For my matrix algebra class I need to prove the following:
If $A^2=0$, prove $\mathrm{rank}(A) \le \frac{n}{2}$. So if A is nilpotent prove $\mathrm{rank}(A) \le \frac{n}{2}$.
I know already how to solve this, but my initial way of solving is false. I am looking for the mistake, but cannot find one. I know there already exists a question where this is asked. I'm just curious about my particular mistake.
Proof:
$A=\begin{bmatrix}a_1&&a_2&& …&&a_n\end{bmatrix}$ and $A=\begin{bmatrix}a^T_1\\a^T_2\\…\\a^T_n\end{bmatrix}$
$AA=\begin{bmatrix}a_1 && a_2 && … && a_n\end{bmatrix}\begin{bmatrix}a^T_1\\a^T_2\\…\\a^T_n\end{bmatrix}$
$=\begin{bmatrix}a_1a^T_1&&…&&a1a^T_n\\…\\a_na^T_1 && …&& a_na^T_n\end{bmatrix}$
$=\begin{bmatrix} a_1 \cdot a_1&&…&&a_1 \cdot a_n\\…\\a_n \cdot a_1 && …&& a_n \cdot a_n\end{bmatrix}$
$=0_{nxn}$
So we know the diagonal is zero thus
$a_i \cdot a_i = 0$
this equals $\vert\vert{a_i}\vert\vert^2$
The square root of this equals the length, therefore the length is equal to 0. The only vector with this property is the zero vector. Herefore all vectors $a$ must be equal to the zero vector.
The last however is not the truth, is there anyone who can spot my mistake?
Best Answer
If $A^2 = 0$, then $A(Ax) = 0$ for all $x\in\mathbb{R}^n$. Therefore, $$\operatorname{col}(A) \subset \operatorname{null}(A) \implies \operatorname{rank}(A) \leq \operatorname{nullity}(A).$$ By the rank-nullity theorem $$ n = \operatorname{rank}(A) + \operatorname{nullity}(A) \geq 2\operatorname{rank}(A). $$