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I have to find a basis for and the dimension of the vector space $$V = \mathrm{span}\{1, \sin^2 x, \cos 2x, \cos^2 x\},$$ where $V \subset C(\mathbb{R})$.

I know that $1 = \sin^2 x + \cos^2 x$, so $1$, $\sin^2 x$, $\cos^2 x$ are not linearly independent. But I don't know whether $\sin^2 x$, $\cos^2 x$, $\cos 2x$ are linearly independent. Can someone help me?

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Yes, remember that $\cos(2x) = 1 - 2\sin^{2}(x)$.

So we have that $\cos^{2}(x) = 1 - \sin^{2}(x)$ from the identity you mentioned (i.e., $\cos^{2}(x)$ can be written as a linear combination of $1$ and $\sin^{2}(x)$), and similarly using the double angle identity I wrote above, $\cos(2x)$ can be written as a linear combination of $1$ and $\sin^{2}(x)$. What does this tell you?

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  • $\begingroup$ Ok, $Dim(V) < 3$ and I guess $Dim(V) = 2$. Am I right? $\endgroup$ – anlex Nov 15 '14 at 20:08
  • $\begingroup$ @anlex Well, we established that $\cos(2x)$ and $\cos^{2}(x)$ are both linearly dependent on $\{1, \sin^{2}(x) \}$. Now the question you have to ask yourself is: is $\{1, \sin^{2}(x) \}$ a linearly independent or dependent set? If independent, then you are right that the dimension of $V$ is $2$. If they are dependent, then the dimension is $1$. So, does there exist scalars $c_{1}, c_{2}$ (not both $0$) such that $c_{1}1 + c_{2} \sin^{2}(x) = 0$ for all $x$? $\endgroup$ – layman Nov 15 '14 at 20:10
  • $\begingroup$ But why bisis of $V$ cannot be $\{cos(2x), cos^2 x\}$? $\endgroup$ – anlex Nov 15 '14 at 20:14
  • $\begingroup$ @anlex Is $\{ \cos(2x), \cos^{2}(x) \}$ a linearly independent set? Yes, which means that is also a basis. Don't forget that we can have more than one basis for the same set. $\endgroup$ – layman Nov 15 '14 at 20:18
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    $\begingroup$ Ok, and there are no non-zero scalars $c1, c2$ such $c_1 1 + c_2 sin^2(x) = 0$ fo all x, because we have $c_1 1 + c_2 0 = 0$ and $c_1 1 + c_2 1 = 0$ which means $c_1 = c_2 = 0$. Right? $\endgroup$ – anlex Nov 15 '14 at 20:23

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