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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?

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  • $\begingroup$ The two different expressions for $A$ that you have are not equal, the second is $A^T$ $\endgroup$
    – clark
    Nov 28, 2016 at 14:02
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    $\begingroup$ The matrix with zeroes on and below the diagonal and ones above the diagonal is nilpotent, but its rank is $n-1$. $\endgroup$
    – Arthur
    Nov 28, 2016 at 14:02
  • $\begingroup$ @clark Wow, thanks I hadn't realized that yet $\endgroup$
    – Marekkk
    Nov 28, 2016 at 14:03
  • $\begingroup$ Do I just leave this here, because it is solved and probably way to easy for MSE? $\endgroup$
    – Marekkk
    Nov 28, 2016 at 14:04
  • $\begingroup$ math.stackexchange.com/questions/441116/… $\endgroup$ Jul 14, 2019 at 13:45

3 Answers 3

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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 $$ n = \operatorname{rank}(A) + \operatorname{nullity}(A) \geq 2\operatorname{rank}(A). $$

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You use the fact that $A=A^T$. Is the matrix $A$ symmetric?

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    $\begingroup$ Why not? The question is: "is there anyone who can spot my mistake?" $\endgroup$ Nov 28, 2016 at 14:03
  • $\begingroup$ You're right. I just don't read, apparently. $\endgroup$
    – Arthur
    Nov 28, 2016 at 14:04
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Pretty late to the party but adding another way of solving since it's different and uses a nice idea, in case anyone stumbles upon this post for help.

We know from the Sylvester inequality $$\mathrm{rank}(XY)\geq\mathrm{rank}(X)+\mathrm{rank}(Y)-n$$

Setting $X=Y=A$ we get $$\mathrm{rank}(AA)\geq\mathrm{rank}(A)+\mathrm{rank}(A)-n$$ or equivalently $$\mathrm{rank}(A^2)\geq2\mathrm{rank}(A)-n$$ and since $$\mathrm{rank}(A^2)=0$$ because $A^2=O_n$, we get $$0\geq2\mathrm{rank}(A)-n$$ or equivalently $$\mathrm{rank}(A)\leq \frac{n}{2}$$ and we are done . $\square$

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