Linear Algebra (Change of Basis)

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Homework Help Overview

The problem involves determining whether the set G={1+x,1-x,1-x²,1-x³} forms a basis for the polynomial space P3, given the standard basis E={1, x, x², x³}. The original poster attempts to find the change of basis matrix S from G to E and is exploring the implications of this matrix.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the validity of the change of basis matrix and its role in proving G is a basis. There are questions about the representation of polynomials as vectors and the requirements for linear independence and spanning.

Discussion Status

Some participants have offered guidance on how to approach proving that G is a basis, suggesting that linear independence should be shown rather than relying solely on the change of basis matrix. Multiple interpretations of the problem are being explored, particularly regarding the representation of polynomials in vector form.

Contextual Notes

There is an emphasis on the dimensionality of P3 and the need to demonstrate that G spans the space, given that it contains four elements. Participants are also questioning the assumptions made about the change of basis matrix's application.

DanielFaraday
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Homework Statement



Let E={1, x, x2,x3} be the standard ordered basis for the space P3. Show that G={1+x,1-x,1-x2,1-x3} is also a basis for P3, and write the change of basis matrix S from G to E.

Homework Equations

The Attempt at a Solution



Here's what I got:

[tex] S_E^G=\left(<br /> \begin{array}{cccc}<br /> 1 & 1 & 1 & 1 \\<br /> 1 & -1 & 0 & 0 \\<br /> 0 & 0 & -1 & 0 \\<br /> 0 & 0 & 0 & -1<br /> \end{array}<br /> \right)[/tex]

Now, to prove that this is also a basis, I just need to show that it has an inverse, right?

Here's the problem. If the above is correct, then when you multiply it by G you should get E, right? After all, it is the "change of basis matrix S from G to E". However, this isn't the case:

[tex] \left(<br /> \begin{array}{cccc}<br /> 1 & 1 & 1 & 1 \\<br /> 1 & -1 & 0 & 0 \\<br /> 0 & 0 & -1 & 0 \\<br /> 0 & 0 & 0 & -1<br /> \end{array}<br /> \right).\left(<br /> \begin{array}{c}<br /> 1+x \\<br /> 1-x \\<br /> 1-x^2 \\<br /> 1-x^3<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{c}<br /> 4-x^2-x^3 \\<br /> 2 x \\<br /> -1+x^2 \\<br /> -1+x^3<br /> \end{array}<br /> \right)[/tex]

Am I doing something wrong, or am I just confused about what a change of basis matrix is supposed to do?
 
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Your vectors in your representation in R4 are going to have just numbers, not polynomials in their entries. So (1+x, 1-x, 1-x2, 1-x3) isn't a vector, instead the vectors are supposed to be representing polynomials, for example 1-x is represented as (1,-1,0,0).

Your change of basis matrix is correct. Don't use it to show G is a basis though... you know P3 has dimension 4, so show G is linearly independent and hence since it has 4 elements it must span the space also
 
I see. so a valid test would be:

[tex] \left(<br /> \begin{array}{cccc}<br /> 1 & 1 & 1 & 1 \\<br /> 1 & -1 & 0 & 0 \\<br /> 0 & 0 & -1 & 0 \\<br /> 0 & 0 & 0 & -1<br /> \end{array}<br /> \right).\left(<br /> \begin{array}{c}<br /> 0 \\<br /> 0 \\<br /> 1 \\<br /> 0<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> -1 \\<br /> 0<br /> \end{array}<br /> \right)[/tex]

And the fact that I get (1,0,-1,0) says that I have 1-x^2, which is what I would expect.

Cool, I think I got it. Thanks!
 
Don't use it to show G is a basis though... you know P3 has dimension 4, so show G is linearly independent and hence since it has 4 elements it must span the space also

Although that is quite valid, in this problem I would find it simpler to show that G spans P3 and thus is linearly independent.
 

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