Consider the subset U ⊂ R3[x] defined as

[Karl Porter]

Homework Statement

Show that U is a subspace of R3[x]

Relevant Equations

U = {p(x) = a3x^3 + a2x^2 + a1x + a0 such that p(0) = 0 and p(−1) = 0}

so to show its a subspace (from definition) I need to prove its closed under addition and multiplication , contains 0 and for every w there is a -w? has it already been proven to contain 0 as p(-1)=0?
also I did sub in -1 and ended up with the equation a1+a3=a2+a0 but I don't know if that is relevant to solving this question?

 

[jbriggs444]

Let me try to get the notation clear first.

The sub-space you are talking about is the space of degree 3 (or lower) polynomials that have x=0 and x=-1 as roots.
The space you are talking about is the space of degree 3 (or lower) polynomials.

You are considering these as vector spaces, yes? So you want to prove closure under vector addition and upon multiplication by a scalar.

So, how would you go about proving either one of those for the sub-space?

 

[Karl Porter]

yea that's right
so for addition the sum of two vectors must belong to R3? when I substitute in -1 I get the equation a1+a3=a2+a0 but that leads me to ask how do i know a2+a0 is part of R3? is it a part of R3 because they are real numbers?

 

[jbriggs444]

yea that's right
so for addition the sum of two vectors must belong to R3? when I substitute in -1 I get the equation a1+a3=a2+a0 but that leads me to ask how do i know a2+a0 is part of R3? is it a part of R3 because they are real numbers?

Remember what you are trying to prove.

You are not trying to prove that the sum of two polynomials with roots at 0 and -1 is a polynomial. You already know that. It follows from the fact that third degree (or fewer) polynomials are a vector space in the first place.

You are trying to prove that the sum of two polynomials, each with roots at 0 and -1 is a polynomial with roots at 0 and -1.

 

[Karl Porter]

Remember what you are trying to prove.

You are not trying to prove that the sum of two polynomials with roots at 0 and -1 is a polynomial.

\You are trying to prove that the sum of two polynomials with roots at 0 and -1 is a polynomial with roots at 0 and -1.

so i would add the equation to itself but one would be x1 and the other x2 and see if its gets me a polynomial with those roots.
however i don't think i can solve that because it would have two unknowns

 

[jbriggs444]

so i would add the equation to itself but one would be x1 and the other x2 and see if its gets me a polynomial with those roots.
however i don't think i can solve that because it would have two unknowns

No. I do not think that you are getting it.

You are adding two polynomials to one another. Polynomials. Not values of the formal parameter x. You are not solving for anything. You are just checking to see if you get a result that has the right property. i.e. that it has roots at 0 and at -1.

The two polynomials would be, for instance, $(a_3, a_2, a_1, a_0)$ and $(b_3, b_2, b_1, b_0)$.

 

[Karl Porter]

No. I do not think that you are getting it.

You are adding two polynomials to one another. Polynomials. Not values of the formal parameter x. You are not solving for anything. You are just checking to see if you get a result that has the right property. i.e. that it has roots at 0 and at -1.

The two polynomials would be, for instance, $(a_3, a_2, a_1, a_0)$ and $(b_3, b_2, b_1, b_0)$.

but what happens to the x values?

 

[jbriggs444]

but what happens to the x values?

What is the sum of $a_3x^3 + a_2x^2 + a_1x + a_0$ and $b_3x^3 + b_2x^2 + b_1x + b_0$ ?

[One step at a time. We will worry about x in a moment.]

 

[Karl Porter]

x^3(a_3+b_3)+x^2(a_2+b_2)+x(a_1+b_1)+(a_0+b_0)

 

[jbriggs444]

x^3(a_3+b_3)+x^2(a_2+b_2)+x(a_1+b_1)+(a_0+b_0)

Perfect. Now two questions:

1. Is this a degree three (or fewer) polynomial?
2. Does it have roots at x=0 and at x=-1?

 

[Karl Porter]

Perfect. Now two questions:

1. Is this a degree three polynomial?
2. Does it have roots at x=0 and at x=-1?

yes degree 3
how do i know if it has those roots? i am left over with the constant values a_0+b_0?

 

[jbriggs444]

yes degree 3
how do i know if it has those roots? i am left over with the constant values a_0+b_0?

Suppose that the two polynomials that you just finished adding together were members of the sub-space.

That means that $a_3x^3 + a_2x^2 + a_1x + a_0 = 0$ when x=0 and when x=-1
That means that $b_3x^3 + b_2x^2 + b_1x + b_0 = 0$ when x=0 and when x=-1.

What do those facts then allow you to prove about the sum you just wrote down?

 

[Karl Porter]

a_0 =0 and b_0=0?

 

[jbriggs444]

a_0 =0 and b_0=0?

More basic.

You have two polynomials that are both zero at x=0. You add them together. What is their sum at x=0?
You have two polynomials that are both zero at x=1. You add them together. What is their sum at x=1?

 

[Karl Porter]

sum is 0

 

[jbriggs444]

sum is 0

Right. So is the sum an element of the sub-space?

1. Is it a polynomial?
2. It is zero at x=0 and at x=-1?

 

[Karl Porter]

Right. So is the sum a member of the sub-space?

1. Is it a polynomial?
2. It is zero at x=0 and at x=-1?

ok so the sum of the two polynomials will equal 0 which is part of the subspace. and since the two polynomials add to give a polynomial to degree 3 its part of R3

 

[jbriggs444]

ok so the sum of the two polynomials will equal 0 which is part of the subspace. and since the two polynomials add to give a polynomial to degree 3 its part of R3

Yes. Though I would state it a bit more carefully. The sum of the two polynomials is not the zero polynomial. The sum of the two polynomials is a polynomial which when evaluated at x=0 and at x=-1 yields a result of zero.

 

[Karl Porter]

Yes. Though I would state it a bit more carefully. The sum of the two polynomials is not the zero polynomial. The sum of the two polynomials is a polynomial which when evaluated at x=0 and at x=-1 yields a result of zero.

would method be the same for proving its closed under multiplication? wouldn't that give a polynomial to degree 6?

 

[jbriggs444]

would method be the same for proving its closed under multiplication? wouldn't that give a polynomial to degree 6?

Look back at post #2. We are not talking about multiplying two polynomials by each other. That is not one of the defined operations for a vector space.

The defined operation for a vector space is the multiplication of a vector by a scalar. i.e. multiplication of a polynomial by a real-valued constant.

On the other hand, perhaps you did not mean to be working with vector spaces. Perhaps you want to work in the ring of formal polynomials or in the ring of polynomial functions over the reals.

 

[Karl Porter]

Look back at post #2. We are not talking about multiplying two polynomials by each other. That is not one of the defined operations for a vector space.

The defined operation for a vector space is the multiplication of a vector by a scalar. i.e. multiplication of a polynomial by a real-valued constant.

righhhttt ok so that would give a polynomial to R3 but what about the zero root?
just same as before at x=0 and x=-1?

 

[jbriggs444]

righhhttt ok so that would give a polynomial to R3 but what about the zero root?
just same as before at x=0 and x=-1?

Assuming that we are working in the vector space of degree 3 polynomials [as opposed to the ring of formal polynomials or the ring of polynomial functions over the reals] then we can proceed as before.

Suppose that we have a real-valued constant $b$ and a polynomial $a_3x^3 + a_2x^2 + a_1x + a_0$ which is a member of the sub-space.

1. What is the product of that scalar and that polynomial?
2. Does that product evaluate to zero at x=-1 and at x=0? [editted to correct typo in original]

 

[Karl Porter]

K(a_3x^3+a_2x^2+a_1x+a_o)
and at x=0 x=-1 its gives 0

 

[jbriggs444]

K(a_3x^3+a_2x^2+a_1x+a_o)
and at x=0 x=-1 its gives 0

Good. So on that basis, is the product a member of the sub-space?

 

[Karl Porter]

OK. So on that basis, is the product a member of the sub-space?

yes!is that all that there is to prove to show its part of the vector space?

 

[jbriggs444]

yes!is that all that there is to prove to show its part of the vector space?

Well, we've proved closure over addition and under multiplication by a scalar.

Let me take a quick trip to Google and see what other axioms we need...

We need the existence of a zero element in the sub-space. This still needs to be proved.

Everything else, it seems that we inherit because we are a sub-space that is closed under addition and multiplication by a scalar.

 

[Karl Porter]

Well, we've proved closure over addition and under multiplication by a scalar.

Let me take a quick trip to Google and see what other axioms we need...

We need the existence of a zero element in the sub-space. This still needs to be proved.

Everything else, it seems that we inherit because we are a sub-space

so multiplying the polynomial by 0 would give a zero vector right?

 

[jbriggs444]

so multiplying the polynomial by 0 would give a zero vector right?

Yes. But you have to make sure that the zero vector is a member of the subspace...

... oh right. Since we've already proved closure under multiplication by a scalar, that one we get for free.

 

[Karl Porter]

the next part asked for the dimension of U
the lack of constants is throwing me off but I put in the values of -1 and 0 and I am left with a matrix
-1 1 -1 1 │ 0
0 0 0 0 │0
so wouldn't the dimension just be 1x4?

 

[jbriggs444]

the next part asked for the dimension of U
the lack of constants is throwing me off but I put in the values of -1 and 0 and I am left with a matrix
-1 1 -1 1 │ 0
0 0 0 0 │0
so wouldn't the dimension just be 1x4?

Can you justify the approach that you are taking here?

Personally, I would be thinking in terms of interpolating polynomials as a way to come up with an alternate basis for the vector space and for the sub-space.

You do understand the relationship between "dimension" and a "basis" for a vector space?

 

[Karl Porter]

Can you justify the approach that you are taking here?

Personally, I would be thinking in terms of interpolating polynomials as a way to come up with a basis for the vector space and for the sub-space.

So changing it to matrix form is to calculate the basis?
I think i went about that the wrong way, did look through some similar questions and they focus on the degree of the power of the polynomial and since every polynomial has degree ≤ 3 it has a dimension of 4?dimension is the size of the vector?
I don't really understand the basis but its spans the vector and is linearly independent

 

[jbriggs444]

So changing it to matrix form is to calculate the basis?

That is not the way I understand things, no.

A basis is a set of vectors, each of which is a member of the space. It has two important properties:

1. The elements of the basis are linearly independent. None of them can be formed as a linear combination of the others.

2. The elements of the basis "span" the vector space. Any vector in the space can be formed as a linear combination of the basis elements.

Is this a definition that you have been exposed to?There is no guarantee of uniqueness. A vector space will not, in general, have a single unique basis. For instance, in our familiar three dimensional space, we can choose a basis of { "one meter east", "one meter north" and "one meter up" }. Or we could choose a basis of { "one foot east", "one furlong northwest", "one kilometer down" }.

 

[Karl Porter]

by linear combination are those under +,-,x,÷ and sqrt

 

[jbriggs444]

by linear combination are those under +,-,x,÷ and sqrt

A "linear combination" is the sum of some scalar multiple of each of the basis vectors.

Going back to our familiar three dimensional space with a basis of "one meter east", "one meter north" and "one meter up". A linear combination would be something like "2 meters east, 3 meters north, 4 meters up" with 2, 3 and 4 being the scalar multiples selected.

If your textbook does not cover this, it should. Wiki does. https://en.wikipedia.org/wiki/Vector_space#Basis_and_dimension

 

[Karl Porter]

rightt so the basis is the 1,1,1?

 

[jbriggs444]

rightt so the basis is the 1,1,1?

A basis is a set of vectors. Is "1" a vector? That is, is it a member of your vector space?

 

[Karl Porter]

A basis is a set of vectors. Is "1" a vector?

no?

 

[jbriggs444]

no?

Correct. Although technically, the constant function f(x) = 1 is a degree zero polynomial, it was pretty clear that you did not intend your "1" to denote that polynomial.

If you want to find a basis for the set of polynomials of degree 3 or less, there is an obvious one to choose:

{ x^3, x^2, x, 1 }

But that is not the only possible basis for the set of polynomials of degree 3 or less.

 

[Karl Porter]

Correct. If you want to find a basis for the set of polynomials of degree 3 or less, there is an obvious one to choose:

{ x^3, x^2, x, 1 }

But that is not the only possible basis for the set of polynomials of degree 3 or less.

what would other possible basis look like?
x^2,x,1?

 

[jbriggs444]

what would other possible basis look like?

For a simple example, you could use:

{ x^3, x^2, x, x+1 }

The four members are still linearly independent. You cannot form anyone as a linear combination of the others. But they still span the entire space of degree 3 or fewer polynomials.

 

[jbriggs444]

what would other possible basis look like?
x^2,x,1?

The proposed basis { x^2, x, 1 } does not work. How can you express $x^3$ as a linear combination of $x^2$, $x$ and $1$?

 

[Karl Porter]

For a simple example, you could use:

{ x^3, x^2, x, x+1 }

The four members are still linearly independent. You cannot form anyone as a linear combination of the others. But they still span the entire space of degree 3 or fewer polynomials.

right so if i wanted to calculate the basis of U which is the vector set and the basis of the subset is { x^3, x^2, x, 1 }. what does that show? The vector space U can have a higher degree than the subset right?

 

[Karl Porter]

The proposed basis { x^2, x, 1 } does not work. How can you express $x^3$ as a linear combination of $x^2$, $x$ and $1$?

so you can't times x^2 by x only a scalar quantity? or added a certain amount

 

[Mark44]

by linear combination are those under +,-,x,÷ and sqrt

so you can't times x^2 by x only a scalar quantity?

No. The only operations allowed in a linear combination are

• multiplication of a vector (or function) by a scalar such as 2, or -3, or 1.5;
• addition of vectors (or functions) of the basis.

Your textbook should have definitions of the terms you're asking about, together with examples.

 

[jbriggs444]

right so if i wanted to calculate the basis of U which is the vector set and the basis of the subset is { x^3, x^2, x, 1 }. what does that show? The vector space U can have a higher degree than the subset right?

You cannot calculate the basis of a vector space. No possible way. It cannot be done. There is no such thing.

You can find a basis of a vector space. It will not be unique.

The dimension of a vector space is the number of elements in a basis for that space.

 

[jbriggs444]

If we are clear on what a "basis" is and how it determines the "dimension" of a vector space we can move on to try to find a basis for the sub-space of third degree polynomials with roots at 0 or -1.

Recall that I had suggested using interpolating polynomials.

Are you ready to proceed? Or would you like to firm up your understanding of the idea of a "basis" and of the "dimension" of a vector space?

 

[vela]

Yes. But you have to make sure that the zero vector is a member of the subspace...

... oh right. Since we've already proved closure under multiplication by a scalar, that one we get for free.

This isn't quite correct. Closure of scalar multiplication says if $p(x) \in U$, then $0\cdot p(x) \in U$. You still need to show that there is in fact a $p(x)$ in $U$.

 

[Karl Porter]

If we are clear on what a "basis" is and how it determines the "dimension" of a vector space we can move on to try to find a basis for the sub-space of third degree polynomials with roots at 0 or -1.

Recall that I had suggested using interpolating polynomials.

Are you ready to proceed? Or would you like to firm up your understanding of the idea of a "basis" and of the "dimension" of a vector space?

we can have different basis but dimension will be the same?
but yes we can move onto interpolating.

 

[jbriggs444]

we can have different basis but dimension will be the same?

Yes. It will turn out that every basis will have the same number of members.

[This stuff is pretty much all reasoned out from my own intuition and stuff picked up over the years. I've never taken a formal course in linear algebra]

but yes we can move onto interpolating.

So here is the idea.

If we have plotted four distinct points on a graph, we can always find a degree 3 (or fewer) polynomial that matches those points.

This is the Lagrange interpolating polynomial for those points. It will be unique.

[My one and only published paper deals with these polynomials. So I have some affinity]

Suppose that we select two x coordinates in addition to -1 and 0. If we assign function values at those points, we can find a polynomial that fits those values. Can you use this idea to come up with two distinct polynomials in our sub-space?

You do not have to write those polynomials down. All you need for now is a proof that they exist and that they are linearly independent.

 

[Karl Porter]

Yes. It will turn out that every basis will have the same number of members.

[This stuff is pretty much all reasoned out from my own intuition and stuff picked up over the years. I've never taken a formal course in linear algebra]So here is the idea.

If we have plotted four distinct points on a graph, we can always find a degree 3 (or fewer) polynomial that matches those points.

This is the Lagrange interpolating polynomial for those points. It will be unique.

[My one and only published paper deals with these polynomials. So I have some affinity]

Suppose that we select two x coordinates in addition to -1 and 0. If we assign function values at those points, we can find a polynomial that fits those values. Can you use this idea to come up with two distinct polynomials in our sub-space?

You do not have to write those polynomials down. All you need for now is a proof that they exist and that they are linearly independent.

yea i lost you there. i was reading through the wiki page, do i need a range? or am i making the range
lets say i pick the points x=1 and x=2 how would i know the function value from these? am I substituting to p(x)