# How to prove that a diagonalizable matrix with eigenvalues of ±1 is equal to its inverse?

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.

• If $D$ is diagonal with entries $\pm 1$ then $D^{-1} = D$. This is because ${1 \over 1} = 1$ and ${1 \over -1} = -1$. – copper.hat Dec 17 '19 at 22:06
• 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. – Karl Dec 17 '19 at 22:33
• 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. – user729424 Dec 18 '19 at 1:58
• @Karl What? Knowing the determinant is $\\pm1$ certainly does not imply $A$ is a rotation or reflection... – David C. Ullrich Dec 18 '19 at 17:20
• @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. – Karl Dec 18 '19 at 18:22

$$A=PDP^{-1}$$ 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 $$A^{-1}=A$$.

• 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 $-1$. – user729424 Dec 18 '19 at 1:56
• 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. – azif00 Dec 18 '19 at 6:48
• In the post. It says suppose the only eigenvalues of $A$ are $\pm 1$ and... – azif00 Dec 18 '19 at 16:04

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.