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Let $A$ be a $3\times 3$ non-zero real matrix and satisfies $A^3 + A = 0$. Then prove that $rank (A) = 2$.

As $A$ is satisfying $A^3 + A = 0$, so $0$ is an eigen value of $A$.So $\operatorname{rank} (A) < 3$. So $\operatorname{rank} (A) = 0,1,\text{or}\, 2$. Clearly $\operatorname{rank}(A) = 0$ is not possible as $A$ is a non-zero matrix.How to show that $\operatorname{rank} (A) = 1$ is also not possible?

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    $\begingroup$ Don't delete and repost questions when you don't like the outcome. $\endgroup$ – Zev Chonoles May 15 '13 at 7:25
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Well, the other two possible eigenvalues are $i$ and $-i$, as they are roots of the polynomial $$x^3+x=0\,.$$ Rank $1$ is possible, if you allow complex matrices. For example $$ A:=\left(\begin{matrix}i&0&0\\0&0&0\\0&0&0\end{matrix}\right) $$ has rank $1$ and satisfies $A^3+A=0$. Otherwise, if you ask for real values matrices, then the set of eigenvalues will be conjugated, i. e. there will be both eigenvalues $i$ and $-i$, therefore the rank must be $2$ or $0$.

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    $\begingroup$ Rank 3 is also possible if you calculate in $\mathbb{F}_2$. Then you can chose $A=E$ and have $A^3+A=E+E=0$ $\endgroup$ – Dominic Michaelis May 15 '13 at 7:17
  • $\begingroup$ True. Good point. $\endgroup$ – Ralph Tandetzky May 15 '13 at 7:25

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