# Homework Help: Direct Sum Problem

1. Oct 25, 2008

### ahamdiheme

1. The problem statement, all variables and given/known data
Let Pn denote the vector space of polynomials of degree less than or equal to n, and of the form p(x)=p0+p1x+...+pnxn, where the coefficients pi are all real. Let PE denote the subspace of all even polynomials in Pn, i.e., those that satisfy the property p(-x)=p(x). Similarly, let PO denote the subspace of all odd polynomials, i.e., those satisfying p(-x)=p(x). Show that Pn=PE$$\oplus$$PO.

2. Relevant equations
Conditions for direct sum.

3. The attempt at a solution

PE=p1x+...+pn-1xn-1
PO=p0+p2x2+...+pnxn
such that n$$\in$${even real numbers}
therefore, PE$$\bigcap$$PO={$$\phi$$}
and PE+PO=p0+p1x+p2x2+...+pn-1+xn-1+pnxn=Pn

is this sufficient to show that PE$$\oplus$$PO=Pn ?

I'm not so sure of what i have done and i know that my notations may be faulty, help please
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 25, 2008

### Dick

Yes, that works. But the intersection of PE and PO isn't {phi}. It's {0}. You've shown that everything in Pn can be written as a sum of something from PE and something from PO. I think the other thing you need to show is that those choices are unique.

Last edited: Oct 25, 2008
3. Oct 25, 2008

### Staff: Mentor

You have the even and odd polynomials switched.
I would represent PE as p0 + p1x + p2x^2 + ... + p2nx^2n
and PO as p1x + p3x^3 + ... + p2n-1x^(2n-1), in both cases where n is in the nonnegative integers.

To show that Pn = PE $$\oplus$$ PO, don't you have to show that any arbitrary polynomial in Pn can be written as the sum of one polynomial from PE and another from PO? I'm a little rusty on this, so there might be some more that you have to show.

4. Oct 25, 2008

### Dick

Yes there is, you want to show that this decomposition is unique. E.g. Pn=Pn+PE. But that's not a direct sum.

5. Oct 26, 2008

### ahamdiheme

I figured i switched them, i guess i was a litle sleepy while typing. Thanks a lot though