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 $V$.
Best Answer
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).