All right, so another subspace example problem-- this one's a little bit different, though,
because the collection of objects that we'll be working with here are actually polynomials.
Remember, the concept of a vector space is very general. It's any collection of things
that have certain axioms or properties that are satisfied. The things are often vectors,
like vectors in R2 or R3 or Rn. But any collection of, quote, things that satisfies the vector
space properties is a vector space. In this example, we're going to be working
with polynomials and show that they satisfy certain vector space axioms to be a subspace.
So first, let's define the overall vector space that we're going to be working with.
We're going to be working with the set of polynomials that we denote as Pn. So Pn is
a collection of objects. And it consists of a lot of things, and it consists of all polynomials
of degree at most n. So n is this arbitrary number, and Pn is this collection of a ton
of polynomials. And it consists of every polynomial you could ever imagine up to degree n.
So for example, if n was 7, the polynomial x cubed plus 4x minus 9 would be in the set
P7 because it's a polynomial of, at most, degree 7 and [INAUDIBLE] degree 3. So it would
be in that set. So any polynomial you could come up with would be in that. So it's quite
a large set for any value n. As an example, and what we write here is we
say that the elements p of x are in P of n. And this single element from P of n we denote
as p of x. And in general, they have the form that looks something like this. In general,
they can go all the way up to x to the n. They have very arbitrary / general coefficients
in front of them. But this is what an arbitrary element from P of n would look like. And depending
on how you want to set a0 a1, all the way up to a n, you can create any polynomial you
want. The example I just gave a second ago, I we
would be able to set the coefficients on the x cubed and the x squared and the a 0 to be
certain values to be any polynomial we mentioned. But in general, these are what the elements
of Pn look like. What we're going to do is we're going to answer
a very specific question. We are going to figure out if P2 is a subspace of Pn for n
greater than or equal to 3. So that's what we're going to work with in this problem.
P2 is the set of all polynomials of at most degree 2. So these will be quadratics and
linear equations and constants. Pn for n greater than or 3 includes polynomials
of degree 3 and higher. So we want to know if P2 is a subspace of Pn. First of all, what
we note is that P2 is a subset of Pn because any polynomial that I have in P2, I can find
in Pn. So P2 is definitely a subset of Pn because second-degree polynomials can be found
in the set of higher order polynomials Pn. So since it's a subset, we know what that
means. Instead of having to check all of the vector axioms or vector properties or vector
space properties, there's really only three things we need to check, the zero element,
the closed under addition, and the closed under multiplication. Let's work through those.
So the first question we have is, does P2 contain the zero vector. So again, zero vector
is a general term. It means the element, such that when I add it to another element, I get
back where I started. So it doesn't necessarily have to be a literal vector. It's just an
element that has a certain property. So first of all, let's go ahead write out
what an arbitrary of P2 looks like. In general, an arbitrary element of P2 looks like this.
It can go up to degree 2, because we're working in P2. And all these coefficients-- a0, a1,
and a2-- can be completely general and arbitrary. The set P2 contains an infinite number of
polynomials as we choose all possible combinations of a0, a1, and a2.
We can figure out if it contains the zero vector by letting a0 equal a1 equal a2. In
that case, what would we have? We would have zero plus zero plus zero equals zero. So this
quantity right here, I'm going to denote it as the polynomial P0 of x, because this polynomial
right here satisfies the zero vector property. If I add the scalar zero to any polynomial
of this form, I get back exactly the polynomial of this form, which, by definition, is the
zero vector property that I need. So no matter what P of x I choose, for all P of x that
I choose, p of x plus p0 of x is just p of x plus 0, which is just p of x. I get right
back where I started. So since this is a legitimate choice, and
this is a polynomial that is in P2, and since it satisfies this property, it is the zero
vector. And it is in P2. So P2 does contain the zero vector.
All right, let's look at the next property. Is P of 2 closed under vector addition? So
let's go ahead and choose two arbitrary polynomials. The first one, I'll call p1. So it's some
polynomial of, at most, order 2 with coefficients a0, a1, and a2. And they can be-- any of these
coefficients can be any value. And I'll choose another arbitrary element
from P2. I'll call this b0 plus b1 x plus b2 x squared. So I've chosen two very arbitrary
elements from P2. Let's see what happens when I add them together. So p1 of x plus p2 of
x is obviously equal to a0 plus b0 plus a1 plus b1 x plus a2 plus b2 x squared, which
I can go ahead and define as c0-- so I'm defining a0 plus b0 as some number c0-- plus c1.
So I've defined c1 to be a1 plus b1 plus-- you can guess what I'm going to do here--
c2, if I've got that equal to a2 plus b2, I end up with this. So this right here looks
just like any other polynomial in P2. It's of most order 2. And these are very arbitrary
coefficients. So this looks just like any other element
in the set P2, so this is in P2. So I've chosen arbitrary elements from the vector space.
I added them together, and I ended up with a vector in the vector space. So it is closed
under vector addition. Again, don't be confused. When we talk about
vector addition in vector spaces, that means very generally these objects we're working
with. The objects we're working with in this particular example are polynomials. So when
I added the polynomials together, I ended up with a polynomial that had the correct
form. And it was in P2, so it was closed under vector / polynomial addition.
All right, so this subset P2 has been a subset. It's been closed under addition. It has had
a zero element. So it's passed the first two tests. There's just one more thing we need
to check. Is it closed under scalar multiplication? So let's see what happens here. Let's go ahead
and pick some arbitrary element again. Let's pick p of x equals this. So I've chosen a
very arbitrary element from the set P2. I'm going to choose a very arbitrary constant
from R. And then I'm going to compute what does c times p of x equal.
Well, if I write this out, I get c times a0 plus c a 1x plus c a 2 x squared. I just multiply
this out. And again, I can go ahead and define this as a new number. I'll call it d0. And
I can define c a 1 as d1. And I can define c a 2 as d2.
And guess what this looks like? This looks like the general form of any arbitrary polynomial
in P2 again. So this is an element in P2. This is the exact form that elements in P2
always have. So I've chosen an arbitrary starting vector.
I chose an arbitrary constant. I formed the scalar multiplication. And I ended up with
an element that is still in the set that I'm working with. So by definition, this is closed
under scalar multiplication. So we have figured out what's going on. Since
P2 is a subset of P of n, and the three properties hold, P of 2 is a subspace of Pn for n greater
than or equal to 3. So that concludes this example. And hopefully, this example kind
of emphasizes the more general concept of vector space, that vector spaces are just
collections of things with addition-- I'm sorry-- yeah, with addition and scalar multiplication
to find that satisfies certain properties. Often, the vector spaces and subspaces are
literally vectors-- you know, R3, R4, Rn-- but sometimes, the collection of objects can
be more general. It can be polynomials. It can be matrices. It can be other quantities.
And that was the purpose of this example, really, was to work with a subspace that was
a subspace of a vector space, but the underlying objects weren't vectors. They were actually
polynomials.