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I would like a verification of this proof:

Prove that $A\in M_n(\mathbb{C})$ is a nilpotent matrix if and only if $0$ is its only eigenvalue.

($\Rightarrow$): if $A$ is nilpotent, then there's a index $k \le n$ such that $A^k = 0$. Let $\lambda$ be an eigenvalue of $A$. Since $Ax = \lambda x$, $A^kx = \lambda^k x = 0$, so $\lambda^k x = 0 \Rightarrow \lambda = 0$.

($\Leftarrow$): argue for the sake of contradition that $0$ is the only eigenvalue of $A$ but $A$ is not nilpotent. Then $A \ne 0, A^2 \ne 0, \dots, A^n \ne 0$. Since $Ax = \lambda x$ with $x \ne 0$, $Ax = 0, A^2x = 0, \dots, A^nx = 0$. A contradition.

But something feels off about this, I'm not sure if this is correct.

B.Li
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  • Use triangular form or Jordan form. – fred goodman Nov 09 '17 at 05:45
  • You can clarify your thoughts by being more explicit about what $x$ is. Before you use the symbol $x$, make sure to introduce it with a quantifier, like "there exists $x$ such that", or "there exists $x\neq0$ such that", or "for all $x$ we have", or similarly. – Chris Culter Nov 09 '17 at 05:48
  • @fredgoodman, got it. In the second proof, put $A$ into jordan form and therefore conclude $A$'s jordan form has $0$ along its diagonal and $1$s or $0$s along the super diagonal, which is a nilpotent matrix. Therefore A is similar to a nilpotent matrix. – B.Li Nov 09 '17 at 05:52
  • I'm reluctant to accept that target as a duplicate. The main reason is that this Question is about "if and only if" (and perhaps about proof verification, but that's a minor aspect if the proof provided is the standard one). A minor aspect is that the target proposed includes a statement that the result is already know for matrices and wants some clarification about operators (although the proof is essentially the same idea). – hardmath Dec 08 '18 at 02:06

2 Answers2

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If $A^k = 0$ and $A v= \lambda v$ then $\lambda^k = 0$ and so $\lambda = 0$.

If all eigenvalues are zero, then the characteristic polynomial is $x^n$ and so Cayley Hamilton gives $A^n = 0$ hence $A$ is nilpotent.

copper.hat
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An $n$-by-$n$ matrix has a minimal polynomial $m(X)$: this means that $f(A)=0$ for a complex polynomial if and only if $m\mid f$. Then $f$ is nilpotent if $m(X)=X^k$ for some $k$. Otherwise $m(X)=(X-\lambda)g(X)$ for some nonzero $\lambda$. Then $g(A)\ne0$. Any vector in the image of $g(A)$ will be an eigenvector of $A$ for $\lambda$.

Angina Seng
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