Diagonalizable matrix with eigenvalues $\pm 1$

Question

Let $A \in \mathbb{R}^{n \times n}$ be a diagonalizable matrix with $1$ or $-1$ as its only eigenvalues. Prove that $A^{2} = I_{n}$.

Could someone help me on this one? I have no idea how to start.

Answer 1

You simply have to use the hypothesis that $A$ is diagonalizable : $A=PDP^{-1}$ where $D$ is a diagonal matrix with non-zero coefficients which are $1$ or $-1$ (and $P$ is invertible). From this you can easily compute $A^2$ and find $I_n$.

Answer 2

Since $A$ is a diagonalizable matrix, by definition, it can be written as $PDP^{-1}$ where $D$ is an $n \times n$ diagonal matrix (only non-zero elements are on the diagonal). Moreover, the entries on the diagonal of $D$ can be comprised only of eigenvalues of $A$ (in this case, $1$ or $-1$).

So,

$ \begin{align*} A^2 &= (PDP^{-1})^2 \\&= (PDP^{-1})\cdot(PDP^{-1}) \\&= PD (P^{-1}P)DP^{-1} \\&= PD( I_n)DP^{-1} \\&= PD^2P^{-1}. \end{align*} $

Now, let's think about what possible matrices $D^2$ could be. Since its entries (remember, only on the diagonal) must be either $1$ or $-1$, and $1^2 = (-1)^2 = 1$, $D^2$ will always be $I_n$ (the $n \times n$ identity matrix). So,

$ \begin{align*} PD^2P^{-1} &= PI_nP^{-1} \\&=PP^{-1} \\&= I_n. \end{align*} $

Thus, $A^2 = I_n$.

Answer 3

Another proof: we have

$$ \mathbb R^n=ker(A-E) \oplus ker(A+E).$$

If $x \in \mathbb R^n$, then $x=y+z$ with $Ay=y$ and $Az=-z$. We get

$Ax=Ay+Az=y-z$, therefore

$$A^2x=Ay-Az=y+z=x=I_nx.$$