Solving Direct Sum Problem: Pn=PE \oplus PO

[ahamdiheme]

Homework Statement

Let P_n denote the vector space of polynomials of degree less than or equal to n, and of the form p(x)=p₀+p₁x+...+p_nxⁿ, where the coefficients p_i are all real. Let P_E denote the subspace of all even polynomials in P_n, i.e., those that satisfy the property p(-x)=p(x). Similarly, let P_O denote the subspace of all odd polynomials, i.e., those satisfying p(-x)=p(x). Show that P_n=P_E$$\oplus$$P_O.

Homework Equations

Conditions for direct sum.

The Attempt at a Solution

P_E=p₁x+...+p_n-1x^n-1
P_O=p₀+p₂x²+...+p_nxⁿ
such that n$$\in$${even real numbers}
therefore, P_E$$\bigcap$$P_O={$$\phi$$}
and P_E+P_O=p₀+p₁x+p₂x²+...+p_n-1+x^n-1+p_nxⁿ=P_n

is this sufficient to show that P_E$$\oplus$$P_O=P_n ?

I'm not so sure of what i have done and i know that my notations may be faulty, help please

 

[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.

 

[Mark44]

You have the even and odd polynomials switched.

PE=p1x+...+pn-1xn-1
PO=p0+p2x2+...+pnxn

I would represent P_E as p₀ + p₁x + p₂x^2 + ... + p_2nx^2n
and P_O as p₁x + p₃x^3 + ... + p_2n-1x^(2n-1), in both cases where n is in the nonnegative integers.

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

 

[Dick]

You have the even and odd polynomials switched.I would represent P_E as p₀ + p₁x + p₂x^2 + ... + p_2nx^2n
and P_O as p₁x + p₃x^3 + ... + p_2n-1x^(2n-1), in both cases where n is in the nonnegative integers.

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

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

 

[ahamdiheme]

You have the even and odd polynomials switched.


I would represent P_E as p₀ + p₁x + p₂x^2 + ... + p_2nx^2n
and P_O as p₁x + p₃x^3 + ... + p_2n-1x^(2n-1), in both cases where n is in the nonnegative integers.

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

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