a) Show that $S=\{v_1, v_2, v_3\}$ is not a basis for $V$.
b) Find a basis for $V$.

a) Identity: $\cos^2u - \sin^2u = \cos2u$
Therefore, $v_3 = v_1 - v_2$, and so linearly dependent set - not a basis.

b) I really need a hint for this one. I'm thinking If I omit $v_3$, the remaining two vectors may form a basis for V. But this i only a guess and i'm not saying this from a point of understanding. I think I'm lacking knowledge about the characteristics of $V$.

  • $\begingroup$ Yeah, so how do I know $v_1$ and $v_2$ span $V$. $\endgroup$ – Bucephalus Aug 28 '17 at 1:15
  • $\begingroup$ I know they can be formed such that $k_1v_1+k_2v_2 = v$ for all k. Is that spanning the vector space $V$? And what is the characteristic of this vector space? For instance I know the characteristics of $R^3$. It is a space formed of three element tuples of real numbers for instance. Should I be looking at the periodicity or something for these functions? $\endgroup$ – Bucephalus Aug 28 '17 at 1:19
  • 1
    $\begingroup$ If $\{v_1, v_2, v_3\}$ span $V$, and $v_3$ is in the span of $\{v_1, v_2\}$, then certainly $v_1$ and $v_2$ span $V$, no? This doesn't automatically mean that they are a basis, but at least they form a spanning set. $\endgroup$ – Xander Henderson Aug 28 '17 at 1:19
  • $\begingroup$ Yes, you're right. That seems so obvious now. Thanks @Xander. $\endgroup$ – Bucephalus Aug 28 '17 at 1:21

Well, a basis for a vector space is a linearly independent spanning set.

Do you know that $\{\cos^2(x), \sin^2(x)\}$ span $V$? Hint: elements of $V$ are of the form $a_1\cos^2(x) + a_2\sin^2(x) + a_3\cos(2x)$. Can that be written as $b_1\cos^2(x) + b_2\sin^2(x)$ for some $b_1, b_2$? If so, $\{\cos^2(x),\sin^2(x)\}$ is a spanning set.

Is it linearly independent? If $c_1\cos^2(x) + c_2\sin^2(x) = 0$, what can you conclude about $c_1$ and $c_2$? Can you prove they must be $c_1=c_2=0$? If so, $\{\cos^2(x),\sin^2(x)\}$ is linearly independent.

If it's linearly independent and also a spanning set, then it's a basis (by definition).

  • 1
    $\begingroup$ hhhmm, thanks @MichaelHartley. I can see the spanning thing. Thanks for pointing that out. I can see that $v_1$ and $v_2$ are linearly independent because neither is a scalar multiple of the other. Thanks for your help. Cheers. $\endgroup$ – Bucephalus Aug 28 '17 at 1:24

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.