I'm working through Axler's "Linear Algebra Done Right". On page 15, it gives this example:
Consider the vector space $P(F)$ of all polynomials with coefficients in $F$ (ie., coefficients taken from the real numbers or complex numbers). Let $U_e$ denote the subspace of $P(F)$ consisting of all polynomials $p$ of the form:
$$p(z) = a_0 + (a_2)z^2 + … (a_{2m})z^{2m}$$
and let $U_0$ denote the subspace of all polynomials $p$ of the form:
$$p(z) = (a_1)z + (a_3)z^3 + … + (a_{2m+1})z^{2m+1};$$
$m$ is a nonnegative integer and the coefficients are from the reals (to keep things simple).
Next, it says I should verify that $P(F)$ is a direct sum of $U_e \oplus U_0$.
Well, to do this I wanted to use a theorem that occurs a few pages later, namely:
if $U_1 … U_n$ are subspaces of $V$. Then $V$ is a direct sum of $U_1 … U_n$ iff:
a. $V = U_1 + … U_n$
b. the only way to write $0$ as a sum $u_1 + … + u_n$, where each $u_j$
is in $U_j$, is by taking all the $u_j$ in $U_j$, is by taking all the
$u_j$'s equal to $0$.
I wanted to use this theorem to show $P(F)$ is a direct sum of $U_e \oplus U_0$.
(a) — from the above theorem — is clearly satisfied. What has me stumped in part (b). It seems to me that that are multiple ways to write the $0$ for the sum of $u_e + u_0$, $u_e \in U_e$, $u_0 \in U_0$. One way to get $0$ is make each coefficient $0$. Or you could make sure that each entry from $U_e$ is paired off with
the corresponding entry from $U_0$, and that together they add up to $0$. When $U_e$ and $U_0$ are added together, this would give a $0$ too.
But this can't be right b/c then $P(F)$ would not be a direct sum of $U_e$ and $U_0$.
Any help?
Best Answer
You cannot add two non-zero elements, one of space $U_{e}$ and the other of space $U_{o}$ to get $0$.
Suppose $p + q = 0$ where $p \in U_{e}$ and $q \in U_{o}$. Then
$$ p(z) = \sum\limits_{k=0}^{n} a_{k}z^{2k} $$
and
$$ q(z) = \sum\limits_{k=0}^{m} b_{k}z^{2k+1} $$
where $a_{k},b_{k} \in \mathbb{F}$.
Then
$$ (p+q)(z) = \sum\limits_{k=0}^{\max{(m,n)}}c_{k}z^{k}$$
where $c_{k} = a_{k}$ if $k$ is even and is $c_{k} = b_{k}$ when $k$ is odd.
Now the only way for $(p + q)(z)\;$ to be zero $\; \forall z \in \mathbb{F}\,$ is if $a_{k} = 0 \quad \forall\; 0\leq k \leq n$ and $b_{k} = 0 \quad \forall \; 0\leq k \leq m$.