An $ n × n $ matrix $A$ satisfies $A^2 = 0 $. Can the rank of $A$ be $n$?

Question

My opinion is that $A^2 = 0 $ then $A$ is also $0$. Is it correct?

No, $A^2=0$ does not imply $A=0$, consider $$A=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}$$ — Dietrich Burde, Jun 24 '20 at 14:12

score 1 · Answer 1 · answered Jun 24 '20 at 14:17

As pointed out in the comments, $A^2=0$ does not necessarily imply that $A=0$.

Regarding your first question, if $A^2=0$ we have $0=\det(A^2)=(\det A)^2$. Since the determinant of $A$ vanishes, $A$ cannot have rank $n$ by definition. And actually the rank of $A$ must be $\leq \frac n2$ (see the comments again).

score 1 · Answer 2 · answered Jun 24 '20 at 14:17

1

Hint:

$A^2=0$ implies the minimal polynomial of $A$ is a divisor of $X^2$.

answered Jun 24 '20 at 14:17

Bernard

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An $ n × n $ matrix $A$ satisfies $A^2 = 0 $. Can the rank of $A$ be $n$?

2 Answers2