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I can see that if $c$ is an eigenvalue of a matrix $A$, then $c^k$ will be an eigenvalue for the matrix $A^k$, but I'm curious about the reverse case. I checked for a simple rotation matrix counterclockwise by $180$ with single eigenvalue $-1$ that the rotation matrix counterclockwise by $\pi/2$ has eigenvalues $\pm i$.

I don't expect it to be the case, though, that if $c$ is an eigenvalue for $A^k$ ($k > 1$), that $c^{1/k}$ are eigenvalues of $A$. The example I think of to contradict this is taking a $2 \times 2$ rotation matrix by $\pi / 3$, $A$, whose cube will have eigenvalue $-1$, which has three cube roots, which exceeds the possible number of eigenvalues of $A$.

Is there, however, a more intuitive way to see this, or a weaker result which holds relating the eigenvalues in the reverse direction?

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  • $\begingroup$ "has three cube roots, which exceeds the possible number of eigenvalues of A" I think if A is positive definite (ie. all eigenvalues are positive), you can just pick the positive n'th root. $\endgroup$ – Navin Oct 3 '12 at 4:46
  • $\begingroup$ I think it's similar to ask if $x^k=a$, can you conclude $x=a^{1/k}$ $\endgroup$ – chaohuang Oct 3 '12 at 4:53
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We prove that if $c$ is an eigenvalue of $A^k$, then at least one of the complex $k$th roots of $c$ is an eigenvalue of $A$.

Suppose $(A^k-cI)v=0$ with $v\ne0$. We have $$A^k-cI=\prod(A-bI)$$ where $b$ runs over the complex $k$th roots of $c$. So $\prod(A-bI)v=0$. So $(A-bI)w=0$ for some $b$ and some $w\ne0$. So at least one of the complex $k$th roots of $c$ is an eigenvalue of $A$.

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  • $\begingroup$ take $k = 3$, how do you see that: $\prod^3 (A - b_iI) = A^3 - (b_1+b_2+b_3)A^2 + (b_1b_2 + b_2b_3 + b_1b_3)A - cI = A^3 - cI$? $\endgroup$ – Palace Chan Oct 4 '12 at 23:11
  • $\begingroup$ The same way I would see it if $A$ were a number and $I$ the number 1. Matrix multiplication is distributive over addition, and associative, and powers of $A$ commute with each other and with $I$. $\endgroup$ – Gerry Myerson Oct 4 '12 at 23:26

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