So let's take a look at another subspace example problem.
In this problem, we're going to be working with a set U.
And U is the set of all vectors that have this form right here.
So it's the set of all vectors of length 3.
And the vector is equal 2r minus s as its first coordinate, 2 as the second coordinate,
and r plus s as the third coordinate.
And r and s are arbitrary, real valued quantities, so we say that r and s are both in r.
So think of every possible value of r and s.
This is a very large set of vectors that has this form.
That's the set U that we're going to be working with.
And what we're asking is is U a subspace of R3?
So U is obviously a subset of R3.
Every single vector in U has this form, so it's obviously some element of R3.
But is it a subspace of R3, and that means something very specific.
It means that we actually need to check for three specific properties.
We know that a set of elements or set of vectors is a subset of some vector space.
To figure out if it's a subspace, there are three things that we need to check.
First, we need to make sure that it contains the zero vector.
We need to make sure that it is closed under vector addition.
And we also need to make sure that it is closed under scalar multiplication.
If this subset of R3 satisfies all three of these properties, then you would be a subspace
of R3.
It turns out this particular subset of R3 is not going to satisfy any of these, so it's
not going to be a subspace.
But we're going to work through each of these properties and show how we can go about showing
that it does not satisfy these properties.
So let's go ahead and take a look at the first property.
Does this subset of vectors U contain the zero vector?
So one thing we can look at is when r and s are equal to 0, then we end up with the
vector 0, 2, 0.
So no matter what we choose for r and s, this second coordinate of equal to 2 is never 0.
So there's nothing we can ever do to make a 0 in the second coordinate.
And because of that, that means that we can never get an all-zero vector.
So this set of vectors U, even though it contains a large number of vectors, it does not contain
the all-zero vector.
And because of that, we could just stop this problem right now and say because it fails
that check, it does not contain the zero vector, U is not a subspace of R3.
So we can go ahead make that conclusion.
We're going to continue working through these other properties, though, just to see how
those play out as well.
So let's go ahead and look at the second one.
If I choose to arbitrary vectors from U-- so I'm going to denote these two arbitrary
vectors as U and V. And you see what I've done here.
I've chosen arbitrary values r1 and s1 to form the vector U.
And I've chosen arbitrary quantities r2 and s2 to form the vector V.
Actually when I just said that there, I probably misspoke.
r1 and s1 are just scalars in r, obviously.
So r1 and s1 are used to construct U, and r2 and s2 are used to construct the vector
v. So I've really chosen two very arbitrary vectors
in the set U.
They both have the form that we know that the vectors in U must have.
Let's go ahead and compute what the U plus a V is equal to.
If I add that up, I am going to end up with U plus a V is equal to 2r1 plus 2r2 minus
s1 minus s2.
So that's the first coordinate of the vector quantity U plus V.
For the second element, I have 2 plus 2 is equal to 4.
And then finally, adding up this plus this, I'm going to have r1 plus r2 plus s1 plus
s2.
And you can see when I wrote down this vector quantity U plus V, I grouped it to keep r's
and s's the same.
It's just nice to write it that way.
This quantity here, 2 times r1 plus r2, I can think of that as a quantity r3, some other
arbitrary number.
Same thing here.
s1 plus s2 is just some number s3.
So if I rewrite this, this vector has the form of 2r3 minus s3, 4, and r3 plus s3.
So the first coordinate is fine.
I have some kind of arbitrary thing right here.
The definition of U always had 2 times r minus an s, So.
That's exactly what it looks like right here, 2 times an r minus an s.
This third coordinate is fine as well.
The definition of U, we always had an r quantity plus an s quantity.
That's exactly what I have here.
But the second dimension does not fit the pattern for U.
Every element of U has a 2 as its second coordinate, and I have a 4 here.
So by picking arbitrary elements or vectors from U and then adding them up, we ended up
with a vector that does not have the proper form.
This vector is not an element of the set U, so it is not closed under addition.
So it fails this property as well.
And then, finally, let's go ahead and look at the third property.
Is this set of vectors closed under scalar multiplication?
So let's just choose some arbitrary vector from the set U again.
And we'll choose an arbitrary scalar c.
And then let's go ahead and compute what the quantities c times U is equal to.
So here is my arbitrary c.
I've chosen an arbitrary vector v from the set U. 2r minus s is the first coordinate,
2 is the second, and r plus s is the third.
If I multiply this out, what do I end up with?
I end up 2rc minus sc as I distributed the c.
I end up with 2c here in the second coordinate because 2 time c, and I end up with rc plus
sc.
So again, this does not fit the pattern for U.
Every vector in you has a 2 as its second coordinate, and I have 2c.
If c is equal to 1, I'm OK.
I still end up with a 2 there.
But for any other value of c, I don't end up with that 2 there.
So it turns out that this set of vector is not closed under scalar multiplication either.
When I choose an arbitrary vector and choose an arbitrary scalar and form the scalar-vector
product c times v, I end up with a vector that is not in the set, so it's not closed
under scalar multiplication.
So that concludes this problem.
We were working with a set of vectors that was obviously a subset of R3, and we were
trying to determine if it was actually a subspace of R3.
When determining that, there were three properties that we had to look at.
If any of them fail, it turns out that you don't have a subspace.
For this particular example, all three of them failed.
So obviously, the set U is not a subspace of R3.