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.
Best Answer
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.