how to show that $r ≤\frac{n}{2}$ .

Question

Let $T : R^n → R^n$ be a linear transformation such that $ T^ 2 = 0$. If $r$ denotes the rank of $T $(that is,$ r = dim(Image(T)))$, then show that $r ≤\frac{n}{2}$

i know that by Rank-nuliity theorem The dimension of $T$ will be $n$ as$ T^2=0$ mean T is a nilpotent ..so dimension of T will be $n-2$.

But here im confused how dimension of $ r = dim(Image(T)))$ will be$ ≤\frac{n}{2}$.

Any Hints or solutions will be appreciated

thank in advance for helping me .

Answer 1

From $T^2=0$ we get $Im(T) \subseteq ker(T)$, hence $\dim ker(T) \ge r$.

It follows that:

$n= r+ \dim ker(T) \ge 2r$.

Answer 2

Since $T^2=$ it follows $im(T)\subset \ker(T)$. Now $$ n = \dim(im T) + \dim (\ker T) = rank(T) + \dim(\ker T) \le 2 \dim(\ker T) $$ Hence $$ \dim(\ker T) \ge n/2 $$ and $$ \dim(im T) = rank(T) \le n/2 $$.