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If a matrix A has the property $A=A^{-1}$, are the only possible eigenvalues 1 and -1 ?

How can the matrices with integer values and the property $A=A^{-1}$ be characterized ?

I found out that if A has the property $A=A^{-1}$, then $-A$, $A^T$ and $B^{-1}AB$ for any invertible B also have the property.

I also think that the theorem of caley-hamilton is useful for my problem.

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A simple necessary condition for $A=A^{-1}$ is det(A)=1 or det(A)=-1. – Peter Jun 13 '14 at 9:30
See also my related question about finding all the nxn-matrices with integers in a given range, which are self-inverse. – Peter Jun 13 '14 at 10:40
up vote 6 down vote accepted

Without making explicit use of characteristic polynomials or the relationship between the eigenvalues of $A$ and $A^{-1}$ we can obtain the result just by using the definition of eigenvalue.

Suppose $v$ is an eigenvector for $A$ with corresponding eigenvalue $\lambda$. Then

$$v = Iv = A^{-1}Av = A^2v = A(\lambda v) = \lambda(Av) = \lambda^2v.$$

It follows that $\lambda^2 = 1$ so $\lambda = \pm 1$.

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Very nice answer! – Peter Jun 13 '14 at 10:15

The eigenvalues of $A$ and $A^{-1}$ are reciprocals. Since $A=A^{-1}$ we have that for any eigenvalue $\lambda = 1/\lambda$ which only works for $1$ and $-1$.

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Why can't the eigenvalues be, for instance, $2$ and $\frac12$? I know they can't be, but I don't think your argument shows it. – TonyK Jun 13 '14 at 10:02
@TonyK, I think you'll agree that $2\neq 1/2$. Which part in particular are you objecting to? – Spencer Jun 13 '14 at 10:08
I think TonyK does not see, why $A=A^{-1}$ implies $\lambda=\frac{1}{\lambda}$. What misses is the prove, that $Ax=\lambda x$ implies $A^{-1}x=\frac{1}{\lambda} x$, in other words, that the eigenvector is the same. – Peter Jun 13 '14 at 10:50
For this reason, I prefer Michaels's answer. – Peter Jun 13 '14 at 10:51
@Spencer: Just to spell it out: I was asking why your argument rules out the possibility that $A$ and $A^{-1}$ both have the eigenvalues $2$ and $\frac12$. – TonyK Jun 13 '14 at 17:11

Set $p(t) = t^2 -1$. Since $A=A^{-1}$ it holds $A^2=I$, which implies $p(A)=0$. The eigenvalues of $A$ are roots of $p$, so your conclusion is valid.

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