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jbriggs444
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A basis is a set of vectors. Is "1" a vector? That is, is it a member of your vector space?Karl Porter said: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 said:rightt so the basis is the 1,1,1?
no?jbriggs444 said:A basis is a set of vectors. Is "1" a vector?
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.Karl Porter said:no?
what would other possible basis look like?jbriggs444 said: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.
For a simple example, you could use:Karl Porter said:what would other possible basis look like?
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 said:what would other possible basis look like?
x^2,x,1?
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?jbriggs444 said: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.
so you can't times x^2 by x only a scalar quantity? or added a certain amountjbriggs444 said: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 said:by linear combination are those under +,-,x,÷ and sqrt
No. The only operations allowed in a linear combination areKarl Porter said:so you can't times x^2 by x only a scalar quantity?
You cannot calculate the basis of a vector space. No possible way. It cannot be done. There is no such thing.Karl Porter said: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?
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##.jbriggs444 said: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.
we can have different basis but dimension will be the same?jbriggs444 said: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?
Yes. It will turn out that every basis will have the same number of members.Karl Porter said:we can have different basis but dimension will be the same?
So here is the idea.Karl Porter said:but yes we can move onto interpolating.
yea i lost you there. i was reading through the wiki page, do i need a range? or am i making the rangejbriggs444 said: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.
OK, Let's pick the points x=1 and x=2.Karl Porter said: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)
x3 x2 x 1jbriggs444 said: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.
Yep. I did lose you there, it seems. Let me try to restore some context and make sure we are straight on the immediate goalsKarl Porter said:x3
and those numbers can be random? px=3,px=4jbriggs444 said: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.
They can be random. However, if it were me, I'd try to choose some simple values.Karl Porter said:and those numbers can be random? px=3,px=4
Actually, you get two equations from the facts that p(0) = 0 and p(-1) = 0, but more about that later.Karl Porter said: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?
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.jbriggs444 said:Perhaps you want to work in the ring of formal polynomials or in the ring of polynomial functions over the reals.
Two points here:Karl Porter said: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?
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.jbriggs444 said: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.
I don't think this will be helpful in what you're trying to doKarl Porter said:but yes we can move onto interpolating.
and then for the basis i sub these a2 and a1 in forms of a3 into p(x)Karl Porter said: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
The part above is OK.Karl Porter said: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
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.Karl Porter said:so for all values of p(x) in u they can be represented by two parameters
(a3-a1)x3+a2x2+a1x
I have no idea what you're trying to say here.Karl Porter said:and then for the basis i sub these a2 and a1 in forms of a3 into p(x)
Basis would be {x^3+x^2,x^3-x}Mark44 said: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.
Yes.Karl Porter said:Basis would be {x^3+x^2,x^3-x}