# Orthogonal Complements

• Apr 20th 2010, 02:16 PM
MatthewD
Orthogonal Complements
This should be easy, I'm not sure what I'm missing!! Any help would be great...

Let W be a finite dim subspace of an inner product space V. Show $\displaystyle \exists$ a projection, $\displaystyle T$ on $\displaystyle W$ along $\displaystyle W^\perp$ such that the nullspace of T, $\displaystyle N(T)=W^\perp$.
Then show $\displaystyle ||T(x)||\leq ||x|| ~~ \forall x\in V$

What I know:
So far, I know that by a theorem, there are unique vectors $\displaystyle u\in W, z\in W^\perp$ such that $\displaystyle y=u+z$

I know by definition $\displaystyle T:V\rightarrow V$ is a projection on W along $\displaystyle W^\perp$ if for $\displaystyle x=x_1+x_2$ with $\displaystyle x_1\in W, x_2\in W^\perp$, we have $\displaystyle T(x)=x_1$

Finally, I know if x and y are orthogonal vectors, that $\displaystyle ||x+y||^2 =||x||^2+||y||^2$

It seems like it should fall into place so easily... I just don't know how to define my T function so it works for each case and still lands me in the complement of W. Please help!!
• Apr 20th 2010, 05:10 PM
NonCommAlg
Quote:

Originally Posted by MatthewD
This should be easy, I'm not sure what I'm missing!! Any help would be great...

Let W be a finite dim subspace of an inner product space V. Show $\displaystyle \exists$ a projection, $\displaystyle T$ on $\displaystyle W$ along $\displaystyle W^\perp$ such that the nullspace of T, $\displaystyle N(T)=W^\perp$.
Then show $\displaystyle ||T(x)||\leq ||x|| ~~ \forall x\in V$

What I know:
So far, I know that by a theorem, there are unique vectors $\displaystyle u\in W, z\in W^\perp$ such that $\displaystyle y=u+z$

I know by definition $\displaystyle T:V\rightarrow V$ is a projection on W along $\displaystyle W^\perp$ if for $\displaystyle x=x_1+x_2$ with $\displaystyle x_1\in W, x_2\in W^\perp$, we have $\displaystyle T(x)=x_1$

Finally, I know if x and y are orthogonal vectors, that $\displaystyle ||x+y||^2 =||x||^2+||y||^2$

It seems like it should fall into place so easily... I just don't know how to define my T function so it works for each case and still lands me in the complement of W. Please help!!

well, you've already defined $\displaystyle T$. for the second part, if $\displaystyle x=x_1+x_2, \ x_1 \in W, \ x_2 \in W^{\perp},$ then $\displaystyle ||T(x)||^2=||x_1||^2 \leq ||x_1||^2+||x_2||^2=||x_1+x_2||^2=||x||^2.$

(note that since $\displaystyle \langle x_1,x_2 \rangle = \langle x_2,x_1 \rangle = 0,$ we have $\displaystyle ||x_1+x_2||^2=\langle x_1+x_2, x_1+x_2 \rangle=||x_1||^2+||x_2||^2.$)