Given a square Matrix $A$ over $\Bbb{R}$ of size $n$ such that $A^2=0$. Show that $\operatorname{rank}(A)\leq\displaystyle\frac n2$.

Question

Let $A\in \mathbb{R}^{n\times n}$ such that $A^2=0$. How could we show that $\operatorname{rank}(A)\leq\displaystyle\frac{n}{2}$?

Answer 1

$rank(A)=dim(Im(A))$, then

$$rank(A)+dim(Ker A)=n$$

$A^2=0$ implies $Im(A) \subset Ker A$ and then $rank(A) \le dim(Ker A)$

Then

$$rank(A) \le \frac{n}{2}$$

Answer 2

We consider the operator $L_A$ on $\mathbb{R}^n$ defined by, for any $x=(x_1,\dots,x_n)^t\in \mathbb{R}^n, L_A(x)=Ax$. Then we have $\operatorname{rank}(L_A)= \operatorname{rank}(A)$.

Now, $A^2=0\Rightarrow {L_A}^2=0$, which gives

$\operatorname{Im}(L_A)\subseteq \operatorname{Ker}(L_A)$

$\Rightarrow \dim(\operatorname{Im}(L_A))\leq \dim(\operatorname{Ker}(L_A)).$

That is, $\operatorname{rank}(L_A)\leq \operatorname{Nullity}(L_A)\dots$ (1)

From Rank-Nullity theorem we have,

$n=\operatorname{rank}(L_A)+\operatorname{Nullity}(L_A)$. Then using (1) we get,

$\Rightarrow n\geq 2\times \operatorname{rank}(L_A).$

$\Rightarrow n/2\geq \operatorname{rank}(L_A)=\operatorname{rank}(A)$.