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Let $A \in {M_n}$ is nonsingular and each eigenvalue of $A$ is either $+1$ or $-1$.Why $A$ is similar to ${A^{ - 1}}$?

Roger
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  • Can you show us what you have attempted so far? Or any thoughts? – Lost in a Maze May 16 '15 at 05:41
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    Hint: We have two facts. (1) There exists a basis consisting of generalized eigenvectors of $A.$ (2) There is a natural correspondence (in the special cases where $\lambda=\pm 1$) and the generalized eigenvectors of $A$ and $A^{-1}.$ – matt biesecker May 16 '15 at 05:58
  • Well, you could put $A$ into Jordan normal form $J$, then compute the inverse of $J$ by blocks, which isn't that hard to do by hand, and check. This is just a explicit form of the above comments. – user148177 May 16 '15 at 06:51
  • What's your basic field? $\mathbb R$ or $\mathbb C$? – Censi LI May 16 '15 at 07:56

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Necessarily, $A$ is invertible and the eigenvalues of $A^{-1}$ are also $\pm 1$. For every positive integer $k$, $rank((A+\epsilon I)^k)=rank(A^{-k}(A+\epsilon I)^k)=rank((A^{-1}+\epsilon I)^k)$ where $\epsilon \in\{\pm1\}$; we conclude using the Jordan test of similarity.