# $M$ is a $n\times n$ orthogonal matrix and has eigenvalues can have either $1$ or $-1$. Show that $M$ is diagonalizable

$$S$$ 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?

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

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 $$\mathbf{R}^n$$.