Direct Sum of Subspaces

I'm going through Axler's book and just got introduced the concept of sums of subspaces and the direct sums.

Here's one of the examples he has.

Let $$P(F)$$ denote all polynomials with coefficients in $$F$$ where F is a field.
Let $$U_e$$ denote the subspace of $$P(F)$$ consisting of all polynomials of the form $$p(z) = \sum_{i=0}^n a_{2i}z^{2i}$$

and let $$U_o$$ denote the subspace of P(F) consisting of all polynomials $$p$$ of the form: $$p(z) = \sum_{i=0}^{n} a_{2i+1}z^{2i+1}$$

You should verify that $$P(F) = U_e \oplus U_o$$

Now the other examples he had were kind of trivial (such as $$\mathbb{R}^2 = U \oplus W$$ where $$U = \{ (x,0) | x \in \mathbb{R} \}$$ and $$W = \{(0,y) | y \in \mathbb{R} \}$$) 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 $$q \in U_e + Uo$$. Then: $$q(x) = \underbrace{a_0 + a_2 x^2 + \cdots + a_{2n}x^{2n}}_{\in U_e} + \underbrace{a_1x + a_3x^3 + \cdots a_{2n+1}x^{2n+1}}_{\in U_o}$$

Then I assume $$a(x)$$ can be represented by different coefficients, say $$b_0, b_1, ..., b_n$$. Since both these representations are equal to $$a(x)$$, 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!)

Office_Shredder
Staff Emeritus
Gold Member
I don't think unique representation of a sum of elements is a good way to go here. Instead you can prove these two things:
$$U_e + U_o = P(F)$$

and

$$U_e \cap U_o = \emptyset$$

Which requires a far smaller combination of handwaving/slogging through ridiculous details. Often (not always) this is the easier way of proving two things are the direct sum of a vector space. Although what you've posted is pretty good also

Oh thanks a lot! I just noticed they had this proved a few pages after. Ah well different means to the same goal I suppose.

Sorry to bring up an old thread. I was just wondering how Office_Shredder's approach works.

Specifically, how do i directly show that $$U_e \cap U_o = \bold{0}$$ ?

Would I suppose $$a(x) \in U_e \cap U_o$$ and come to the conclusion that the only way this could occur was if the coefficients were equal to 0?

$$a_0 + a_2x^2 + a_4x^4 + \cdots + a_{2n}x^{2n} = a_1 + a_3x^3 + \cdots + a_{2n+1}x^{2n+1}$$

On the LHS, the odd-superscripted coefficents is equal to 0 so the RHS must be equal to 0 as well. And similarly for the even-superscripted coefficients.

Would that be the way to go?

jbunniii
$$a_0 + a_2x^2 + a_4x^4 + \cdots + a_{2n}x^{2n} = a_1 + a_3x^3 + \cdots + a_{2n+1}x^{2n+1}$$