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Let $\alpha$ and $\beta$ be nonzero elements in $\mathbb{F}(n,1)$. Then $A = \alpha\beta^T \in \mathbb{F}(n,n)$ Prove that a necessary and sufficient condition for $A$ to be nilpotent is $\alpha^T\beta=0$. If so, what is the degree of $A$.

Solution: Suppose $A$ is nilpotent, then we know that all of its eigenvalues must be $0$. Using this idea, I think I should be able to relate it to the conclusion that $\alpha^T\beta=0$. Moreover I know that $\text{rank}(\alpha\beta^T) \leq \min({\text{rank}(\alpha),\text{rank}(\beta})) \Rightarrow \text{rank}{(\alpha\beta^T)} = 1$

Not sure how to begin with the other direction.

Any help is appreciated!

Please provide directions only (if possible), no solutions!

Git Gud
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AAP
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1 Answers1

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Hint: Try using induction to show that $A^{n}=(\beta^{T}\alpha)^{n-1}A$, and notice that $\beta^{T}\alpha=\alpha^{T}\beta$.

user84413
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