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If $A$ is a $ \displaystyle 10 \times 10 $ matrix such that $A^{3} = 0$ but $A^{2} \neq 0$ (so A is nilpotent) then I know that $A$ is not invertible, but why does at least one eigenvalue of $A$ have to be equal to zero? How would one show that all eigenvalues of $A$ are equal to zero?

David
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4 Answers4

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If $v$ is a non-zero eigen vector corresponding to an eigenvalues $\lambda$ we have, by definition, $Av=\lambda v$. Then $A^2v= A( Av)= A(\lambda v)= \lambda^2v$. It easily follows that $\lambda^n$ is an eigenvalue for $A^n$ but the latter is the zero matrix, for which all eigenvalues are zero, hence $\lambda=0$.

P Vanchinathan
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    Wow! I'm impressed. 5 up-votes in 27 seconds! – TravisJ May 04 '15 at 02:39
  • Travis, you are right. But I wrote the words "all eigenvalues are zero" for the zero matrix, thinking it requires no justification. However disregarding nilpotency if $B=p(A)$, with $p$ a polynomial, then it is true that $\lambda$ is an eigenvalue of $A$ iff $f(\lambda)$ is an eigenvalue for $B$. – P Vanchinathan May 04 '15 at 02:43
  • eigenvectors are by definition nonzero. – abel May 04 '15 at 02:45
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    @PVanchinathan Yes, see my comment above. Your latter statement is not true, however: If we take $A := -I$, $p(x) := x^2$, then $B = P(A) = I$ but $1$ is not an eigenvalue of $A$ whereas $p(1) = 1$ is an eigenvalue of $B$. – Travis Willse May 04 '15 at 02:47
  • @Travis: thanks for enlightening me and pointing out my mistake in the comment with an (embarrassingly) simple example. – P Vanchinathan May 04 '15 at 02:54
  • Ah I see thanks, that makes a lot of sense! Does that mean A can be diagonalized? – David May 04 '15 at 02:55
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    @David: think about it. If it can be diagonalized what can be that diagonal matrix? – P Vanchinathan May 04 '15 at 03:02
  • @PVanchinathan Yes, it seems it's a good morning for mistakes if my (now-erased) comment is any indication! :) I think the statement one wants is that if $\lambda_1, \ldots, \lambda_k$ are the eigenvalues of $A$, in particular counting multiplicity, then $\lambda_1^n, \ldots, \lambda_k^n$ are the eigenvalues of $A^n$ counting multiplicity, and (since there are $n$ eigenvalues in the list) this accounts for all of the eigenvalues of $A^n$. – Travis Willse May 04 '15 at 03:08
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Suppose $\lambda$ is an eigenvalue of the nilpotent matrix $A,$ and $u$ its associated eigenvector. Then $$Au = \lambda u, u \neq 0$$ multiplying by $A$ on the right shows $$A^2u = \lambda Au = \lambda^2 u$$ and by induction $$A^n u = \lambda^n u$$

If $A$ is nilpotent, then $A^k = 0$ for some $k>0.$ that implies $\lambda^k = 0\to \lambda = 0.$

mgus
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abel
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Alternatively, do the contrapositive. If $A$ has a non-zero eigenvalue, $\lambda,$ then $A^{k} \neq 0 $ for all $k.$

Proof: there exists $v \neq 0,$ such that $$ Av = \lambda v, $$ so $$ A^{k} v = \lambda^{k} v \neq 0. $$ Done.

Mark Joshi
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Suppose $n$ is the smallest integer for which $A^n=0$. Since $A$ is non-zero $\implies$ $n \gt 1$

let the characteristic polynomial for $A$ be

$$ A^m+c_1A^{m-1}+\cdots + c_m = 0 $$ here $m \le n$. multiplying the equation by $A^{n-1}$ gives $$ c_mA^{n-1} =0 $$ hence $c_m=0$. this procedure can be repeated to show that all coefficients except the first are zero, hence the equation is simply: $$ A^m=0 $$ and we must have $m=n$

since the eigenvalues are roots of the equation: $$ x^m = 0 $$ it follows that they must all be zero

Ketan
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David Holden
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  • I think $n$ should be greater than $1$ (first line of your post). How do you assure $m \leq n$, i.e. how do exclude the case $m \gt n$? – el_tenedor May 04 '15 at 06:17