What is the the rank of a $4 \times 5$ matrix whose null space is three dimensional?
Wouldn't this mean that I would have something like
$\begin{bmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{bmatrix}$
So wouldn't it's rank = (the # of pivot columns) - the null space dimension? so rank $= 4 - 3 = 1$?
On
The number of pivot columns is the rank: row-reducing gives you pivots, which in turn gives you a basis for the column space. In terms of matrices, the rank-nullity theorem says that the total number of columns is equal to the sum of the nullity and the rank.
You seem to be confusing the total number of columns with the dimension of the column space. Take for example the $1 \times 2$ matrix $\begin{pmatrix}1 & 1 \end{pmatrix}$. Evidently the total number of columns is $2$, but these columns are linearly dependent. A basis for the column space consists of a single column $\begin{pmatrix}1 \end{pmatrix}$, and hence the rank = dimension of column space = $1$.
The rank-nullity theorem states that for a matrix $A$ $$ \DeclareMathOperator{rank}{rank}\rank(A)+\DeclareMathOperator{nullity}{nullity}\nullity(A)=\#\text{ columns of }A $$ The rank of $A$ is the dimension of the column space of $A$ and $\nullity(A)$ is the dimension of the null space of $A$.
Your question asks for the rank of a $4\times 5$ matrix $A$ whose null space is three-dimensional. The rank-nullity theorem immediately implies $$ \rank(A)=\#\text{ columns of }A-\nullity(A)=5-3=2 $$
The example you give is $$ A = \left[\begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right] $$ This matrix has $\rank(A)=4$ and thus $\nullity(A)=4-3=1$. It is thus not a relevant example of your problem.
A relevant example would be $$ A = \left[\begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$ This matrix has nullity three and thus has rank two.