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This question already has an answer here:

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?

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marked as duplicate by rschwieb, C. Falcon, Ken Duna, Alex M., user223391 Dec 18 '16 at 18:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Conceptually, it's because the matrix is collapsing that dimension down irreversibly. $\endgroup$ – Mehrdad May 4 '15 at 5:57
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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 $\lambda=0$.

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    $\begingroup$ Wow! I'm impressed. 5 up-votes in 27 seconds! $\endgroup$ – TravisJ May 4 '15 at 2:39
  • $\begingroup$ 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 $B$. $\endgroup$ – P Vanchinathan May 4 '15 at 2:43
  • $\begingroup$ eigenvectors are by definition nonzero. $\endgroup$ – abel May 4 '15 at 2:45
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    $\begingroup$ @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 $B$. $\endgroup$ – Travis May 4 '15 at 2:47
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    $\begingroup$ @David: think about it. If it can be diagonalized what can be that diagonal matrix? $\endgroup$ – P Vanchinathan May 4 '15 at 3:02
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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 $\lambda^k = 0\to \lambda = 0.$

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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.

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suppose $n$ is the smallest integer for which $A^n=0$. since $A$ is non-zero $n \gt 0$

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

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  • $\begingroup$ 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 $m \gt n$? $\endgroup$ – el_tenedor May 4 '15 at 6:17

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