Main content

### Course: Linear algebra > Unit 3

Lesson 1: Orthogonal complements- Orthogonal complements
- dim(v) + dim(orthogonal complement of v) = n
- Representing vectors in rn using subspace members
- Orthogonal complement of the orthogonal complement
- Orthogonal complement of the nullspace
- Unique rowspace solution to Ax = b
- Rowspace solution to Ax = b example

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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 .(12 votes)
- 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 :)(15 votes)

- So basically, this video shows that V = (V^perp)^perp ?(16 votes)
- 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 (=(5 votes)- 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.(12 votes)

- Are there subspaces for which we cannot find an orthogonal complement?(3 votes)
- 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:

To add one other thing that might be helpful:

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.(3 votes)

- I understand that he proves both ways, but, in this particular case, isn't just one case enough?(2 votes)
- 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.(4 votes)

- How do you know to assume v = x + w in the first ?(3 votes)
**x**=**v**+**w**was proven in one of the previous videos.(2 votes)

- 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.(2 votes)
- 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.(3 votes)

- 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!(2 votes)- 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 .(2 votes)

- Following this logic it looks like that ((V^(perp))^(perp))^(perp) is actually V^(perp). That is, for any odd number of 'perps' you have V^(perp), and for any even number of 'perps' you have V.(2 votes)
- how can any vector in R^3 be represented by two vectors?(2 votes)
- hey emesdg, check out the last lecture -- especially the comments. BrianLHouser explains how this works. someone explained it to me using the walls of my room. if you take the floor and the wall, the two are orthogonal complements [that one point where they meet -- that's the zero vector!]. anyway. if you were to combine the two bases of the wall and the floor, it feels intuitive to me that you could get anywhere in R3.

not a rigorous proof -- but helped me visualize it.(1 vote)

## Video transcript

Let's say I have some subspace
of rn called v. Let me draw it like this. So that it is r n. And I have some subspace of it
we'll call v right here. So that is my subspace v. We know that the orthogonal
complement v is equal to the set of all of the
members of rn. So x is a member of rn. Such that x dot v is equal to 0
for every v that is a member of r subspace. So our orthogonal complement of
our subspace is going to be all of the vectors that
are orthogonal to all of these vectors. And we've seen before that they
only overlap-- there's only one vector that's
a member of both. That's the zero vector. It's right there. Let's take the orthogonal
complement. Let's say it's this set right
here in pink, so that's the orthogonal complement. Fair enough. Now, what if we were to think
about the orthogonal complement of the orthogonal
complement? So, that's the orthogonal
complement in pink. We want the orthogonal
complement of that. So this is going to be all
of the x's-- let's just write it like this. All of the x's that are members
of rn such that x dot w is equal to 0. For every w that is a member
of the orthogonal complement of v. That's what that thing
is saying. So it's all of the vectors in
rn that are orthogonal to everything here. Obviously, all of the things in
v are going to be a member of that because these guys are
orthogonal to everything in these guys. But maybe this is just a subset
of the orthogonal complement of the orthogonal
complement. So maybe this thing in blue
right here looks like this. Maybe it's a slightly
larger set than v. Maybe there are some things,
these things that I'm shading in blue, maybe there are some
vectors that are orthogonal to the orthogonal complement of v
but that are outside of v. We don't know that yet. We don't know whether this
area right here exists. Or maybe the orthogonal
complement of the orthogonal complement. Maybe that takes us back to v. Maybe it's like the transpose
or an inverse function where it just goes back to our
original subspace. Let's see if we can
think about that a little bit better. Let's say that I have some
member of the orthogonal complement of the orthogonal
complement. So let's say I have some vector
x that is a member of the orthogonal complement of
the orthogonal complement. Now, we saw on the last video
that any vector in rn can be represented by a sum of some
vector in a subspace and the subspace's complement. So we know that x can be
represented-- we can say that x can be represented as the
sum of two vectors. One that's in v and one that's
in the orthogonal complement of v. So one, let's call that the
vector that's in v and let's call w the vector that's
in the orthogonal complement of V. Let me write it like this. Where v is a member of the
subspace v and the vector w is a member of the orthogonal
complement of v. Right? So this is some member. It could be some guy out here. It could be some
guy over here. He's a member of the orthogonal
complement of the orthogonal complement. Which is this whole area here. Which v is a subset of, but
we're not sure whether v equals that thing. But we say, look, anything
that's in the orthogonal complement of your orthogonal
complement, is going to be a member of rn. And anything in rn can be
represented as a sum of a vector in v and a vector
in the orthogonal complement of v. So that's all I wrote
right there. Now, what happens if I take the
dot product of x with w? What is this going
to be equal to? This is the orthogonal
complement of the orthogonal complement. Would you take the dot product
of any vector in this with any vector in the orthogonal
complement, which this vector is. Right? It's a member of the orthogonal
complement. You're going to get
0 by definition. These are all of the vectors. This factor is definitely
orthogonal to anything in just v perp right? Anything in v perp perp
is orthogonal to anything in v perp. So, this thing is going
to be equal to 0. But what's another way
of writing x dot w? We could write it like this. This is the same thing
as v plus w dot w. Which is the same thing as
v dot w plus w dot w. Now, what is v dot w? v is a member of our
original subspace. And if you take the dot product
of anything in our original subspace anything in
its orthogonal complement, you're going to get 0. So this term right here is going
to be 0, and you're just going to get this term which
is the same thing as the length of our vector
w squared. Now, that has to equal 0. Remember we just
wrote x dot w. x is a member of the orthogonal
complement of your orthogonal complement. So, you dot that with anything
in the orthogonal complement, that's got to be equal 0. But, if we write it the other
way, if we write it as the sum of v plus w and distribute this
w, we say that's the same thing as the magnitude
of w squared. So the magnitude of w squared
has got to be equal to 0. The magnitude of w squared, or
the length of w squared, has got to be equal to 0. Which tells us that w
is the zero vector. That's the only factor in rn
when you take its length and, especially when you square
it, you get 0. But you could just
take its length. So what does that mean? That means that our original
vector x is equal to v plus w. But w is just equal to 0. So that implies that
our original vector x is equal to v. And v is a member of
our subspace v. Right? So that tells us that x is a
member of our subspace v. So we just showed that if
something is a member of the orthogonal complement of the
orthogonal complement then that same vector has
to be a member of the original subspace. So there is no such thing as
something being in the orthogonal complement of the
orthogonal complement and not being a member of our
original subspace. All of this has to be inside
of this right there. So there is no outside
blue space like that. All of that is our original
subspace if you want to view it that way. Now I just at the beginning of
the video, anything in our subspace is going to be a
member of our orthogonal complement. And then you can kind of reason
that in your head. Let's use the same argument to
just be a little bit more rigorous about it. Right now we say if anything
is in the orthogonal complement of the orthogonal
complement, then it's going to be the original subspace. Let's go the other way. Let's say that something
is in the original subspace just like that. Let me draw another graph
right here because this might be useful. Let me draw rn again. Let me draw all of
rn like that. Now, we have the orthogonal
complement. Let me just draw that first.
So v perp And then you have the orthogonal complement of
the orthogonal complement which could be this
set right here. Right? This is v perp. I haven't even drawn
the subspace v. All I've shown is, I have some
subspace here, which I happen to call v perp. And then I have the orthogonal
complement of that subspace. So this means that anything in
rn can be represented as the sum of a vector that's here
and a vector that's here. So, if I say that w-- let
me do it in purple. If I say the vector w-- let
me write it this way. The vector v can be represented
as the sum of the vector w and the vector x where
w is a member of the orthogonal complement of v or
v perp And x is a member of its orthogonal complement. Notice, all I'm saying, I could
have called this set s. And then this would have been
s and its orthogonal complement. And we learned that anything in
rn could be represented as the sum of something in a
subspace and the subspace's orthogonal complement. So it doesn't matter that v is
somehow related to this. It can be represented as
a sum of a vector here plus a vector there. Fair enough. Now, what happens if
I dot v with w? I'm doing the exact same
argument that I did before. Well, if you take anything
that's a member of our original subspace, and you dot
it with anything in its orthogonal complement, that's
going to give us 0. What else is that going
to be equal to? If we write v in this way, v
dot w is the same thing as this thing dot w. So w plus x dot-- and this is
going to be equal to w dot w plus x dot w. And what's x dot w? x is in the
orthogonal complement of your orthogonal complement. And w is in the orthogonal
complement. So if you take the dot
product, you're going to get 0. They're orthogonal
to each other. So this is just equal to w dot
w or the length of w squared. And since since has to equal
0, we just have a bunch of equals here, that tells us that
once again the vector w has to be equal to 0. So that tells us v is
equal to w plus x. But if w is equal to 0, then v
is going to be equal to x. So we've just shown that if v is
a member of the subspace v, then v is a member of the
orthogonal complement of the orthogonal complement. Right? v is equal to x, which is
a member of the orthogonal complement or the orthogonal
complement. So we've proven it both ways. If you look at the original
statement, we wrote here that if you're a member of the
orthogonal complement of the orthogonal complement,
you're a member of the original subspace. So, we've proven this and,
earlier in the video, we proved that if x is a member of
the orthogonal complement of the orthogonal complement,
then x is a member of our subspace. So these two things
are equivalent. Anything that's in
the subspace is a member of v perp perp. Anything in v perp perp is
a member of our subspace. So, our subspace in v perp
perp are the same set. And of course it overlaps. This equals this. And of course it overlaps with
V perp and its orthogonal complement only at the zero
vector right there.