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Let $A∈M_{n\times n}(\mathbb R)$. Suppose the only eigenvalues of $A$ are ±1 and $A$ is similar to a diagonal matrix. Show that $A^{-1}=A$.

Does this question mean that the eigenvalues are all either +1 or -1, or would it be true with both +1 and -1s.

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  • $\begingroup$ If $D$ is diagonal with entries $\pm 1$ then $D^{-1} = D$. This is because ${1 \over 1} = 1$ and ${1 \over -1} = -1$. $\endgroup$ – copper.hat Dec 17 '19 at 22:06
  • $\begingroup$ I'm not confident but I think that the determinant of $D$ would be $+-1$ which represent rotation or reflection. This makes geometric sense as to why successive transformations result in the identity. I'm not 100% sure but thought I'd add it if not for it to be corrected. $\endgroup$ – Karl Dec 17 '19 at 22:33
  • $\begingroup$ It doesn't matter how many of the diagonal entries are $1$ and how many are $-1$. It will still be the case that $A^{-1}=A$. I show this in my solution below. $\endgroup$ – user729424 Dec 18 '19 at 1:58
  • $\begingroup$ @Karl What? Knowing the determinant is $\\pm1$ certainly does not imply $A$ is a rotation or reflection... $\endgroup$ – David C. Ullrich Dec 18 '19 at 17:20
  • $\begingroup$ @David C.Ullrich. thanks I am mistaken. I was referring to the $D$ matrix of eigenvalues and not $A$ thinking there should be a connection but regardless I am incorrect. $\endgroup$ – Karl Dec 18 '19 at 18:22
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$A=PDP^{-1}$ for some invertible $n\times n$ matrix $P$ and some diagonal matrix $D$ where every entry on the diagonal of $D$ is $1$ or $-1$. Note that $D^2=I$, the identity matrix. It follows that

$$A^2=PDP^{-1}\cdot PDP^{-1}=PD^2P^{-1}=PP^{-1}=I.$$

So $A^2=I$. Hence $A^{-1}=A$.

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  • $\begingroup$ The question also asked if we need to assume that the diagonal entries are all $1$'s, or all $-1$'s. Note that we did not make any assumptions about how many of the diagonal entries were $1$ and how many were $-1$. So $A=A^{-1}$ holds regardless of how many of the diagonal entries were $1$ and how many were $-1$. $\endgroup$ – user729424 Dec 18 '19 at 1:56
  • $\begingroup$ If we have a zero diagonal entry, then $0$ is an eigenvalue of $A$, but $\pm 1$ are the only eigenvalues of $A$. Also, if $0$ is an eigenvalue of $A$, this cannot be invertible, meaning that this question it would makes no sense. $\endgroup$ – azif00 Dec 18 '19 at 6:48
  • $\begingroup$ In the post. It says suppose the only eigenvalues of $A$ are $\pm 1$ and... $\endgroup$ – azif00 Dec 18 '19 at 16:04
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We have $C^{-1}AC=D$, where $D$ is diagonal and $C$ invertible. As $A$ only has $\pm 1$ as eigenvalues, $D$'s diagonal consists only of $\pm 1$. The inverse of an invertible diagonal matrix $$M=\left(\begin{array}{cccc}{a_1} & {0} & {\ldots} & {0} \\ {0} & {a_2} & {\ldots} & {0} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {0} & {0} & {\ldots} & {a_n}\end{array}\right)$$ equals $$M^{-1}=\left(\begin{array}{cccc}{\frac{1}{a_1}} & {0} & {\ldots} & {0} \\ {0} & {\frac{1}{a_2}} & {\ldots} & {0} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {0} & {0} & {\ldots} & {\frac{1}{a_n}}\end{array}\right)$$

Hence, the inverse of $D$ must be itself (since $\frac{1}{\pm 1}=\pm 1$). Thus, taking the inverse of the equation $C^{-1}AC=D$ gives: $C^{-1}A^{-1}C=D$. Therefore, we have $C^{-1}AC=C^{-1}A^{-1}C$. Multiplying this by $C$ on the left and $C^{-1}$ on the right gives $A=A^{-1}$.

To answer your other question: $D$ can also have both $1$ and $-1$ on its diagonal.

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