Linear Algebra (Change of Basis)

[DanielFaraday]

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:

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

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:

<br /> \left(<br /> \begin{array}{cccc}<br /> 1 &amp; 1 &amp; 1 &amp; 1 \\<br /> 1 &amp; -1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -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)<br />

Am I doing something wrong, or am I just confused about what a change of basis matrix is supposed to do?

 

[Office_Shredder]

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

 

[DanielFaraday]

I see. so a valid test would be:

<br /> \left(<br /> \begin{array}{cccc}<br /> 1 &amp; 1 &amp; 1 &amp; 1 \\<br /> 1 &amp; -1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -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)<br />

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!

 

[CoCoA]

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.