In this video, we're to work another problem related to working with subspaces.
In some ways, it's very similar to the Example number 3 that we just worked.
We're going to be working with a set U, but in this case, the set U's just a little bit
different.
Some of the things that we'll look at with respect to set U will be the same, but some
of the cases and things that fail or don't fail with respect to being a subspace are
definitely a little bit different from the previous video.
So the set that we're working with, U, consists of all vectors of the form x and y, so these
are always two-dimensional vectors, and the condition is that both x and y are greater
than or equal to 0.
Unioned with the set of vectors x and y such that x is less than 0 and y is greater than
or equal to 0.
So there's really kind of these two pieces.
We have this part, and we have this part.
And together, if you sketch these-- and that's one of the first things we're going to do--
you can see what this looks like.
And essentially, this is the first quadrant of the plane, and this is the second quadrant
of the plane.
So really, what we have here is kind of the top half of the 2D plane.
So similar to last time, let's first just be able to visualize what this collection
of points looks like.
So let's sketch this set of vectors by sketching the 2D plane, and then shading in all of the
points in the 2D plane that are either in this set or this set.
And we're unioning these two things together to form the set U. Apologize for the notation.
This is the capital letter U and this is the union symbol between the two different sets
here.
They look kind of similar, but they are different.
So if we sketch in all of these points in the plane where both x and y are greater than
0, that's the first quadrant.
And then if we sketch in all the points in the 2D plane where x is less than 0 but y
is greater than 0, we get these points.
And actually, these are inclusive greater thans and less thans.
So it actually includes this line right here.
So that is the set of points U.
We have sketched this set.
So like I already noted, U is really just the first and second quadrants of the 2D plane.
So it consists of all the points in R2 in the first and the second quadrants.
In the second part, what we're going to try to do is we're trying to figure out, is U
a subspace of R2?
And we know that subspace means something very specific.
Obviously, we have that U is a subset of R2, because R2 is the whole plane.
And what we shaded in here is a subset of it.
So if we want to check that if U is a subspace, there are three things that we need to check.
We need to check to make sure the 0 vector is present.
We need to check to make sure that this set is closed under vector addition, and we need
to make sure that it is closed under scalar multiplication.
So there are three things that we need to check.
The first one we're going to work with right here is with respect to seeing if the 0 vector
is present.
So the 0 vector we write as this bold 0, and the 0 vector has the property that when you
add it to any vector in the set that you're working with, nothing changes about the thing
that you added it to.
So after you add the 0 vector, nothing has changed.
So obviously for this set, the 0 vector is actually a two-dimensional vector whose elements
are 0 and 0.
Because no matter which point I pick from U, if I add 0 0 to it, nothing changes about
the point that I picked from the set U.
So that, by definition, makes it the 0 vector.
Well, looking at the sketch up here, we said that this line was included, so that means
that the origin right here, 0 0, is included.
So obviously, the vector 0 0 is an element of the set U.
So this checks out.
This set of points, U, does contain the 0 vector.
So the first of three criteria has passed.
So this set has a chance of being a subspace of R2 because it's a subset of R2, and it
contains the 0 vector.
There are still two more things that we need to look at.
The other thing that we need to check is to see if it is closed under vector addition.
So to see if it is close under vector addition, I need to pick two arbitrary vectors from
U and add them together and make sure the resultant summation is still in U.
So I have added, or selected, the arbitrary vector x1 y1, which is in U, and the arbitrary
vector x2 and y2 which is in U.
If I add these together, obviously, I get a vector x1 plus x2, and y1 plus y2, which
if we want to, we can define as x3 y3, and you can see what can happen here.
When I choose x1's and x2's, this is the x-coordinates of these arbitrary vectors that I've chosen,
I'm free to choose any value between minus infinity and positive infinity.
If you go back to our sketch, our sketch extended from all the way on the left, all the way
on the right.
Any value along the x-axis was legitimate.
So no matter what values I choose for x1 and x2, they're always going to be between plus
or minus infinity.
So that means after I add them together, I'm going to end up with an x3 that is between
somewhere between minus infinity and positive infinity.
And that's OK, because that still is an element of the set U.
It's going to be somewhere along the x-axis that's in a spot that's OK.
What about the points y1 and y2?
Well, when I pick these arbitrary vectors x1 y1 and x2 y2, y1 always has to be greater
than or equal to 0.
That's a constraint to being in the set U.
So y1 has to be greater than or equal to 0, and the same thing when I chose the vector
x2 y2. y2 has to be greater than or equal to 0 if it's going to be in the set.
So I must have these conditions be true.
So y1 is greater than or equal to 0, y2 is greater than or equal to 0.
So that means when I add them together here to get y3, I'm adding two numbers that are
greater than or equal to 0.
So that means y3 must be greater than or equal to 0.
So when I'm done with this, I end up with a vector whose first coordinate, x3, can be
any value, really.
It can be anything between minus infinity to positive infinity, and I end up with a
vector whose second coordinate, y3, has to be greater than or equal to 0.
Well obviously, that's going to be a point somewhere in the shaded region.
So that means x3 y3 has to be in either the first quadrant or the second quadrant.
And because of that, that means at this point, x3 y3, is going to be in the set U.
So since I've chosen two arbitrary vectors from U, added them together and ended up with
a vector that is still in the set U, this is closed under vector addition.
So the second property for this problem checks out.
All right, finally, we need to look at the third and final property.
We need to figure out if this set of vectors U is closed under scalar multiplication.
So for this one, I'm actually just going to pick a particular point in the set U, and
I'm actually going to show that it is not closed under scalar multiplication.
So if you want to show that something is true, you only need to have this one example to
show that it fails in general.
So by finding one counterexample, we can show that it is not closed under scalar addition--
I'm sorry, scalar multiplication.
And the point that I've chosen is the point 1 2, and the constant I'm going to choose
to multiply by is the scalar and negative 1.
So with the vector 1 2 and the scalar negative 1, we can form the product C times 1 2, so
c times the vector 1 2.
And obviously, this is equal to a negative 1 times the vector 1 2, which is equal to
a negative 1 and negative 2.
This point, at negative 1 negative 2, is not in the set U.
This is something in the third quadrant.
Part of it has a negative x dimension, which is OK, but its y dimension is negative.
And we can't have any second components that are negative in the set U.
So this point here that I got by choosing an element from U and then multiplying by
some scalar that I chose, we end up with a point that is outside the set U.
So that means that this is not closed under a scalar multiplication.
And since this property failed, that means that U cannot be a subspace of R2.
So this problem's a little bit different.
We had a set of points U.
We wanted to know if it was a subspace.
It was obviously a subset.
We worked through the remaining three properties.
The first two checked out, but the third one failed.
If any of them fail, then we cannot be a subspace of the vector space that we're working with.