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I have to determine the dimension and base of this vector space:

$\langle 2,t,\sin^2(t),\cos^2(t)\rangle $

I did this:

$\langle 2,t,\sin^2(t),\cos^2(t)\rangle =\langle 2,t,\sin^2(t)+\cos^2(t),\cos^2(t)\rangle =\langle 2,t,1,\cos^2(t)\rangle =\langle 1,t,\cos^2(t)\rangle $

So this is the base and the dimension is $3$. Am I right? How can I prove those 3 are independent of each other?

Regards, Kevin

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    What's the domain of these functions? If it is $\mathbb{R}$, use the fact that $\alpha+\beta t$ has an image being the whole $\mathbb{R}$ if $\beta\neq0$. Then this combination cannot be equal to $\cos^2t$.2012-01-19

2 Answers 2

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Implicit in this answer is the assumption that $t \in \mathbb R$

To prove linear independence, you need to prove that,

$\lambda +\mu t+\nu \cos^2 t=0 \implies \lambda=\mu=\nu=0$ But note that, the equation is true for all values of $t$. Take $t=0$ and $t=2\pi$. You'll have $\lambda+\nu=0$ $\lambda+2\pi \mu+\nu=0$ This proves that $\boxed{\mu=0}$. Nextly, set $t=\dfrac{\pi}{2}$ and you'll have that, $\boxed{\lambda=0}$ and hence, $\boxed{\nu=0}$.

This completes the picture.

I posted a completely idiotic answer. Thanks are due to OP for pointing out, I have bluffed!

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    Oh Sorry! I'll edit it right away.2012-01-19
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You are on the right track. It remains to show that $\cos^2$ is not an affine function. There are many ways of doing this.

For instance, notice that an affine function is uniquely determined by its output for two different values of the input variable $t$.

If $\cos^2$ was an affine function then $\cos^2(\pi/2)=0$ and $\cos^2(3\pi/2)=0$ would imply $\cos^2(t)=0$ for every $t\in\mathbb R$. But $\cos^2(0)=1$.