0
$\begingroup$

I'm having some trouble in solving some exercises related to vector spaces, and I can't even start the solution.

I need to check if the sets given span the same subset of the vector space $V$:

(i) $S_1 = ({\sin^2\theta, \cos^2\theta, \sin\theta \cos\theta})$ and $ S_2 = ({1, \sin \theta, \cos \theta})$, when $V = C(\mathbb R)$ (all the continuous functions).

(ii) $S_1 = (1,t,t^2,t^3)$ and $S_2 = (1, 1+t,1-t,1-t-t^2)$, when $V = P_3(\mathbb R)$ (all the polynomials with degree $\leq 3$).

Thanks in advance.

$\endgroup$
4
$\begingroup$

A positie answer can be obtained by showing (in both directions!) that the generators of one span can be written as linear combinat6ions of the other generators.

In (ii) it is immediate that $1,1+t,1-t,1-t-t^2$ can be written as linear combination of $1,t,t^2,t^3$. In the other direction for example $t=1-(1-t)$ or $t=\frac12(1+t)-\frac12(1-t)$. However, $t^3$ cannot be written as linear combination of $1,1+t,1-t,1-t-t^2$; the reason is that a linear combination cannot have higher degree than the maximal degree of the summands.

In (i) we have for example $\sin^2\theta+\cos^2\theta = 1$, but one needs a somewhat tricky observation to show that the spans differ in the end: We have $f(\theta)=f(\theta+2\pi)$ for all $f\in S_2$; and we have (why?) $f(\theta)=f(\theta+\pi)$ for all $f\in S_1$. As $\sin\theta$ does not have period $\pi$, we conclude that it is not in $S_1$.

$\endgroup$

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.