# Diagonalizable Matrix with Eigenvalue 1, -1

## Problem 37

Suppose that $A$ is a diagonalizable $n\times n$ matrix and has only $1$ and $-1$ as eigenvalues.
Show that $A^2=I_n$, where $I_n$ is the $n\times n$ identity matrix.

(Stanford University Linear Algebra Exam)

See below for a generalized problem.

## Hint.

Diagonalize the matrix $A$ so that the diagonal matrix has $\pm 1$ on diagonal entries.

## Proof.

Since $A$ is diagonalizable, there exists an invertible matrix $P$ such that $P^{-1}AP=D$, where $D$ is a diagonal matrix.
Since $A$ has only $\pm 1$ as eigenvalues, we can choose $P$ so that the diagonal entries of $D$ are either $\pm 1$.

Then we have $A=PDP^{-1}$ and $\require{cancel}$
\begin{align*}
A^2=(PDP^{-1})(PDP^{-1})=(PD\cancel{P}^{-1})(\cancel{P}DP^{-1})=PD^2P^{-1}.
\end{align*}
Note that $D^{2}=I_2$ and thus we have $A^2=I_2$ as required.

### Generalization

As a generalization, consider the following problem.

Problem.
Suppose that $A$ is a diagonalizable $n\times n$ matrix and the eigenvalues of $A$ are $r$-th roots of unity for some positive integer $r$.
Show that $A^r=I_n$, where $I_n$ is the $n\times n$ identity matrix.

The proof of Problem 2 is a straightforward generalization of the proof of Problem 1. So give it a shot.

## Related Problem.

For a similar problem, see the post ↴
Finite order matrix and its trace.

This problem is the converse of the current problem in some sense.

If $L:\R^2 \to \R^3$ is a linear transformation such that \begin{align*} L\left( \begin{bmatrix} 1 \\ 0 \end{bmatrix}\right) =\begin{bmatrix} 1 \\...