I'm going to take a look at a subspace in this problem.
A vector subspace example number one.
U is going to be the set of all vectors that have this form.
So it's going to be the set of all vectors whose first element can be written as 2 times
r minus s.
Whose second element is three times r.
And who's third element is r plus s.
These values r and s are all real-valued numbers.
So we can see that U is-- you know, it looks kind of like R3, but it contains just a subset
of the vectors of R3 because everything has been restricted to this form.
There are certain vectors of R3 that aren't in this set.
So the question we have, given that we've constructed this subset of vectors from the
vector space R3, is U actually a proper subspace of R3?
So is U itself a vector space?
That's the question that we're asking here.
So one thing that we can note is that we can actually rewrite this vector in a certain
way.
We can rewrite this vector as r times the vector 2 because there's a 2 right here.
As the second element, we can put a 3 there.
And 1 is the third element.
And then we can add s times the vector negative 1, 0, and s.
There is a negative here, in terms of how many S's there are, there are zero s's here,
and there is 1 s right here.
So one thing to note is that this vector that we're working within this general form can
be factored as this linear combination in terms of r times the vector plus s times the
vector.
And one thing that's interesting to note is that both of these vectors, 2, 3, 1, and 1,
0, negative 1, both of these vectors are actually elements of R3.
And that fact alone helps us solve this problem in terms of determining is U a subspace of
R3?
Another thing to note is since we've been able to write U in this form, we've been able
to write U as r times a vector plus s times a vector for all values of r and s.
By definition, we've been able to write this set U as the span of these vectors.
Because remember what the span means, span means take all linear combinations of the
vectors.
Well, that's exactly what we have here.
R and s are arbitrary scalars from the reals.
So it's very clear that we can write U as the span of these two vectors 2, 3, 1, and
1, 0, negative 1.
So how does this information help us establish that U as a subspace of R3?
Well, there's a theorem that helps us do this.
There's a theorem that says, if you have vectors v1, v2, all the way up to vn, some set of
vectors, and they are in the vector space V-- so you're working on some vector space
V-- and the span of those vectors is a subspace of V.
So there's a theorem that says, given some finite collection of vectors from a vector
space, their span is always a subspace of the vector space.
So what does that mean for this problem?
Well, that means for this problem, what do we have?
We established that we have two vectors in R3.
Those vectors were 2, 3, and 1, 1, 0, negative 1.
They're in R3.
And we were able to show that U is the span of those two vectors, just like it says in
the theorem.
So that means that U, by this theorem, is a subspace of R3.
This is a very useful theorem anytime that we're working with a vector space.
If we have vectors from that vector space, and we take the span of them, the set of all
linear combinations, we know that we always end up with a subspace of the upper vector
space, V.
For this particular problem, that means that U is a subspace of R3 because we were able
to factor our original vector into these two vectors.
We noted that both were in R3, and we had the fact that U was the span of these vectors.
So that establishes the fact that U as a subspace of R3 by this theorem.
Another way to work this problem is, actually just go back and establish the properties
of the subspace.
Remember, when we're working with a subspace.
Or a subset of vectors from a vector space, instead of having to check all 10 properties,
there's just three properties that we need to check.
We need to check that the zero vector is present, that it's closed under addition, and it's
closed under scalar multiplication.
That'd be another completely legitimate way to work this problem is to work through each
of those three properties and show that they're all true.