0
$\begingroup$

So I have the question Let F be a field and let L be the set of all polynomials f(x) element of F[X] satisfying the condition that deg(f) is even. Is L a subspace of F[X]?

I would say that L is not a subspace of F[X].

Taking the condition f(-a)=-f(a)

if we took f(x)=x^2 -2, which has an even degree, then

f(-a)=(-a)^2 -2 = a^2 -2 which does not equal -(x^2 -2)

Is my approach correct?

$\endgroup$
  • $\begingroup$ Where did the condition $f(-a)=-f(a)$ come from? That's what it means for the polynomial $f$ to be an odd function, but actually being an even function ($f(-x)=f(x)$) isn't the same as having even degree. $\endgroup$ – Nate Eldredge Jun 18 '14 at 18:31
  • $\begingroup$ You need to distinguish between the concepts of even degree and even as a function; the former means the leading coefficient of $f(x)$ is divisible by $2$; that latter means $f(-a) = f(a)$ for any $a \in F$. $\endgroup$ – Robert Lewis Jun 18 '14 at 18:32
  • $\begingroup$ @NateEldredge: whoops! Corrected! Thanks for pointing that out! $\endgroup$ – Robert Lewis Jun 18 '14 at 18:34
  • $\begingroup$ @RobertLewis, thank you, yes I understand the difference now, silly mistake. $\endgroup$ – cele Jun 18 '14 at 19:07
1
$\begingroup$

Even degree polynomial is NOT the same as even function. For example: $x^2+x$ is an even degree polynomial but not an even function. So if your set $L$ has even degree polynomials then it is NOT a subspace because $f(x)=x^2+x$ and $g(x)=-x^2$ are both in $L$ but $f+g \not\in L$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.