# Prove the intersection of two orthogonal subspaces is {0}

1. Jan 23, 2013

### SithsNGiggles

1. The problem statement, all variables and given/known data

Let $A$ and $B$ be two orthogonal subspaces of an inner product space $V$. Prove that $A\cap B= \{ 0\}$.

2. Relevant equations

3. The attempt at a solution

I broke down my proof into two cases:

Let $a\in A, b\in B$.

Case 1: Suppose $a=b$. Then $\left\langle a,b \right\rangle = \left\langle a,a \right\rangle = 0$, which implies $a=b=0$. Thus $0 \in A\cap B$.

Case 2: Suppose $a \not= b$. Then $b \not\in A \wedge a \not\in B$, so $a,b \not\in A \cap B$. This implies $A \cap B = \emptyset$.

Therefore $A\cap B = \{ 0\}$.

My main question is if my second case works. It took me quite some time to convince myself that it was, but now I'm doubting myself again. Thanks

2. Jan 23, 2013

### VantagePoint72

No, it doesn't work—at least, not without more explanation. Try a proof by contradiction for this step: assume $b \in A$, see what that implies. Then do the same for $a \in B$. Together, these conclusions will contradict $a \not= b$.

Generally speaking, if you need to think very hard to convince yourself that a step in your proof is correct, then your proof isn't complete. All the convincing should be done on the page.

3. Jan 23, 2013

### HallsofIvy

Very well said!

4. Jan 23, 2013

### SithsNGiggles

Yeah, I considered doing a proof by contradiction, but the way I did it involved (essentially) the same "logic" I used in the first case of what I posted earlier.

Here goes:
...
Case 2: Suppose $a\not =b$.

Let $b \in A$. Then, $b \bot b \Rightarrow \left\langle b,b \right\rangle = 0 \Rightarrow b = 0$.

Let $a \in B$. Then, $a = 0$ by similar reasoning.

Thus $a = b$, contradicting the assumption. Therefore $A \cap B = \{ 0 \}$

Thanks for the tip.

I also have a slightly irrelevant question. Since $A$ and $B$ are orthogonal sets, does that necessarily mean that $A = B^{\bot}$ (orthogonal complement)? Just wondering. The definitions my professor provided weren't quite clear on this.

5. Jan 23, 2013

### Dick

You really don't need to split into cases or choose two vectors. Choose one vector a that is in AnB. Since a is in A and a is in B a must be perpendicular to a. So a=0 using your argument. And no, in three dimensional space the x-axis is perpendicular to the y-axis, but the orthogonal complement of the x-axis is the y-z plane. A is a subset of the orthogonal complement of B, but it's not necessarily equal to it.