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This is a proof question and I am not sure how to prove it. It is obviously true if you start with $A = 0$ and square it.

I was thinking:

If $ A^2 = 0 $

then $ A A = 0 $

$ A A A^{-1} = 0 A^{-1}$

$I\,A = 0 $

but the zero matrix is not invertible and that it was not among the given conditions.

Where's a good place to start?

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    $\begingroup$ Hint: Nilpotent matrices. $\endgroup$ Commented Apr 15, 2015 at 18:12
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    $\begingroup$ You were not given that $A$ is invertible, so $A^{-1}$ may not make sense. In fact, $A^2=0$ proves it is not invertible, because the deteminant of $A^2$ is the square of the determinant of $A$, so the determinant of $A$ is zero. $\endgroup$ Commented Apr 15, 2015 at 18:22

3 Answers 3

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HINT: Consider $A = \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$

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Given two nonzero orthogonal vectors $u, v \in \mathbb{R}^n$. Let $A = vu^T$, then

$$ A^2 = vu^Tvu^T = 0 $$

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consider $$A=\begin{bmatrix} 2 & 1\\ -4 & -2 \end{bmatrix}$$

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