Lin Alg: Representing vectors in Rn using subspace members

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Lin Alg: Representing vectors in Rn using subspace members : Showing that any member of Rn can be represented as a unique sum of a vector in subspace V and a vector in the orthogonal complement of V.

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Let's say I have some subspace V, that is a subset of Rn.
And let's say that we also have its orthogonal
complement, we write that as V perp.
That'll also be a subset of Rn.
A couple of videos ago, it might have even been the last
video if I remember properly, we learned that the dimension
of V, plus the dimension of the orthogonal complement of
V, which is also another subspace, is going
to be equal to n.
Remember dimension is just the number of linearly independent
vectors you need to have a basis for V.
And the dimension here is the number of linearly independent
vectors you need to have a basis for the orthogonal
complement of V.
Now given this, let's see if we can come up with some other
interesting ways in which these two subspaces relate to
each other.
Or how they might relate to all of the vectors in Rn.
So the first question is, do these two subspaces have
anything in common?
Are there any vectors that are in common with the two.
And to test whether there is, let's just assume there is,
and see what the properties of that vector would have to be.
Let's assume right here that I have some vector x that is a
member of my subspace V.
Let's also assume that x is a member of the orthogonal
complement of V.
Now what does this second statement mean?
Membership in the orthogonal complement means that x dot v,
for any v that is a member of our subspace, is going to be
equal to 0.
Let me write it this way actually.
x dot v is equal to 0 for any v that is a
member of our subspace.
That's what it means to be a member of V's orthogonal
complement.
Now we assume that x is also a member of V.
So that means that we can stick x here as well.
For any member of V.
x is also a member of V.
So that implies that x dot x is equal to 0.
Another way to write that is that the length of x squared
is equal to 0.
Or the length of x is equal to 0.
And that's only true for one vector.
You can even try it out with the different
constituents of x.
The only vector that that's true for is the 0 vector.
So x has to be equal to the 0 vector.
That's the only vector in Rn that when you dot it with
itself you get 0, or whose the square of its
length is equal to 0.
And we've shown that many, many, many videos ago.
What this tells us is at that the intersection between V and
V complement-- this kind of upside down U just means
intersection, it just means where do these two sets
overlap-- the only place that these overlap is with a subset
of the 0 vector.
So if I were to draw all of Rn like this.
Let's say that this is Rn.
And let's say I draw the subspace V.
And let's say I draw the orthogonal complement to V.
It's all of these vectors right here.
This is the orthogonal complement to V right there.
So this is V perp.
These are all of the vectors that when I dot it with any
vector here, I'm going to get equal to 0.
So this is V perp.
The intersection, their overlap, the only vector that
is a member of both is the 0 vector.
That's their only intersection.
So that's fair enough.
The only vector that's a member of a subspace and its
orthogonal complement is the 0 vector.
Nothing too profound there.
Let's see if we can come up with some other interesting
relations between the subspace and its orthogonal complement.
Maybe some arbitrary vectors in Rn.
So let's just write down-- well we know that the
dimension of our subspace V is equal to k.
If its equal to k, we know that its dimension plus its
orthogonal complement has to be equal to n, because we're
dealing in Rn.
And we also know that the orthogonal complement of V is
a subset of Rn, I drew it right here.
The dimension of V is equal to k.
That's a k right there.
And what's the dimension of the orthogonal complement of V
going to be?
Well, when you add them together-- I wrote that up
here-- they have to equal n.
So this guy's going to have to be n minus k.
If you have k here.
This guy's dimension is k, this guy's dimension right
here if, it's n minus k, that when you add these two up, k
plus n minus k is going to be equal to n.
So this guy will have a dimension of n minus k.
Now what does dimension mean?
It means that that's the number of linearly independent
vectors you need to form a basis.
I have k vectors as a basis for V.
I have v1, v2, all the way to vk.
And this is a basis for V, which just means they're all
linearly independent.
And they span V.
Any member of V right here can be represented as a linear
combination of these vectors.
Now the dimension of the orthogonal complement of
V is n minus k.
So we could have n minus k vectors.
Let's call them w1, w2, all the way to wn minus k.
We have n minus k of these characters.
And this set is a basis for the orthogonal
complement of V.
So any vector in here can be represented as a linear
combination of these guys right here.
And all of these guys are linearly independent.
So you don't have any redundant vectors there.
Now let's explore.
And I'll tell you where I'm trying to go.
I'm trying to see if I combine these two sets, whether I get
a basis for all of Rn.
That's what I'm trying to understand.
Let's just say that for some constants c1 times v1 plus c2
times v2 plus all the way to ck times vk plus-- for the
constants on these guys I'll use d-- plus d1 times w1 plus
d2 times w2, all the way to plus dn minus k times the
basis vector wn minus k.
Let's say that I'm curious about setting this sum equal
to 0, equaling the 0 vector for some scalers.
The scalers are these c's and these d's.
And we know that there's at least one solution set of
scalers for which this is true.
We could multiply all of these constants-- c1, c2, ck, d1,
d2, all the way to dn minus k.
They could all be 0.
Or there might be more than one solution.
In fact, if the only solution is that all of these constants
have to be equal to 0, then we know that all of these vectors
are linearly independent with respect to each other.
And if they're all linearly independent with respect to
each other, then we know that they can be a basis for Rn.
But we don't know that yet.
We don't know that the only solution to this is all of the
constants being equal to 0.
So let's see if we can experiment with
this a little bit.
If we take this equation, which I just wrote down, we
know that one solution is all of the constants, the c's and
d's equalling to 0, but we don't know that
that's the only one.
Let's just subtract all of the w vectors from both sides of
this equation.
So what are we going to get?
We're going to get c1, v1 plus c2, v2 all the
way to plus ck, vk.
And we're going to subtract this from both
sides of the equation.
It's going to be equal to the 0 vector.
Which is really just 0, I don't even have to write it
down, but maybe I'll write it down there just so you
understand.
I'm just taking this equation, I'm subtracting these guys
from both sides.
So 0 vector minus d1, w1 plus d2, w2 plus all the way to dn
minus k, wn minus k.
All I did is I subtract these terms right here from both
sides of this equation.
I don't even have to write this is 0 here,
that's a bit redundant.
So what I have here is I have some combination of the basis
vector of V.
So if I look at this, this is some linear combination of the
basis vectors in V.
If I call this a vector-- let me call this some vector x.
Let's say x is equal to c1, v1 plus c2, v2 all
the way to ck, vk.
We know that it's a linear combination of our basis
vectors of V, so x is a member of V.
By definition, any linear combination of the basis
vectors for a subspace is going to be a
member of that subspace.
Well, similarly, what do we have on the right-hand side of
this equation?
On the right-hand side of this equation, I have some linear
combination of the basis vectors of V complement You
could put just put a minus all along that, but that won't
change the fact that this is some linear combination of V
complement's basis vectors.
So this vector over here is going to be a member of-- we
could also call this x.
So x is equal to this, but its also going to be equal to
this, and since it can be represented as a linear
combination of the orthogonal complement of V's basis
vector, or V perp's basis vector, we know that this also
has to be a member of V perp.
Let me just review this, because it can be
a little bit confusing.
I just set up this equation right here.
We know that there's at least one solution-- all of the
constants equalling 0.
Anyone could do this.
Now I subtracted all of the yellow terms from both sides,
and I got this equality.
The left=hand side of this equality is linear
combinations of the basis vectors of V.
So any linear combinations of the basis vectors of V is
going to be a member of V.
That's the definition of basis vectors.
So if I set x equalling to this letf-hand side, I can say
that x is a member of V.
Well, if x is equal to the left-hand side, it's also
equal to the right-hand side.
The right-hand side is some linear combination of V perps,
or the orthogonal complement of V's basis vectors.
Which tells us that x is also a member of V perp.
Well, what does that mean?
That means that x must be equal to 0.
I just showed you at the beginning of the video, the
only vector that's a member of a subspace and its complement
is the 0 vector.
So we know that because these are orthogonal complements, we
know that x must be equal to 0.
So just to reiterate, we know 0's has to equal both of these
sides of the equation.
And these are the same constants that we
had to begin with.
But what do we know about these two sets?
We know that the 0 vector has to be equal to this.
That's the only vector in Rn that's a member both of V and
of the orthogonal complement of V.
Now, this is a 0 vector and we have this linear combination
of V's being set equaling to the 0 vector.
What do we know about these constants?
What does c1, c2, all the way to ck have to be?
We know that v1 through vk is a basis for V.
That tells us that they span V and that they are linearly
independent.
Linear independence by definition means that the only
solution to this equation right here is that all of the
constants have to be 0.
So linear independence tells us that c1, c2, all the way
through ck must be 0.
All of these guys right here are 0.
Which is the same as all of these guys.
All of these guys must be 0.
Now let's look at the right-hand
side of this equation.
We could put the minus all the way, but the
same argument holds.
This linear combination of V perp's basis
vectors is equal to 0.
The only solution to this-- because each of these w1's,
w2's, and wn minus k's are linearly independent-- being
equal to 0 is all of the constants have
to be equal to 0.
That falls out of linear independence.
If this negative is confusing you a bit, if it makes it look
different than that, you could just multiply this negative
out and say minus d1 would have to be equal to 0, minus
d2 would have to be 0, minus d and minus k
would have to be 0.
But it's the exact same argument.
Linear independence, which falls out of the fact that
this is a basis set, implies that the only solution to this
being equal to 0 is each of the constants
being equal to 0.
Well, that means that d1, d2, all the way to dn
minus k must be 0.
Let's go back to what I wrote up here.
This was the original equation that we were
experimenting with.
Just manipulating this equation a bit and
understanding that the only intersection between V and V
perp is a 0 vector.
And that you only have linear independence if the only
solution to these vectors equalling 0 is all of their
constants equalling 0.
Then we know that all of these terms, c1 through ck, d1
through dn minus k, they all have to be equal to 0.
That's the only solution to this larger equation that I
wrote up here.
Well, the only solution to this large equation that I
wrote up here is that all of the constants are equal to 0.
That implies that if I were it take the set right here of v1,
v2, all the way to vk, and I were to augment that with the
basis vectors of V perp, which are w1, w2, all the way to wn
minus k, that this is a linearly independent set.
And I know that because the only solution to this equation
is each of these constants having to be equal to 0.
That's what linear independence means.
This implies this.
Linear independence implies that.
We used the fact that linear independence implies that all
of these equal 0 to get the fact that c1 all the way to ck
was equal to 0.
And then we use it again when we set this thing also being
equal to the 0 vector.
We knew that all of the d's had to be equal to 0.
I don't know if you remember, the 0 vector came out from the
fact that that was the only vector that is a
member of both sets.
I know I'm being a little bit repetitive, but I really want
you to understand that this proof isn't some type of
circular proof.
That we just wrote this equation, we wondered about
what the solution set is to it, we rearranged it, we said
hey both sides of this equation are members of both V
and V perp.
The only vector that's a member of
both is the 0 vector.
So both of these sides of the equation have
to be equal to 0.
The only solution to that is all of these constants being
equal to 0, because each of these are
linearly independent sets.
So therefore all of these constants have
to be equal to 0.
And then this augmented set, where if you combined all of
the basis vectors, that is going to be linearly
independent.
Now many, many, many, many, many videos ago, we learned
that if we have some subspace with dimension n, and we have
n linearly independent vectors that are members of your
subspace, then those n linearly independent vectors,
or the set of your n vectors, is a basis for the subspace.
Now Rn is a subspace of itself.
Rn is an n dimensional subspace.
We could write the dimension of Rn is equal to n.
Now we have n linearly independent vectors in Rn.
So that tells us that these guys right here
are a basis for Rn.
We have n linearly independent vectors.
We have n minus k that are coming from V perp.
We have n that are coming from V from their
basis for those subspace.
So now we have a total of n vectors.
They're linearly independent.
They're all members of Rn.
So they are a basis for Rn.
Which tells us that any vector in Rn can be represented by
linear combinations of these guys, which is fascinating.
So this is a basis for Rn.
So that tells us that we can take any vector-- let's say a
is a member of Rn, some vector.
That means since this is a basis for Rn that a can be
represented to some linear combination of
all of these guys.
So it can be represented as C1 times V1 plus C2 times V2 all
the way to plus Ck times with Vk.
Let me use a different letter just to make sure that you
understand that this is a different equation that I'm
writing than I wrote earlier in the video.
So I can write this, and then I can have some other
constants that say plus e1 times our V perp basis vector
1 plus e2 times this guy plus all the way to en minus k
times the n minus k basis vector for V perp.
I can represent any vector in Rn this way.
Or another way to say it.
What is this?
This is some vector that is a member of our subspace V.
And then this is some vector over here that is a member of
the orthogonal complement of V.
This is just a linear combination of V
perp's basis vectors.
This is just a linear combination
of V's basis vectors.
So given that all of these characters are a basis for Rn
tells us that any member of Rn can be represented as a linear
combination of them.
But that means that any member of Rn can be represented as a
sum of a member of our subspace V plus a member of
your subspace V perp.
This is a member of V, and this is a member of V perp.
And that's a really, really interesting idea.
You give me a subspace and then we can figure out its
orthogonal complement.
Any other vector in Rn can be represented as a combination,
or sum, of some vector in our subspace and some vector in
its orthogonal complement.
Now the next question you might be asking, is this
representation unique?
So is this unique?
Well let's test it out by assuming it's not unique.
So that means that I have for some vector a that is a member
of Rn, I can represent it two ways.
I can represent it as equalling some member of my
subspace V, plus some member of the orthogonal
complement of V.
I can represent it that way.
Or I could represent it as some other member of my
subspace V plus some other member of my orthogonal
complement.
So x1, x2 are members of V perp.
And then v1 and v2 are members of V.
If we assume it's not unique, there's two ways
that I could do this.
And I'm representing it as these two vectors.
Now clearly this side of this equation is equal to that.
These are both representations of a.
So we can rearrange this a little bit.
We could say that v1 minus v2-- if I subtract v2 from
both sides, I get v1 minus v2 is equal to-- that's
subtracting v2 from both sides, and if I subtract x1
from both sides-- x2 minus x1.
These are both members of the subspace V.
And any subspace is closed under addition and
subtraction, which is really a special case of addition.
The vector v1, let me write it this way, let me call my sum
vector z, being equal to both of these guys which are equal
to each other.
z is the vector V1 minus V2.
Any subspace is closed under addition, if you take two
vectors and find their difference in a subspace, then
that resulting difference is also
going to be in a subspace.
z is going to be a member of our subspace V.
This vector right here-- which is also the same thing we just
said that to be equal our vector z-- is going to be a
member of our V perp.
Why?
Because both x1 and x2 are members of the subspace V's
orthogonal complement.
And that is a subspace as well.
So it is closed under addition and subtraction.
So this is also going to be a member of your subspace.
So we could also say that z is a member of V perp or the
orthogonal complement of V.
Well, we've done this multiple times.
This was the first thing we showed in the video.
The only vector that's a member of a subspace and its
orthogonal complement is the 0 vector.
So z has to be equal to the 0 vector.
So this is equal to the 0 vector.
Well, if both of these are equal to the 0 vector, we know
that v1 minus v2 is equal to the 0 vector, which implies
that v1 must be equal to v2.
And we also know that x2 minus x1 is equal to the 0 vector.
Or x2 is equal to x1.
So we try to say that, hey, there's two ways to construct
some arbitrary vector a that's in Rn.
And we wrote that down.
But then we found out that no, v1 must be equal to v2 and x1
must be equal x2.
So there's only a unique way to write any member of Rn as a
sum of a vector that's in our subspace V and a vector that
is in the orthogonal complement of V.