<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] The rank of a matrix is the number of non empty rows once it is une the reduced echelon form. Your matrix could be reduced by doing the following operations $3L_2+2L_1$ then $7L_3-4L_2$.
$$\begin{pmatrix}3&2\\ -2&1\\ 0&4\end{pmatrix}\sim\begin{pmatrix}3&2\\ 0&7\\ 0&4\end{pmatrix}\sim\begin{pmatrix}3&2\\ 0&7\\ 0&0\end{pmatrix}$$
The matrix has two non empty lines, so $\text{rank}{A}=2$. The function is injective.
There is an other way to know the function is injective: if the kernel has only one element.
The kernel is the set of all matrix from the domain that are maps to the $0$ of the image. We need to find all value of $r_1$, $r_2$, $r_3$ and $r_4$ such that
$$\begin{pmatrix}3r_1+2r_3&3r_2+2r_4\\ -2r_1+r_3&-2r_2+r_4\\ 4r_3&4r_4\end{pmatrix}=\begin{pmatrix}0&0\\ 0&0\\ 0&0\end{pmatrix}$$
This gives us six equations with four unknows. Starting with the bottom line, $r_3=r_4=0$ once those are known, the other equations give $r_1=r_2=0$. Only the null matrix is part of the kernel.
To find a basis for the image, we take a basis of the domain and apply the function. It will give us four linearly independant matrix that generate the image of $f$. A basis of $\Bbb R^{2\times2}$ could be
$$e_1=\begin{pmatrix}1&0\\ 0&0\end{pmatrix}, e_2=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}, e_3=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}, e_4=\begin{pmatrix}0&0\\ 0&1\end{pmatrix}$$
We apply the function to each matrix.
$$f(e_1)=\begin{pmatrix}3&0\\ -2&0\\ 0&0\end{pmatrix}, f(e_2)=\begin{pmatrix}0&3\\ 0&-2\\ 0&0\end{pmatrix}, f(e_3)=\begin{pmatrix}2&0\\ 1&0\\ 4&0\end{pmatrix}, f(e_4)=\begin{pmatrix}0&2\\ 0&1\\ 0&4\end{pmatrix}$$
These four matrix are the basis of the image.
</s>