# Linear Algebra (Change of Basis)

1. Jul 20, 2009

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

Here's what I got:

$$S_E^G=\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right)$$

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:

$$\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right).\left( \begin{array}{c} 1+x \\ 1-x \\ 1-x^2 \\ 1-x^3 \end{array} \right)=\left( \begin{array}{c} 4-x^2-x^3 \\ 2 x \\ -1+x^2 \\ -1+x^3 \end{array} \right)$$

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

2. Jul 20, 2009

### Office_Shredder

Staff Emeritus
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

3. Jul 20, 2009

I see. so a valid test would be:

$$\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right).\left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \right)=\left( \begin{array}{c} 1 \\ 0 \\ -1 \\ 0 \end{array} \right)$$

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!

4. Jul 21, 2009

### CoCoA

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