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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 $V$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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).

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    $\begingroup$ 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

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