dim(Col(A)) ≤ dim(Nul(A))

Question

Prove or provide a counter example:

If $A\in M_{n\times n}(\mathbb R)$ s.t. $A^2 = 0$ then $\dim(Col(A)) \leq \dim(Nul(A))$.

After trying out some examples, I have the intuition it's true but not really sure how to prove it. I've managed to show that $\det(A) = 0$ which brings me to the conclusion that $Nul(A)$ is non-trivial, if that helps.

Answer 1

It is just that $Col(A)\subseteq Nul(A)$ because, for $x\in Col(A)$ we have $x=Ay$ for some column $y$, but then $Ax=A^2y=0$ (since $A^2=0$), so $x\in Nul(A)$.

As $Col(A)\subseteq Nul(A)$, the statement $\dim(Col(A))\le \dim(Nul(A))$ follows directly.