# Thread: [SOLVED] Hard Orthogonal Complement Proof

1. ## [SOLVED] Hard Orthogonal Complement Proof

At least, I think it is hard. I don't know how to start this at all.

Let $V$ be a vector space equipped with a positive definite inner product. Let $W$ be a subspace of $V$. Let $W'$ be its orthogonal complement. Define a map $\pi$: $V$ -> $V$ as $\pi(v) = v_1$ where $v = v_1 + v_2$ with $v_1$ in $W$ and $v_2$ in $W'$. Show that:

a) The map $\pi$ is welldefined, linear with range $W$ and that $\pi^2 = \pi$. The map $\pi$ is called the projectiion map onto the subspace $W$.

b) Define $v := I - \pi$. Show that $v$ is the projection map onto $W'$ i.e. show that $v$ is a linear map with range $W'$ and that $v^2 = v$.

2. Originally Posted by Zalren
At least, I think it is hard. I don't know how to start this at all.

Let $V$ be a vector space equipped with a positive definite inner product. Let $W$ be a subspace of $V$. Let $W'$ be its orthogonal complement. Define a map $\pi$: $V$ -> $V$ as $\pi(v) = v_1$ where $v = v_1 + v_2$ with $v_1$ in $W$ and $v_2$ in $W'$. Show that:

a) The map $\pi$ is welldefined, linear with range $W$ and that $\pi^2 = \pi$. The map $\pi$ is called the projectiion map onto the subspace $W$.

b) Define $v := I - \pi$. Show that $v$ is the projection map onto $W'$ i.e. show that $v$ is a linear map with range $W'$ and that $v^2 = v$.
very staightforward. note that $V=W \oplus W'.$ let $v_1=w_1+w_1', \ v_2 =w_2 + w_2'$ be in $V,$ where $w_j \in W, \ w_j' \in W'.$

to show $\pi$ is well-define suppose that $v_1=v_2.$ then $w_1=w_2$ and so $\pi(v_1)=w_1=w_2=\pi(v_2).$ so $\pi$ is well-defined.

to show $\pi$ is linear let $c$ be a scalar. then $\pi (cv_1 +v_2)=\pi (cw_1+w_2 + cw_1' + w_2')=cw_1+w_2=c \pi(v_1) + \pi(v_2).$

also $\pi^2(v_1)=\pi(\pi(v_1))=\pi(w_1)=w_1=\pi(v_1)$ and so $\pi^2=\pi.$ the rest of the problem is as easy as this.