Let $A$ be an $n \times n$ matrix. We know that $A^{23} = 0$. What are the eigenvalues of $A$?
I think it's just $0$, but I'm not sure. How should I do this problem?
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Rodrigo de Azevedo
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Max
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Let $\lambda \in \mathbb{R}$ be an eigenvalue of $A$. Let $x \in \mathbb{R}^n$ be a corresponding eigenvector. Then $$Ax = \lambda x \quad \Rightarrow \quad \lambda^{23}x = A^{23}x = 0x = 0.$$ Since $x \neq 0$, $\lambda^{23} = 0$ and thus $\lambda = 0$.
Matias Heikkilä
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4When you don't know how to prove something, it's usually worth a shot to, among the first things, refer to the definition and try if you can get something out of that. – Matias Heikkilä Dec 15 '16 at 09:40
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Yeah, that's exactly what I did! I was just entirely unsure of it, thank you! – Max Dec 15 '16 at 09:44
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For my own curiosity, how are you sure $x$ doesn't equal 0? – Stephan Dec 15 '16 at 13:58
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2@Stephan An eigenvector is nonzero by definition. – M.B. Dec 15 '16 at 15:10
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Needless to say that $n\ge 23$ so that $A$ is a nilpotent matrix. What about the eigenvalues of a nilpotent matrix?
Nitin Uniyal
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the polynomial vanishing on A has only one root , which is $0$ which is also the eigenvalue of $A$ there is also a well known result saying that if A is nipotent ovec an algebric closed field its only eigenvalue is $0$
Seginus
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No need for an algebraically closed field. A nilpotent matrix (of nonzero size) has $0$ as its unique eigenvalue, period. – Marc van Leeuwen Dec 15 '16 at 13:28
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Pedantically speaking, $0$ is an eigenvalue only if $n>0$ (since the $0\times 0$ matrix has no eigenvalues at all). But there are definitely no other eigenvalues than $0$, since if $P$ is any annihilating polynomial (here $P=X^{23}$), any eigenvalues must be a root of$~P$ (since $P[A]$ acts on an eigenspace for$~\lambda$ by the scalar $P[\lambda]$).
Marc van Leeuwen
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