Show that R (reflection operator) is diagonalizable.

Question

Let be $V$ a real vector space (finite dimension) with standard scalar product, $S$ an $V-$subspace and $R: V\rightarrow V$ the reflection operator on $S$. Show that $R$ is a diagonalizable operator.

I can think of how to do it on the plane - using eigenvalues and eigenvectors. But this question, I honestly do not know how to begin.

Answer 1

HINT: Decompose $V$ as $S\oplus S^{\perp}$. What do $R\vert_S$ and $R\vert_{S^{\perp}}$ look like?

Answer 2

Since A is reflection operator so it will satisfy the equation $A^2=I$. So it can be easily seen that the minimal polynomial of A will annihilate the polynomial $(x^2-1)$ so minimal polynomial of A is either (x-1) or (x+1) or (x-1)(x+1). In each case minimal polynomial have only linear factors. Therefore the operator A is diagnolisable.