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$S$ is subspace of $\mathbb{R}^n$ and $P_S$ is matrix of orthogonal projection onto $S$. $$M = I-2P_S$$

We know that $M$ is orthogonal and can have eigenvalues can be either $1$ or $-1$. I need to show that $M$ is diagonalizable.

I know that for a matrix to diagonalizable, it needs to take form $M = PDP^{-1}$ where $P$ is eigenvector matrix and $D$ is the diagonal matrix of eigenvalues. Also we know that if $M$ is diagonalizable, then $$\operatorname{tr}(M) = \sum_{i=1}^n \lambda_i$$

But I'm not really sure how to approach this problem. We can't use determinants. Can anyone help?

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    $\begingroup$ every orthogonal matrix is diagonalizable, isn't it ? $\endgroup$ Oct 7, 2019 at 22:06
  • $\begingroup$ But I have to prove it.... $\endgroup$
    – donnyan
    Oct 7, 2019 at 22:29

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You should think about it like this: $M$ is diagonalizable if there is a basis of $\mathbf{R}^n$ consisting of eigenvalues of $M$.

For example, for $P_S$, every vector in $S$ or in $S^\perp$ is an eigenvector since $P_S(x) = x$ if $x \in S$ and $P_S(x) = 0$ if $x \in S^\perp$. Taking bases for $S$ and $S^\perp$ gives you an eigenbasis for $\mathbf{R}^n$.

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