$A \in \mathbb{R}^{n\times n}$, with $A^2 = 1$ and $A\ne\pm1$
Show that the only eigenvalues of $A$ are $1$ and $-1$.
$A \in \mathbb{R}^{n\times n}$, with $A^2 = 1$ and $A\ne\pm1$
Show that the only eigenvalues of $A$ are $1$ and $-1$.
Hint: if $\lambda$ is an eigenvalue of $A$, what can you say about eigenvalues of $A^2$?
We have that:
$$\mathbf{A}^{2} = \mathbf{I} \implies \mathbf{A} = \mathbf{A}^{-1}$$
But we know that if the eigenvalues of $\mathbf{A}$ are $\lambda_{1},\dots,\lambda_{n}$, then the eigenvalues of $\mathbf{A}^{-1}$ are $\lambda_{1}^{-1},\dots,\lambda_{n}^{-1}$, but if:
$$\lambda_{i}=\lambda_{i}^{-1} \qquad \forall i \in \{1,\dots,n\}$$
Then:
$$\lambda_{i}=\pm 1 \qquad \forall i \in \{1,\dots,n\}$$
$$ Av = \lambda v \Rightarrow A^2v = \lambda Av = v $$
So $$ \lambda Av = v \Rightarrow \lambda \lambda v = \lambda ^2v = v $$
This shows $\lambda = 1$ or $\lambda = -1$.