Show that orthogonal projection is diagonalizable Let $S \subset V$ be a subspace of finite dimensional vector space $V$. Show that the orthogonal projection $P_S:V \to S$ is diagonalizable. I don't know how to start this question. Can someone help me start with a hint?
 A: Let $u_1,\ldots,u_d$ be an orthonormal basis of $V$ so that the first $k$ basis vectors lie in the subspace $S$.  Then $P_S(u_j)=u_j$ for $j\le k$.  Also, $P_S(u_j) = 0\cdot u_j$ for $j > k$.
More details: A linear transformation $T:V\rightarrow V$ is diagonalizable if there is a basis of $V$ consisting of eigenvectors of the transformation.  An orthogonal projection $P_S$ acts as the identity on the subspace $S$ and maps any element of $S^\perp$ (the vectors orthogonal to $S$) to $0$.  $P_S$ is defined by $P_S^2=P_S$ and $P_S^*=P_S$.  The image of the orthogonal projection $P_S$ will be $S\subset V$ and the kernel will be $S^{\perp}$.
Because we know that $\dim(S)+\dim(S^{\perp}) = \dim(V)$, and we know that $P_{S}$ acts as the identity on $S$ and acts as $0$ on $S^{\perp}$, we can diagonalize $P_{S}$ by any basis $u_1,\ldots, u_d$ with the first $\dim(S)$ elements in $S$ and the last $\dim(S^{\perp})$ elements in $S^{\perp}$.  Such a basis always exists, for instance by extending a basis of $S$ to a basis of $V$, then applying the Gram Schmidt process.
Note that $P_S$ is actually unitarily/orthogonally diagonalizable, since we can diagonalize it with an orthogonal basis.
A: The orthogonality condition is redundant. Every projection on $V$, be it orthogonal or not, is diagonalisable.
Apporach 1: the minimal polynomial of $P$ must divide $x(x-1)$.
Approach 2: as $P$ is a projection, we have $V=PV\oplus\ker(P)$ and you can construct an eigenbasis of $P$.
