# If A is a square matrix, and $A^3$=I, how can I prove that eigenvalues of A are either 1 or -1?

If $A$ is a square matrix, and $A^3$=I, how can I prove that eigenvalues of $A$ are either $1$ or $-1$?

Thanks for all your input, I figured it out! :)

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You can't conclude that $A = I$ or $-I$, as it's not true! Take $$A = \begin{pmatrix} -1 && 0 \\ 0 && 1\end{pmatrix}$$ for example – Alex J Best Apr 5 '13 at 1:20
@lawlipop: The question went from $A^2$ to $A^3$? Was it stated wrong the first time? – Amzoti Apr 5 '13 at 1:36
Why did you change the question after you got an answer? – Potato Apr 5 '13 at 1:36

Consider the action of $A$ on an eigenvector $x$. If $Ax = \lambda x$, so $A^2x = \lambda^2 x = x$, since $A^2 = I$.

This implies that $\lambda^2 = 1$, so $\lambda = \pm 1$.

For the new (changed) question, we can use the same reasoning. $A^nx = \lambda^n x = x$, so the possible values of $\lambda$ are the $n^{th}$ roots of unity: for real numbers, this is just $1$ if $n$ is odd and $\pm 1$ if $n$ is even.

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+1: Guess I should have read the post more carefully before putting up my answer. :S – Cameron Buie Apr 5 '13 at 1:28

The polynomial $x^2-1$ is certainly satisfied by $A$ and so the minimal polynomial of $A$ divides it. The roots of the minimal polynomial give the eigenvalues.

Update: For the new question just consider the roots of $x^3-1$. It is guaranteed that the eigenvalues belong to the set of such roots. The same reasoning can be extended to any matrix $A$ with $A^n=I$ for any $n$. In fact if A satisfies any polynomial, the eigenvalues will always belong to the set of its roots.

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This isn't actually correct. One can have $A = \left[ \begin{array}{cc} 1 & -1 \\ 3 & -2 \end{array} \right]$, so that $A^3 = I$, but the characteristic polynomial of $A$ is $\lambda^2 + \lambda + 1$, whose roots are complex.

What is guaranteed is that the eigenvalues will be roots of the polynomial $\lambda^3 - 1,$ one root of which is $\lambda = 1$, and the others of which have the product $\lambda^2 + \lambda + 1$.

This is as Andrew Salmon points out above (in general, the eigenvalues are the complex $n$th roots of unity), but I thought I'd note explicitly that the assertion you asked for proof of is actually false.

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