Prove that $P_{M^{\perp }}=P_M^{\perp }$.

Question

Result(i) Let $H$ be a hilbert space. $M$ be a closed subspace. Let $v\in H\setminus M$ and define $d=\inf\{||v-w|| |w\in M\}$. Then there exists an unique point $w_o \in M$ such that $||v-w_o||=d.$

Let $P_M^{\perp }$ and $P_M$ are operators defined using the Orthogonal Projection on to a closed subspace $M$ given by

$P_Mv=\Big\{ \begin{array}{lr} v, & \text{if } v\in M\\ w_0, & \text{if } v\in H\setminus M \end{array}$

$P_M^{\perp }v=\Big\{ \begin{array}{lr} 0, & \text{if } v\in M\\ v-w_0, & \text{if } v\in H\setminus M \end{array}.$

Prove that $P_{M^{\perp }}=P_M^{\perp }$.

My attempt:-

By definition, we can write $P_{M^{\perp }}v=\Big\{ \begin{array}{lr} v, & \text{if } v\in M^{\perp }\\ w^\star, & \text{if } v\in H\setminus M^{\perp } \end{array}$ where $w^\star$ is a unique point in $M^{\perp }$(Using Result(i)). $P_M^{\perp }v=\Big\{ \begin{array}{lr} 0, & \text{if } v\in M\\ v-w_0, & \text{if } v\in H\setminus M \end{array}.$ How these two operators are same?

Answer 1

A basic theorem in Hilbert space theory says $H=M\oplus M^{\perp}$. From this it follows that $P_{M^{\perp}}=I-P_M$. Now it is clear that $P_{M^{\perp}}=P_M^{\perp}$.