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

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

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

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

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,

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

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