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Dr. Robert van de Geijn: So let's have a look at subspaces.
What are subspaces?
So we're going to look at the question, what subsets of Rn are vector spaces.
We're going to call such sets subspaces of Rn.
What we're going to try to do is build up some intuition
through a sequence of exercises.
And the vehicle that we're going to use is Homework 9.4.2.1.
So what I'm going to do is walk you through this.
And what you're going to want to do is periodically stop this video,
so that you can go and think for yourself and do the homework exercise.
And then come back to this discussion.
So the first question is what if we have the set of all vectors that
look like this, except that the first component always has to be equal to 0?
So that gives us vectors of this form, except that you
can pick your chi 1 and chi 2 any way you want.
Is that set a subspace of r3?
So go and take a break, and do the homework, and then come back.
So the answer is yes.
And what I failed to do was to start by asking the question,
is the zero vector in this set?
And it's not hard to see that by picking chi 1 to be 0, and chi 2 to be 0,
you end up with the zero vector.
And that vector satisfies this right here.
So we move on and ask the question if I take arbitrary
x and y that are in this set, is the vector x plus y is also in the set?
OK.
Well, an arbitrary x would look like this for arbitrary chi 1 and chi 2.
And an arbitrary y would look like this for arbitrary psi 1 and psi 2.
What happens if we add these two vectors together?
We get a vector that looks like this.
But notice that the first component is, again, equal to 0,
and therefore it is in set.
We can conclude that x plus y is indeed in this set, if x and y are.
So let's move on.
The other thing we need to show is that if we have a scalar alpha
and we have an x in the set, then alpha times x is also in the set.
So you pick an arbitrary x in the set, and an arbitrary alpha in R.
And you look at the alpha times x.
Well, alpha times x has this form, because x was in this set.
Therefore x had this form.
Therefore, if you do alpha times x, you get this right here.
But you again have the first component equal to 0,
and therefore, the result is the set.
So now we know that 0 is in the set.
We know that if x and y are in the set, then x plus y is in the set.
And we know that if alpha is a real and x is in the set, then alpha times
x is in the set.
Thus, this is a subspace.
OK, well, what about all of the vectors that
have the property that the first element is equal to 1?
So that's the set of all vectors that look like this.
Go down to the homework question about this and then come back.
So this time the answer is no.
And why this is not the case?
Well, notice that the zero vector is not in this set,
because the zero vector doesn't have a first component equal to 1.
What about the set of all vectors in R3 that have three components.
And if you multiply the first and the second component together, you get 0.
Now that's quite a bit harder.
Why don't you go have a look at that?
So, if you have trouble with that one, don't feel bad.
This is not so easy.
The fact is that this is not a subspace.
Why is this not a subspace?
Well, it turns out that we can find an x and the y that are in this set,
namely, the vector (1, 0, 0) and another vector y (0, 1, 0).
And notice that if you multiply the first and the second component of x
together, you get 0, so it satisfies this.
If you multiply the first and the second component of y together,
you satisfy this.
But once you add these two vectors together, you get the vector (1, 1, 0).
And if you multiply 1 times 1, you get 1,
and therefore you have a vector that is not in this set.
OK, what about this one?
What in the world is this?
m this actually turns out to be the set of all vectors that look like beta 0
for the first component, beta 0 plus beta1
for the second component and beta1 times 2 for the third component.
We recognize this as a linear combination of these two vectors.
So it's the set of all linear combinations of the vector (1, 1, 0)
and (0, 1, 2).
Go and do that one.
OK, so it turns out that this is a subspace.
And how would we show that?
If we call this set S, notice that 0 is in S. How do we know that?
Well, pick beta0 equal to 0, pick beta1 equal to 0,
and you get the zero vector.
Now, what if we have an x that's in S and we have a y that's in S?
Well, let's see.
Then you know that there is some kind chi 1 times (1, 1, 0)
plus some chi 2 times (0, 1, 2), that is equal to x.
We know that there is some psi 0 times the first vector
plus some psi 1 times the second vector, because otherwise y
would not be in this set.
And if you look at x plus y, what you get is this plus that,
but if you worked it out, that is the vector (chi 0-- whoops,
that should've been (0, 1, 1), right?
Come on.
chi 0 plus psi 0 times the vector (1, 1, 0) and the scalar chi 1 plus
psi 1 times the vector (0, 1, 2).
And what you notice is that x plus y is also a scalar times this first vector
plus a scalar times the second vector.
And therefore, it satisfies this.
OK, well, what if we have x in S, and we have some scalar alpha that is in R?
Then, notice that x has the form chi 0 times (1, 1, 0)
plus chi 1 times (0, 1, 2).
And if we look at alpha times x, it's not
hard to see that that becomes alpha times chi 0 times (1, 1, 0),
plus alpha times chi 1 times (0, 1, 2).
And then if you put our parentheses around that,
then you'll see that, again, you get a scalar times the first vector
plus a scalar times a second vector, and therefore this is too in the set S.
What about this one?
We'll go think about it for a bit.
This one is a little harder to figure out.
Well, no, not, really.
So first of all, notice that 0 in this set.
So let's call this set S, and how do we see that?
Well, notice that if you plug in (0, 0, 0), then this is indeed equal to 0.
So again if this is S, and now we take x in S and we take y in S, then
what do we know?
We know that x is equal to (chi 0, chi 1, chi 2) and chi 0 plus -chi 1
plus 3 times chi 2 is equal to 0.
And we know that y is equal to (psi 0, psi 1,
psi 2 ) and psi 0 plus -psi 1 plus 3 times psi 2 is equal to 0.
Now, if you look at x plus y, that's just
a vector where you add these components to those components.
But if you add this right here, you get chi 0 plus psi 0 plus -chi1 plus -psi1.
Whoops, that should've been a minus here all along.
And then plus 3 times chi 2 plus psi 2 is equal to 0.
And if you look at the components of x plus y,
that's exactly this this, and that.
Anyway you do the same thing with x in S and alpha.
A maybe number, and all of these things all fall in place.
You're starting to see the pattern, though?
You pick arbitrary x and y.
You write what the two components are.
You add them together.
You see whether it satisfies the conditions of the set,
et cetera, et cetera, et cetera.
What about the empty set?
Is it a subspace of Rn?
Well, go have a look.
Go think about this.
And the answer is no.
Why?
Because 0 is not an element of the empty set.
The empty set has nothing in it.
OK, well what about the set that only has the zero vector in it?
And the answer, go think about that one.
OK, you're back.
And the answer this time is yes!
Why because certainly 0 is an element of S, if we call this set S.
And if x and y are an element of S, and then it is the case that x plus y
is equal to 0 plus 0, because x and y both must be the zero vector.
It's the only vector in the set, and that's
equal to 0, which then itself is back in the set that
only has the zero vector in it.
And if you have x, an element of S and alpha, a real number, then alpha times
x is the same as alpha times the zero vector,
because that's the only element in the set.
And that's just a zero vector, which is in the set.
OK?
What about this one?
Now notice that that's the set of all vectors.
Let's just restrict ourselves to n equal to 2.
OK, so that's the set of all vectors that
are in the ball that has a radius 1.
Well.
The problem there is if you have some vector x here,
then alpha times x could very easily point out here.
So you have a vector x here.
It's very easy to construct a case or choose an alpha such
that this is outside.
And therefore, this is not a subspace.
Anyway, you can go do the rest of the homeworks
and might gain some experience.