In this video, we're going to take a look at the concept of a subspace.
We're actually going to be working with a polynomial vector space in this problem.
So we're going to work with the vector space P3 specifically.
We know what a vector space is just a collection of objects.
So in this case, the collection of objects is all polynomials of order three.
And the operations plus and scalar multiplication are just the typical operations that we use
in working with polynomials.
So we've defined the collection of objects.
Just two sets of polynomials of order three, the operations plus and the scalar multiplication
operations.
That's all we need to define a vector space.
We've talked about polynomial vector spaces, before so we know that P3 is indeed a vector
space.
What we're going to work with here is we're going to work with the set of all polynomials
of this form.
So p of x equals ax cubed.
So this includes just a subset of the polynomials in P3, because, remember, P3 includes polynomials
of order 3.
So even things like 1 plus 2 times x squared is part of this set, as is 1 plus 2x plus
9x cubed.
That's also part of this vector space.
What we've done here is we've restricted ourselves to just polynomials of this form, so they
can only contain an x cubed term times a scalar number a from the reals.
So this set right here of all possible ax cubes is much smaller than the elements in
the vector space P3.
Our question is, is this set of polynomials PX equals x cubed, is this a subspace P3?
To determine if it's a subspace I need to check just three things.
So I'm already working with a vector space P3, I know P3 is a vector space.
And I've chosen a subset of elements from it-- this subset right here.
Because of that, instead of having to check all 10 vector space axioms to establish that
this new set is a vector space, I only need to check three things.
I need to check that our subspace contains the zero vector.
I need to check that it's closed under additionl.
And I need to check that it is closed under scalar multiplication.
If it has these three properties, then this set is a subspace of the vector space P3.
And since it's a subspace, it's itself a vector space.
All the other properties, the other seven properties that we won't check explicitly,
those are all inherited from P3 in a trivial way that we don't even need to check for.
So let's go ahead and see if this set of polynomials of the form ax cubed for arbitrary values
of a in the reals is indeed satisfying these three properties.
So let's go through them one by one.
So first of all, does this collection of vectors contain the zero vector?
Note that I get to choose any value that I want in a.
So for all a in the reals.
That means I can let a equal zero.
So that means that my set contains the polynomial p of x equals 0.
This is one of the vectors in this set.
That means that for any p of x that I want to choose, so for some arbitrary p of x in
my set, if you want to think of that, is bx cubed or 1.2x cubed or a2 x cubed, I can add
0 to it and end up back where I started.
So this element, p of x equals 0, satisfies exactly the zero property that I need.
Any arbitrary p of x that I choose, when I add this zero element to it, I get back where
I started.
So this is part of the set, and it satisfies the zero property.
So I can put a checkbox by the zero vector property.
What about being closed under addition?
Let's choose two arbitrary polynomial polynomials from the set that I'm working with.
So let's choose a1 x cubed and a2 x cubed.
So these are both elements of my set.
And I've chosen them very arbitrarily, letting a1 be an arbitrary number and a2 be an arbitrary
number from the reals.
Let's see what happens when I add them together.
So let's compute a1 x cubed plus a2 x cubed.
Well, if I add those together, I get a1 plus a2 x cubed, and this looks like just another
element from my set, right?
My set is of the form ax cubed for all values a.
Well, this just looks like another value a right here.
It matches up right there.
So when I add two arbitrary polynomials from my set, I end up with another element from
my set.
So that means that I am closed under vector addition.
And, finally, let's check to see if I'm closed under scalar multiplication.
If I choose an arbitrary element from my set, I'm going to call it a1 x cubed.
So I've chosen some arbitrary number a1.
And let's also choose some arbitrary scalar from the reals.
I'm just going to call it b.
What happens if I multiply this scalar times my vector?
Well, I'm going to get b times a1 x cubed.
And, again, this is an element of my set of polynomials x cubed because I have a number
times x cubed.
Well, my second consists of all polynomials of the form ax cubed, so the number a1 is
in the set of reals.
So this is closed under scalar multiplication as well.
So for this subspace, I'm working with a set of polynomials taken from a larger set of
polynomials.
I have a subset.
I have shown that it has a zero vector element.
It satisfied this property.
I've shown that arbitrary elements selected from my set, when added together, yield another
element from my set, so it's closed under addition.
And I've also shown that when I choose an arbitrary element from my set and multiply
by an arbitrary scalar, I end up with another element from my set, so I'm closed under scalar
multiplication.
So having satisfied these three properties, we know that the set of polynomials in this
problem is indeed a subspace of P3.