Dimension of the null space if the rank is the number of columns. Maybe I'm just a bit confused on what all these terms mean, but a google search tells me that the null space can never be empty as the zero vector is always in it. However, by the rank-nullity theorem, if all the columns are independent, then the rank is the number of columns and the nullity must then be 0. But isn't the zero vector in the null space?
 A: Oh! I get it. You have faced the same problem that I have faced in my college days.
A dimension of a vector space $V$ over some field  $F$ $=$ The number of linearly independent vectors of $V$ that spans $V$ $=$ Cardinality of the Basis of $V$.
Now, let $T:V^n(F)\to V^n(F)$ be a linear map, where $V^n(F)$ denotes a vector space $V$ of dimension $n$ over the field $F$. Then the kernel of $T$ is denoted as $Ker(T)$ and defined as $Ker(T)=\{v\in V:T(v)=0\}$. Here $0$ means $Null$ vector of $V$. Dimension of $Ker(T)$ is called the nullity of $T$.
You know that every linear map $T$, maps null vector to null vector i.e $T(0)=0$ for all $T\in\mathcal L(V,V)$, where $\mathcal L(V,V)=\{\psi|\psi:V\to V\text{ is a linear map}\}$.
Hence, in our case $Ker(T)\ne \emptyset$ as $0\in Ker(T)$.
Now suppose that, $Ker(T)=\{0\}\ne \emptyset$. Then what is the dimension of $Ker(T)$ i.e. what is nullity of $T$?
Obviously, $dim(Ker(T))=nullity(T)=Cardinality(\text{Basis of $Ker(T)$})=$ Number of linearly independent vectors in $Ker(T)$.
Now, how many linearly independent vectors are present in $Ker(T)$? The answer is NONE, because $Ker(T)$ contains no nonzero vector, the only vector it contains is $0$, which is the zero vector, so clearly the basis for $Ker(T)$ is Empty, and since $Cardinality(\emptyset)=0$ so $dim(Ker(T))=0$.
Hope this works.
