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### Course: Linear algebra>Unit 3

Lesson 1: Orthogonal complements

# Orthogonal complement of the orthogonal complement

Finding that the orthogonal complement of the orthogonal complement of V is V. Created by Sal Khan.

## Want to join the conversation?

• Hi. Shouldn´t the "Areas" around V and comp(V) fill up all R^n ? I mean V union comp(V) = R^n .
• I think the confusion is that when we make a Union, that doesn't include making all linear combinations of the things that we have "united". So let's say that I have R3 and take 3 separate vectors i=[1 0 0] j=[0 1 0] k[0 0 1]. If I take the spans separately span(i), span(j), span(k), they form 3 separate subspaces(3 separate lines). Of course all of them are orthogonal to the other 2. Taking the Union of any 2: Union( span(i), span(j) ) is just the union of 2 lines, but not R2, although inside R2 they are each others orthogonal complements. To span the whole R2, you could write Span( Union( span(i), span(j) ) ), that would not just be 2 lines, but all linear combinations of them, so the whole R2. Of course you would be more efficient by leaving out the initial spans, not to generate the initial lines so Span( Union(i, j) ) would do the same thing. Please note that the inner Union(i, j) are just {i, j}, just a set of 2 vectors, the same with the lines: union of 2 lines does not yet make a new, higher subspace, you have to take their span; the union i and j as 2 vectors do not make a new subspace, you have to take their span to do that.
To sum up the key idea: The Union is just taking a set at putting the thing into it. What you guys were thinking of as Span(Union(V, Vcomp))=R3 and that would be correct. So Sal's blob drawings are correct. I hope that made it clear and wasn't very circular :)
• So basically, this video shows that V = (V^perp)^perp ?
• I have the same question as user onceltuca which was "Hi. Shouldn´t the "Areas" around V and comp(V) fill up all R^n ? I mean V union comp(V) = R^n ." , & because I'm still confused after reading the answers, I'm posting this (=
What I understand so far:
In the last video, it was shown that:
basis for V + basis for Vperp = basis for R^n
but as V and Vperp are both subspaces (closed under addition & scalar multiplication), so:
V = span(basis for V) & Vperp = span(basis for Vperp)
& R^n = span(basis for R^n) => span(basis for V + basis for Vperp)
does this then => R^n = span(basis for V) + span(basis for Vperp)? Because if so then doesn't that show V union Vperp = R^n?
If not, please can someone explain why not. Thanks.
Sorry about the long convoluted question (=
• Let me take the simplest example possible to show you why V union V perp IS NOT R^n :)
Let n = 2, so our R^n is a plane.
V = span ( [1, 0] ), so V is just the x-axis and all the points that are on it.
Vperp = span ( [0, 1] ), so to Vperp belong all points on y-axis.
Now if you take only basis for V you can only "achieve" points on the x-axis, like [0, 0], [1, 0], [123, 0], [-4/5, 0]. You are constrained to move only to right and left.
Similarly with Vperp - only points on y-axis are "achievable", e.g. [0, 0], [0, 1], [0, 22], [0, -57]. Therefore you can only move up and down.
V n Vperp = [0, 0] as it's the only point on the x-axis and y-axis - belongs to V and Vperp.
V union Vperp = only points on axes, on the x-axis from V and on the y-axis from Vperp. You don't have there [2, 3], because it's neither in V or Vperp.
But when you combine span for V with span for Vperp "magic" can start to happen :) Now you can move in all four directions. For example to get to [2, 3] you can take 2 times [1, 0] vector and 3 times [0, 1] vector. Now you have span([1, 0], [0, 1]) which is our whole space R^n, in this case an R^2 plane.
• Are there subspaces for which we cannot find an orthogonal complement?
• As a rather degenerate answer, technically yes because not every vector space is an inner product space in the first place (i.e. not every vector space has a notion of orthogonality).

However, if a vector space has an inner product, then no, every subspace has an orthogonal complement. Such orthogonal complement might not be computable though, in the sense that you might not be able to find an explicit basis for it.

This extreme case would only happen in a vector space that is not finite dimensional.

This answer is a little complicated honestly so if you have questions let me know.

Edit:
If one of the subspaces has an orthogonal complement, then all other subspaces do (the thing I said about computability still applies though), and

If one of the subspaces doesn't have an orthogonal complement, then none of them do, and this would be the situation where the vector space does not have the notion of an inner product.
• I understand that he proves both ways, but, in this particular case, isn't just one case enough?
• I think you need both parts, but I could be wrong.

The first part shows that every element in V perp perp is in V, ie, V perp perp as a blob is at most as big as V. The second part shows that every element in V is in V perp perp. As a blob, V perp perp must enclose at least V. Therefore, it must enclose exactly V. It's a little like the squeeze theorem.
• How do you know to assume v = x + w in the first ?
• x = v + w was proven in one of the previous videos.
• How does Sal know that v, a member of V can be written as a sum of w in V^perp and x in (V^perp)^perp? I watched the last video, but I do not see the reasoning here.
• V⊥ is a subspace, so that means that any vector can be written as the sum of a member of V⊥ and a member of (V⊥)⊥, as shown in the last video. That includes v, a member of V.
• When Khan represented all of R^n to be v + w (where v and w are orthogonal complements), shouldn't all of R^n be av + w (where a is a scalar value)? If v is a vector in set V, v could still be restricted in size length. The set of W won't be, however, since W is the set of ALL vectors orthogonal to the set V, but if V is a set restricted in size, then won't R^n only be defined as av + w?

Thanks!
• I belive there is n restriction to the size length of vector v because a scalar 'c' times v would still be in the subspace of V .