In the following practice problem I am trying to give an example of a $4 \times 5$ matrix $A$ with $\dim(\mathrm{null}(A)) = 3$
Workings:
Since the $\dim(\mathrm{null}(A))= 3$ we know that the $\mathrm{rank}(A) = 5 - 3 = 2 $. Since rank means the leading $1's$. Would a example of a $ 4 \times 5$ matrix be:
1 0 0 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
Sorry I do not know the proper formatting for matrices.
Yes the example you gave is true. In general a $4 \times 5$ matrix has at most rank $= 4$, which means at most the $dim(Null(A))$ is 1 (in the case where all the rows are linearly independent).
Now if 2 rows are linearly dependent, then $dim(Null(A))$ is 2 and rank $A$ is 3.
Also if 3 rows are linearly dependent, then $dim(Null(A))$ is 3 and rank $A$ is 2. (which is what you say)
\begin{equation}\label{eq13} A= \begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4\\ \end{pmatrix} \end{equation} where $x_k$ are $1 \times 5$ vectors. Let $x_1$ be linearly independent of $x_2$, $x_3$ and $x_4$. Also let $x_2$, $x_3$ and $x_4$ be linearly dependent, then rank $A = 2$ and $dim(Null(A))= 3$. Many counterexamples you can have such as:
$x_1 = [a_1 ,a_2, a_3, 0]$ for at least one $a_k \neq 0$
$x_2 = [0 ,0, 0, b]$ for $b \neq 0$
$x_3 = [0 ,0, 0, c]$ for $c \neq 0$
$x_4 = [0 ,0, 0, d]$ for $d \neq 0$