What should the maximum possible rank of a matrix be under certain conditions?

Question

I have been trying questions in Linear Algebra and I couldn't correctly solve this particular question.

Let $A$ be a non-zero $4\times 4$ complex matrix such that $A^2= 0$. What is the largest possible rank of $A$?

All eigenvalues must satisfy $\lambda^{2} = 0$ and so all eigenvalues are $0$. I am unable to think what results I should use now. Can anyone please tell how I should proceed. Any help would be really appreciated.

Answer:

$2$

Answer 1

We can construct one of rank $2$: $$A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 &0 \\ 0& 0&0&1 \\ 0&0&0&0\end{bmatrix}$$ fits the bill. Clearly $A$ can't have rank four (since then it would be invertible), but why can't $A$ have rank $3$? If it had rank $3$, then there could only be one linearly independent vector in the zero eigenspace (i.e. nullspace). That would have to mean that there is only one Jordan block in the Jordan Normal Form of $A$ (number of Jordan blocks for a given eigenvalue $=$ number of linearly independent eigenvectors for that eigenvalue). Thus, $A$ would have to be similar to a $4 \times 4$ matrix that is a single zero Jordan block, i.e. $$A = S \cdot \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ 0& 0&0&1 \\ 0&0&0&0\end{bmatrix} \cdot S^{-1}$$ for some invertible $S$. But clearly $$A^2 = S \cdot \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1 \\ 0& 0&0&0 \\ 0&0&0&0\end{bmatrix} \cdot S^{-1}$$ cannot be zero, so we conclude it must be impossible for $A$ to have rank $3$. $\square$

Answer 2

Consider a space of dimension $N$. Here $N=4$, but the question is more interesting for odd dimensions.

Upon applying the mapping $A$, $n$ vectors in a basis fall off, so $\mathrm{rank}(A)=N-n$ remain.

Upon applying $A$ again, all these $N-n$ fall off.

However, $N-n\le n$ since the second set of vectors are in the null space of dimension $n$. So $2n\ge N$; the minimum value of $n$ occurs when $n=N/2$ for $N$ even. This is achieved in fact for the evenly split matrix $\begin{pmatrix}O&O\\I&O\end{pmatrix}$.

For odd dimensions, the minimum of $n$ is $n=\lceil N/2\rceil$, again achieved by $\begin{pmatrix}O_{n\times m}&O_{n\times n}\\I_{m\times m}&O_{m\times n}\end{pmatrix}$. The rank is then $m=\lfloor N/2\rfloor$.