Consider the subset U ⊂ R3[x] defined as

[jbriggs444]

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)

OK, Let's pick the points x=1 and x=2.

You do not need to know the function values at x=1 and x=2. Make some up.

Make up two function values for your first proposed basis member to use.
Make up two function values for your second proposed basis member to use.

 

[Karl Porter]

OK, Let's pick the points x=1 and x=2.

You do not need to know the function values at x=1 and x=2. Make some up.

Make up two function values for your first proposed basis member to use.
Make up two function values for your second proposed basis member to use.

x³ x² x 1
2x³ 2x² 2x 1
?

 

[jbriggs444]

x³

Yep. I did lose you there, it seems. Let me try to restore some context and make sure we are straight on the immediate goals

We want to find a basis for our vector sub-space of degree 3 polynomials with roots at 0 and -1.

We know that the full space of degree 3 polynomials has ${x^3, x^2, x, 1}$ as a basis. Four vectors in the basis, so the dimension of the full space is 4.

There are many other possible sets of 4 polynomials that also qualify to be a basis for the full space. We do not particularly care. As long as we have one set of linearly independent vectors that spans the space, we know we have a basis and we know that the dimension of the space is the number of vectors in that basis.

But now we want to know the dimension of the sub-space. That basis that we had in hand (${x^3, x^2, x, 1}$) will not help us out. That set of vectors is not a basis for the sub-space because none of the vectors in that supposed basis are elements of the sub-space.

We want to find some polynomials that are elements of the subspace.

The suggestion is to use Lagrange Interpolating Polynomials.

In order to come up with a Lagrange Interpolating Polynomial for degree three polynomials, we need four x coordinates and four corresponding values for p(x).

If we want the resulting polynomials to have roots at 0 and at -1, two of the four x coordinates are already chosen for us:

x = 0 p(x) = 0
x = -1 p(x) = 0 [whoopsie, had that value as 1 initially]

Now we have to find two more points and values. You've suggested x=1 and x=2. So we have two more table entries to fill.

x = 1 p(x) = ?
x = 2 p(x) = ?

We need two different polynomials. So you'll need to fill in those two table entries twice. We need four numbers in total.

 

[Karl Porter]

Yep. I did lose you there, it seems. Let me try to restore some context and make sure we are straight on the immediate goals

We want to find a basis for our vector sub-space of degree 3 polynomials with roots at 0 and -1.

We know that the full space of degree 3 polynomials has ${x^3, x^2, x, 1}$ as a basis. Four vectors in the basis, so the dimension of the full space is 4.

There are many other possible sets of 4 polynomials that also qualify to be a basis for the full space. We do not particularly care. As long as we have one set of linearly independent vectors that spans the space, we know we have a basis and we know that the dimension of the space is the number of vectors in that basis.

But now we want to know the dimension of the sub-space. That basis that we had in hand (${x^3, x^2, x, 1}$) will not help us out. That set of vectors is not a basis for the sub-space because none of the vectors in that supposed basis are elements of the sub-space.

We want to find some polynomials that are elements of the subspace.

The suggestion is to use Lagrange Interpolating Polynomials.

In order to come up with a Lagrange Interpolating Polynomial for degree three polynomials, we need four x coordinates and four corresponding values for p(x).

If we want the resulting polynomials to have roots at 0 and at -1, two of the four x coordinates are already chosen for us:

x = 0 p(x) = 0
x = -1 p(x) = 1

Now we have to find two more points and values. You've suggested x=1 and x=2. So we have two more table entries to fill.

x = 1 p(x) = ?
x = 2 p(x) = ?

We need two different polynomials. So you'll need to fill in those two table entries twice. We need four numbers in total.

and those numbers can be random? px=3,px=4

 

[jbriggs444]

and those numbers can be random? px=3,px=4

They can be random. However, if it were me, I'd try to choose some simple values.

Remember, you need to pick four numbers.

One number for the value of the first polynomial evaluated at x=1.
One number for the value of the first polynomial evaluated at x=2.

One number for the value of the second polynomial evaluated at x=1.
One number for the value of the second polynomial evaluated at x=2.

 

[Mark44]

IMO, the thread has gone astray from the original intent in post #1, which I have copied below.

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?

Actually, you get two equations from the facts that p(0) = 0 and p(-1) = 0, but more about that later.

Perhaps you want to work in the ring of formal polynomials or in the ring of polynomial functions over the reals.

I doubt very much that rings of polynomials or polynomial functions are germane to this question, something that you (@jbriggs444) suspected in a post that followed this quote.

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?

Two points here:
1) the first row above is correct, but the second is not.
From the equation p(0) = 0, you get $a_3 \cdot 0 + a_2 \cdot 0 + a_1 \cdot 0 + a_0 = 0$, or more simply, $1a_0 = 0$. This means that the second row of your matrix needs to be 0 0 0 1 | 0.
2) The rank of the corrected matrix is 2. If you row-reduce that matrix, you get a basis for your function space.

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.

This might be an approach, but there is one that is much simpler, based on the matrix derived from the equations p(-1) = 0 and p(0) = 0.

The corrected matrix is
$\begin{bmatrix} -1 & 1 & -1 & 1 & | & 0 \\ 0 & 0 & 0 & 1 & | & 0\end{bmatrix}$
We can simplify this a bit by multiplying the top row by -1, to get:
$\begin{bmatrix} 1 & -1 & 1 & -1 & | & 0 \\ 0 & 0 & 0 & 1 & | & 0\end{bmatrix}$
By adding the 2nd row to the first, we get the matrix in reduced row-echelon form.
$\begin{bmatrix} 1 & -1 & 1 & 0 & | & 0 \\ 0 & 0 & 0 & 1 & | & 0\end{bmatrix}$

The first and second rows represent these equations:
$a_3 - a_2 + a_1 ...= 0$
$.....a_0 = 0$
So $a_3$ depends on $a_2$ and $a_1$, which means that $a_2$ and $a_1$ can be chosen arbitrarily. $a_0$ depends on no other parameter, and is zero.

So
$a_3 = 1a_2 - 1a_1$
$a_2 = 1a_2 + 0a_1$ (It's arbitrary.)
$a_1 = 0a_2 +1a_1$ (Also arbitrary.)
$a_0 = 0a_2 + 0a_1$
If you squint your eyes a bit, you might be able to spot the coordinates of two functions that form a basis for U.

but yes we can move onto interpolating.

I don't think this will be helpful in what you're trying to do

 

[pasmith]

You can write a polynomial of degree at most 3 with roots at 0 and -1 as x(1+x)p(x) where p is a polynomial of degree at most 1. Closure under vector addition and scalar multiplication then follows straightforwardly: x(1+x)p(x) + \alpha x(1+x)q(x) = x(1+x)(p(x) + \alpha q(x)).

 

[Karl Porter]

ok so i went through a few more examples and ended up with Dim 2

at x=0 i ended up with the equation a₀=0
at x=-1 i get a₂=a₁+a₃
so for all values of p(x) in u they can be represented by two parameters
(a₃-a₁)x³+a₂x²+a₁x

 

[Karl Porter]

ok so i went through a few more examples and ended up with Dim 2

at x=0 i ended up with the equation a₀=0
at x=-1 i get a₂=a₁+a₃
so for all values of p(x) in u they can be represented by two parameters
(a₃-a₁)x³+a₂x²+a₁x

and then for the basis i sub these a₂ and a₁ in forms of a₃ into p(x)

 

[Mark44]

ok so i went through a few more examples and ended up with Dim 2

at x=0 i ended up with the equation a₀=0
at x=-1 i get a₂=a₁+a₃

The part above is OK.

so for all values of p(x) in u they can be represented by two parameters
(a₃-a₁)x³+a₂x²+a₁x

No. Take a closer look at post #56. I've laid it all out for you, including how to get a set of basis functions.

and then for the basis i sub these a2 and a1 in forms of a3 into p(x)

I have no idea what you're trying to say here.

 

[Karl Porter]

The part above is OK.
No. Take a closer look at post #56. I've laid it all out for you, including how to get a set of basis functions.
I have no idea what you're trying to say here.

Basis would be {x^3+x^2,x^3-x}

 

[Mark44]

Basis would be {x^3+x^2,x^3-x}

Yes.
Let's call these $p_1(x)$ and $p_2(x)$, respectively.
As a check that these functions are in U, it would be good to verify that $p_1(0) = 0, p_1(-1) = 0$ and that $p_2(0) = 0, p_2(-1) = 0$.