If $A$ is a square matrix and $A^2 = 0$ then $A=0$. Is this true? If not, provide a counter-example.

Question

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?

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. — Ross Millikan, Apr 15 '15 at 18:22

score 18 · Accepted Answer · answered Apr 15 '15 at 18:13

18

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

answered Apr 15 '15 at 18:13

Adhvaitha

score 4 · Answer 2 · answered Sep 04 '20 at 21:14

4

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

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

answered Sep 04 '20 at 21:14

Null_Space

score 3 · Answer 3 · edited Dec 23 '17 at 13:25

3

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

edited Dec 23 '17 at 13:25

choco_addicted

answered Dec 23 '17 at 10:56

dragon

3 Answers3