthogonal to both $(2,-1)$ and $(1,2)$?
