# Vector space base and dimension

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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You're right as far as you have gone. – user21436 Jan 19 '12 at 17:07
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 $\cos^2t$. – Damian Sobota Jan 19 '12 at 17:12

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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So you have proved that any pair of three given elements is linearly independent, but it doesn't mean that all they three together are linearly independent. Compare: $\mathbb{R}^2$ and vectors: $[1,0],[0,1],[1,1]$. – Damian Sobota Jan 19 '12 at 17:28
Oh Sorry! I'll edit it right away. – user21436 Jan 19 '12 at 17:29

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 $\cos^2(0)=1$.

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