If $A$ is an $n \times n$ matrix and $A^2 = 0$, then $\text{rank}(A)\le n/2$.

Was revising for a Linear Algebra when I came across this question.

"Given a Matrix $A$ over $\mathbb{R}^{n \times n}$, and $A^2 = 0$, show that rank($A$) $\leq n/2$"

My attempt:

I was thinking of the theorem from Cayley-Hamilton but the matrix is a $n \times n$ matrix hence that's probably invalid.

Is there perhaps a significance of $A^2 = 0$? apart from a special case of the nilpotent matrix?

• Consider Sylvester's inequality: $\mathrm{rank}(A)+\mathrm{rank}(B)-n\le\mathrm{rank}(AB)$. Commented Jul 11, 2013 at 9:38

$$\dim \ker A+rank A=n.$$ In addition, $A^2=0$, hence $im A\subset \ker A$, so $$\dim im A=\operatorname{rank} A\le \dim ker A.$$ These two expressions allow to conclude.

Hint: Let $k = \dim\ker A$. What can you say about $\dim\ker A^2$? Now use $\dim\ker A^2 + \mathop{\rm rank} A^2 = n$.

• Yea got it thanks! I realized applying Sylvester's inequality just made the whole problem solvable in 2 steps. :) Thanks also to @dentisDark .. Commented Jul 11, 2013 at 10:12

it's characteristic equation is $$x^2=0$$ which means the eigenvalues of $A$ are all zero.

Besides, $A$ is a nilpotent matrix and all the Jordan blocks have an order no more than 2. This observation leads to its rand no more than $n/2$.

• "and all the Jordan blocks have an order no more than 2". How does it follow? Commented Jun 12, 2016 at 11:11
• @jaggu Larger Jordan blocks don't satisfy $x^2 = 0$. Commented Jul 14, 2019 at 13:44

Let $$N = \ker A$$. Choose a basis $$e_n, ..., e_{r+1}$$ of $$N$$, where $$n-r = \dim N$$. Take a completion of that basis to obtain a basis $$e_1, ..., e_r, e_{r+1}, ..., e_n$$ of $$K^n$$. Denote $$M$$ as the subspace spanned by the vectors $$e_1, ..., e_r$$.

We can then write $$K^n = M\oplus N$$. Suppose that $$\text{rank}(A) = r = \dim M>n/2$$, then $$\dim N = n-r\leq \lfloor n/2\rfloor$$. Then $$Ae_1$$, ..., $$Ae_r$$ all belong to $$N$$, but since there's $$\geq\lceil n/2 \rceil$$ amount of them (in case $$n/2$$ is an integer, we want to use the inequality $$r>n/2$$), some must be linearly dependent, say $$Ae_r = a_1Ae_1+...+a_{r-1}Ae_{r-1} = A(a_1e_1+...+a_{r-1}e_{r-1})$$. But $$A$$ is 1-1 on $$M$$, so we must have $$e_r = a_1e_1+...+a_{r-1}e_{r-1}$$, which is a contradiction with the fact that $$e_1, ..., e_r$$ is a basis of $$M$$. Hence we must have $$\dim M\leq n/2$$.

$$A$$ is injective on $$M$$, because if it were $$Ax = Ay$$, then $$A(x-y) = 0$$ hence $$x-y\in M\cap N = \{0\}$$, so that $$x = y$$.

• Can you please explain about the one-one ness of $A$ on $M$ and why ? Commented May 4, 2020 at 18:37
• @LAMDA Thank you for finding my mistake. By $M$ I didn't mean the image of $A$, although I wrote that I did. Commented May 5, 2020 at 14:02