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

• SithsNGiggles
In summary, the proof shows that if A and B are two orthogonal subspaces of an inner product space V, then their intersection is simply the zero vector. The first case of the proof shows that if a and b are the same vector from A and B respectively, then their inner product is zero, implying that a=b=0. The second case shows that if a and b are different vectors from A and B, then both a and b must be zero, leading to a contradiction. Therefore, A intersect B is just the zero vector.
SithsNGiggles

## Homework Statement

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

## 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

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.

LastOneStanding said:
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!

LastOneStanding said:
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.

SithsNGiggles said:
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.

## What does it mean for two subspaces to be orthogonal?

Two subspaces are said to be orthogonal if every vector in one subspace is perpendicular to every vector in the other subspace.

## What is the intersection of two subspaces?

The intersection of two subspaces is the set of all vectors that belong to both subspaces.

## Why is the intersection of two orthogonal subspaces always {0}?

If two subspaces are orthogonal, it means that all vectors in one subspace are perpendicular to all vectors in the other subspace. This means that the only vector that belongs to both subspaces is the zero vector, which has a magnitude of 0 and is perpendicular to all vectors.

## How can I prove that the intersection of two subspaces is {0}?

To prove that the intersection of two subspaces is {0}, you can show that the only vector that belongs to both subspaces is the zero vector. This can be done by showing that the dot product of any two non-zero vectors in the subspaces is equal to 0.

## Can the intersection of two subspaces be a non-zero vector?

No, the intersection of two subspaces can only be a non-zero vector if the two subspaces are not orthogonal. If the subspaces are orthogonal, the only vector that belongs to both is the zero vector.

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