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Here's one of the examples he has.

Let [tex]P(F)[/tex] denote all polynomials with coefficients in [tex]F[/tex] where F is a field.

Let [tex]U_e[/tex] denote the subspace of [tex]P(F)[/tex] consisting of all polynomials of the form [tex]p(z) = \sum_{i=0}^n a_{2i}z^{2i}[/tex]

and let [tex]U_o[/tex] denote the subspace of P(F) consisting of all polynomials [tex]p[/tex] of the form: [tex]p(z) =

\sum_{i=0}^{n} a_{2i+1}z^{2i+1}[/tex]

You should verify that [tex]P(F) = U_e \oplus U_o[/tex]

Now the other examples he had were kind of trivial (such as [tex]\mathbb{R}^2 = U \oplus W[/tex] where [tex]U = \{ (x,0) | x \in \mathbb{R} \}[/tex] and [tex]W = \{(0,y) | y \in \mathbb{R} \}[/tex]) since we simply showed the uniqueness of each component.

For this one, I get rattled up in notation and not sure where to head. Here's what I was thinking

let [tex]q \in U_e + Uo[/tex]. Then: [tex]q(x) = \underbrace{a_0 + a_2 x^2 + \cdots + a_{2n}x^{2n}}_{\in U_e} + \underbrace{a_1x + a_3x^3 + \cdots a_{2n+1}x^{2n+1}}_{\in U_o}[/tex]

Then I assume [tex]a(x)[/tex] can be represented by different coefficients, say [tex]b_0, b_1, ..., b_n[/tex]. Since both these representations are equal to [tex]a(x)[/tex], then the two representations are equal. Then we equate the coefficients and we arrive at a contradiction.

Sound good? Sorry for the long winded post. I appreciate all the help so far! (And that last thread .. that was pretty silly of me to mess up!)