If $T\in M_n(\mathbb{F})$ is a matrix such that $T^2=0$, then $rk(T)\le\frac{1}{2}n$.

Question

If $T\in M_n(\mathbb{F})$ is a matrix such that $T^2=0$, then $rk(T)\le\frac{1}{2}n$.

I know that $rk(T)$ is the dimension of the subspace spanned by the vectors of $\mathbb{F}^n$ given by the columns of $T$. I just don't know how to use the fact $T^2=0$ to calculate the subspace and it's dimension.

Answer 1

Hint: If $T^2 = 0$, then the image (column-space) of $T$ is contained in the kernel (null-space) of $T$. Apply the rank-nullity theorem.