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.
 A: 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.
A: 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.
A: You can choose between calculating the rank or the kernel since
$$\ker(f) = \{0\} \Leftrightarrow f \text{ injective}$$
for linear maps. Let's calculate the kernel. The kernel is just the set of all matrices $A \in \mathbb{R}^{2 \times 2}$ which get send to $0 \in \mathbb{R}^{3 \times 2}$ by $f$. So you need to solve a system of linear equations:
$$\begin{align} 
3r_1&+&&&2r_3&&&=0\\
&&3r_2&+&&&2r_4&=0\\
-2r_1&+&&&r_3&&&=0\\
&&-2r_2&+&&&r_4&=0\\
&&&&4r_3&&&=0\\
&&&&&&4r_4&=0
\end{align}$$
which solves to $r_1 = r_2 = r_3 = r_4 = 0$. So the only matrix $A \in \mathbb{R}^{2\times 2}$ which gets send to $0$ by $f$ is $0$ and thus the kernel is $\{0\}$. So $\emptyset$ is a basis of the kernel. 
Now that you know that $f$ is injective you know that a linear independent set gets send to a linear independent set. So the standard basis gets send to a linear independent set $\{f(e_1), f(e_2), f(e_3), f(e_4)\}$ which is contained in the range of $f$ for obvious reasons. These are 4 matrices. Because of the rank-nullity-theorem you know that $\dim(\text{range}(f)) = 4$ and because of this you know that the 4 matrices actually form a basis of the range.
A: First note that
$$\begin{pmatrix}-1&-2&1\\-2&-3&2\end{pmatrix}\begin{pmatrix}3&2\\-2&1\\0&4\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$
Let's let $A=\begin{pmatrix}-1&-2&1\\-2&-3&2\end{pmatrix}$ and $B=\begin{pmatrix}3&2\\-2&1\\0&4\end{pmatrix}$. Let $I$ be the $2\times2$ identity matrix.
Note that our function $f:\Bbb{R}^{2\times2}\to\Bbb{R}^{3\times2}$ is defined by letting $f(X)=BX$ for all $X\in\Bbb{R}^{2\times2}$.
First, let's show that the kernel is $\{0\}$. Let $X$ be in the kernel. So we have that
$$BX=f(X)=0.$$
Since $BX=0$, we have that
$$X=IX=(AB)X=A(BX)=A0=0.$$
So $X=0$. So the kernel is trivial, and $f$ is injective.
Let $\{v_1,v_2,v_3,v_4\}$ be your favorite basis for $\Bbb{R}^{2\times2}$. Since $f$ is injective, $\{Bv_1,Bv_2,Bv_3,Bv_4\}$ will be a basis for the range.
A: Let's call
$$\begin{pmatrix}
r_1  &r_2 \\ 
 r_3&r_4 
\end{pmatrix}$$
X and $$\begin{pmatrix}
3 & 2\\ 
 -2&1 \\ 
 0& 4
\end{pmatrix}$$
A for simplicity. $f(X)=A\times X=0$, e.g., X is in the kernel of f if and only if the range of X is a subspace of the kernel of A. But what is the kernel of A? To be in the kernel of A, a vector $\begin{bmatrix}v_1&v_2\end{bmatrix}^T$ would have to simultaneously satisfy $4v_2=0$ and $3v_1+2v_2=0$, in other words, it would have to be the zero vector. So for X to be in the kernel, the range of X would have to be the zero space, making X the zero matrix. So the basis of the kernel is the zero matrix. 
As for the range, that will just be all matrices in $\mathbb{R}^{3\times2}$ whose range is a subspace of the range of A. A basis for the range of A would be any two vectors that are linearly independent ($\begin{bmatrix}3&-2&0\end{bmatrix}^T$ and $\begin{bmatrix}2&1&4\end{bmatrix}^T$ are not co-linear, so they work).Then, you take every map from a basis element of $\mathbb{R}^{2}$ to an element of the basis of A and you have a basis for the range of f. 
A: Hint: Consider bases for $\mathbb{R}^{2 \times 2}$ and $\mathbb{R}^{3 \times 2}$. 
Write the matrix representation of $f$ with respect to this pair of bases (since $f$ is a linear transformation).
Now we reduced the problem to calculating the kernel and range of a matrix, which is a straightforward task (this can always be done, for any linear transformation between finite dimensional vector spaces).
