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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?

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    $\begingroup$ What "corresponding entry" are you pairing up with? The powers of $z$ available in the different subspaces are disjoint. $\endgroup$ Commented Aug 23, 2013 at 1:21
  • $\begingroup$ Here is what I had in mind: when you add the first p(z) polynomial listed to the second p(z), you can add a_0 to the first entry of the second p(z) and so on. By doing this you can make the "corresponding entries" add up to zero. $\endgroup$ Commented Aug 23, 2013 at 4:17
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    $\begingroup$ You mean matching up $a_0$ with $a_1$? But if $a_0 = -a_1 = 1$ then we get $p(z) = 1 - z + ``o(z^2)"$ which is clearly not $0 \in P(F)$. No matter how you choose $a_0$ and $a_1$, they're never going to cancel when you add the polynomials because $a_1$ is multiplying a power of $z$. $\endgroup$ Commented Aug 23, 2013 at 4:20
  • $\begingroup$ I am really not getting this. Can you reword this. Where is the p(z) formula coming from? The last sentence doesn't make any sense to me. $\endgroup$ Commented Aug 25, 2013 at 13:46

2 Answers 2

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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 $b_{k} = 0 \quad \forall \; 0\leq k \leq m$.

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  • $\begingroup$ I still don't see why the first thing you said is true. $\endgroup$ Commented Aug 23, 2013 at 4:08
  • $\begingroup$ I still don't see why the first thing you said is true. In fact, it seems patently false. Let z = -1. Let a_0 = -1. Let a_1 = -1, a_2 = 1, a_3 = 1. Doing something like that you can get pairs of -1 and 1. You get a total of 0. $\endgroup$ Commented Aug 23, 2013 at 4:14
  • $\begingroup$ @LarryJohnson But you want it to work for all $z$. $\endgroup$ Commented Aug 23, 2013 at 5:42
  • $\begingroup$ So maybe I'm not understanding. I thought if there was more than one way to arrive at the 0 vector, then all bets are off. $\endgroup$ Commented Aug 23, 2013 at 16:19
  • $\begingroup$ @LarryJohnson: the zero vector in the vector space $P(F)$ is the zero polynomial $p(z)=0$. We're not choosing a particular $z$ at any stage. $\endgroup$ Commented Aug 25, 2013 at 14:09
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There is a very simple way to justify what Vishal is saying. Think of a basis for Pe and a basis for P0, the standard one using monomials. One uses even powers and the other uses odd powers. Together they form a basis for the whole space . But they are linearly independent, so how can they add up to be Zero without all coefficients being zero? impossible! cheers,

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