<s>[INST] Consider the linear function$ f : \mathbb{R}^{2\times2}\to \mathbb{R}^{3\times2}$ defined as follows: Consider the linear function $f : \mathbb{R}^{2\times2} → \mathbb{R}^{3\times2}$ defined as follows:
$$\begin{pmatrix} r_1 & r_2 \\  r_3 & r_4 \\  \end{pmatrix} \in \mathbb{R}^{2\times2}\mapsto f\begin{pmatrix} r_1 & r_2 \\  r_3 & r_4 \\  \end{pmatrix} :=\begin{pmatrix} 3 & 2 \\  -2 & 1 \\ 0 & 4 \end{pmatrix} \begin{pmatrix} r_1 & r_2 \\  r_3 & r_4 \\  \end{pmatrix}$$
Check whether $f$ is injective and/or surjective. If it is bijective, find its inverse function. Finally, find bases for its Kernel and its Range.
Attempted solution:
This matrix multiplication evaluates to:
$$\begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\  -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix}$$
$f$ cannot be surjective since $m>n$. However, it can be injective if $n=\text{rank}$. Therefore, it is not bijective.
Now I get stuck. I don't understand the technique to find the bases for its kernel or range nor can I calculate its rank. I know we can row reduce this but it isn't in the proper form so I'm unsure of how to proceed.
 [/INST] Let's calculate the kernel first. We are searching for all matrices $R=\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix}$ satisfying
$$f(R)=\begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\  -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix} =0.
$$
So, from the last line of the above matrix we obtain $r_3=r_4=0$, and from the first line we obtain $r_1=r_2=0$. So the only matrix on the kernel is the zero matrix. Therefore, your map is injective.
Now, let's calculate the range. 
$$f(R) = \begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\  -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix}.
$$
This matrix can be written as:
$$f(R) = r_1\begin{pmatrix} 3 & 0 \\  -2 & 0 \\ 0 & 0 \end{pmatrix}
+r_2\begin{pmatrix} 0 & 3 \\  0 & -2 \\ 0 & 0 \end{pmatrix}
+r_3\begin{pmatrix} 2 & 0 \\  1 & 0 \\ 4 & 0 \end{pmatrix}
+r_4\begin{pmatrix} 0 & 2 \\  0 & 1 \\ 0 & 4 \end{pmatrix}.
$$
Therefore, the range of $f$ is generated by the matrices
$$\begin{pmatrix} 3 & 0 \\  -2 & 0 \\ 0 & 0 \end{pmatrix}
,\quad
\begin{pmatrix} 0 & 3 \\  0 & -2 \\ 0 & 0 \end{pmatrix}
,\quad
\begin{pmatrix} 2 & 0 \\  1 & 0 \\ 4 & 0 \end{pmatrix}
,\quad
\begin{pmatrix} 0 & 2 \\  0 & 1 \\ 0 & 4 \end{pmatrix}.
$$
Furthermore, from the injectivity of $f$ you actually obtain these matrices are linearly independent, and therefore they form a basis for the range.
</s>