Prove the intersection of two orthogonal subspaces is {0}

[SithsNGiggles]

Homework Statement

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

Homework Equations

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

 

[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.

 

[HallsofIvy]

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.

Very well said!

 

[SithsNGiggles]

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.

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.

 

[Dick]

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.

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.