# Orthoganlity question

1. Apr 17, 2010

### talolard

1. The problem statement, all variables and given/known data
Hello,
Please pardon me if my terms are off, i'm studying in hebrew and might lacka few of the english words.
Anyway this isnt homewokr but just curiosity.
We learnt in class that if V is a vector space and U is a subspace of V then $$U\oplus U^{\bot}=V$$
But then it seems to me that this implies that every vector that is in V but not in U is orthogonal to every vector in U. i.e.

This just strikes me as odd and counterintuitive. Is it correct or am I issing something.
Thanks
Tal

3. The attempt at a solution

Last edited: Apr 17, 2010
2. Apr 17, 2010

### jeppetrost

I don't find it counterintuitive.. for instance, let the subspace U be some plane in a 3d-V. Then all you need to get V is U and its normal.
And it doesn't say that every vector in V but not in U is orthogonal to U, but rather that with U and the vectors orthogonal to U spans V.
I think the U= some plane in 3d-V-space is the best and most intuitive example i can think of.

3. Apr 17, 2010

### HallsofIvy

No, that's not what it is saying. $U\oplus U^{\bot}$ is NOT a union of sets. In particular, it does NOT mean that every vector in V is in one or the other of those. It means, rather, that every vector in V is a unique sum of a vector in U and a vector in $U^{\perp}$.

Suppose V is the xy-plane and U is the x-axis. Then $U^{\perp}$ is the y- axis. Every vector in V, $a\vec{i}+ b\vec{j}$ is a unique sum of a vector in U and a vector in V- here, just $a\vec{i}$ and $b\vec{j}$, respectively.

A slightly harder example: With V as before, let U= {(x,y)|y= x}, the line y= x. Then $U^{\perp}$ is {(x,y)| y= -x}. Given a vector $p\vec{i}+ q\vec{j}$ how would we write it as a sum of vectors in U and $U^{\perp}$? Well, any vector in U is of the form $a\vec{i}+ a\vec{j}$ for some number a and every vector in $U^{\perp}$ is of the form $b\vec{i}- b\vec{j}$ for some number b so we would need to find a and b such that $(a\vec{i}+ a\vec{j})+ (b\vec{i}- b\vec{j}= p\vec{i}+ q\vec{j}$.

That is, $(a+ b)\vec{i}+ (a- b)\vec{j}= p\vec{i}+ q\vec{j}$ so we must have a+ b= p and a- b= q. Adding the two equations, 2a= p+ q so a= (p+ q)/2. Subtracting the two equations, 2b= p- q so b= (p- q)/2.

That is, any vector in R2 can be written, in a unique way, as the sum of a vector in U and a vector in $U^{\perp}$.

4. Apr 17, 2010

### talolard

Thanks for clearing that up.