$M = \text{Identity matrix} (I) - 2P_S$. Show that its eigenvalue must be $1$ or $-1$, and $M$ is diagonalizable. $S$ a subspace of $\Bbb R^n$, $P_S$ is the matrix of orthogonal projection onto $S$. $M = \text{Identity matrix}(I) - 2P_S$. Show that if $\lambda \in \Bbb R$ is an eigenvalue of $M$, then $\lambda = \pm 1$ 
and $M$ is diagonalizable.
 A: Since $P_S$ is an orthogonal projection, by definition
$P_S^T = P_S = P_S^2; \tag 1$ 
then 
$M = I - 2P_S \tag 2$
is a symmetric matrix, for
$M^T = I^T - 2P_S^T = I - 2P_S = M;  \tag 3$
thus $M$ may be diagonalized by some orthogonal matrix $O$:
$D = OMO^T = OMO^{-1}, \tag 4$
where
$O^TO = OO^T = I, \tag 5$
so that
$O^{-1} = O^T; \tag 6$
the diagonal entries of $D$ are the eigenvalues of $M$; furthermore,
$D^2 = OMO^T OMO^T = OMIMO^T$
$= OM^2O^T = OIO^T = OO^T = I; \tag 7$
therefore the diagonal entries $\mu_j$ of $D$, which are the eigenvalues of $M$, all satisfy
$\mu_j^2 = 1; \tag 8$ 
hence every
$\mu_j = \pm 1. \tag 9$
Note Added in Edit, Monday 7 October 2019 12:36 PM PST:  It has been called to my attention in the comment of to this answer by Qazwsx199 that I have neglected to show why 
$M^2 = I, \tag{10}$
where $M$ is as in (2); observe, using (1):
$M^2 = (I - 2P_S)^2$
$= I - 4P_S + 4P_S^2 = I - 4P_S + 4P_S = I. \tag{11}$
End of Note.
A: Since a projection matrix $P$ satisfies $P^2=P$ by definition, all its eigenvalues must satisfy $\lambda\in\{0,1\}$. Thus $-2P$ has eigenvalues in $\{0,-2\}$. Finally, adding the identity matrix to $-2P$ (or any matrix in general) has the effect of shifting the eigenvalues up by $1$, so $I-2P$ has eigenvalues in $\{1,-1\}$.
