A question on an assignment asks to find the rank and nullity of the linear map B, from the set of all matrices of size $2\times 3$ with complex entries, given by;
$$
B\begin{pmatrix}
a & b & c\\
d & e & f
\end{pmatrix}
=\begin{pmatrix}
a & 0 & 2c\\
0 & e-d & f
\end{pmatrix}
$$
Am I right in thinking that:
$$
\mathrm{ker}(B) = \begin{pmatrix}
0 & x & 0\\
y & y & 0
\end{pmatrix}
$$
And that therefore nullity $=$ dim(ker($B$)) $= 2$
And using the Rank-Nullity Theorem,
$$
\mathrm{rank}(B) =
\dim(
\begin{pmatrix}
a & b & c\\
d & e & f
\end{pmatrix}
)-2 = 6-2 = 4$$
Best Answer
$$B\begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \qquad \Longrightarrow \qquad \begin{pmatrix} a & 0 & 2c \\ 0 & e-d & f \end{pmatrix}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
Therefore, $a=c=f=0$ and $e=d$. Thus $b,d$ are free to take on any real value. So, yes.
$$ \mathrm{ker}(B) = \left\{ b\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} + d\begin{pmatrix} 0 & 0 & 0 \\ 1 & 1 & 0 \end{pmatrix} \;{\Huge |}\; b,d \in \mathbb{R} \right\}$$
Clearly the kernel is 2-dimensional. The domain being 6-dimensional means that the rank must be $6-2=4$. You are correct.
You can also see that the rank is 4 directly. Notice that
$$\left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\}$$
are all elements in the range of $B$ [set $a,\dots,f=0$ with the exception of (1) $a=1$, (2) $c=1/2$, (3) $e=1$, and (4) $f=1$.]
Clearly these elements span the range and are independent. Thus the dimension of the range is 4 (so rank=4).