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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.

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    $\begingroup$ If $A$ is diagonalizable then $A = Q \Lambda Q^{-1}$, hence $A^2 = Q \Lambda Q^{-1} Q \Lambda Q^{-1} = Q \Lambda^2 Q^{-1}$. Since your $\Lambda$ is diagonal with $\pm 1$ on its diagonal, $\Lambda^2 = I_n$, hence $A^2 = Q Q^{-1} = I_n$. $\endgroup$ – Florian Jan 10 '17 at 12:06
  • $\begingroup$ Right, I get it, thank you! $\endgroup$ – simp Jan 10 '17 at 12:11
  • $\begingroup$ Interesting. This question is a partial converse of Is the matrix $A$ diagonalizable if $A^2=I$ $\endgroup$ – BCLC Oct 24 '18 at 5:51
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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 $I_n$.

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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, $A^2 = I_n$.

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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

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

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