Linear Independence/Basis for the span of a set Any hints for this question?
Let V be the complex vector space of continuous functions $f: \mathbb{R}\rightarrow \mathbb{C}$ and let
$v_{1}(t)=e^{2it}, v_{2}(t)=\cos^2 t, v_{3}(t) =\sin 2t, v_{4}(t) = 1+e^{-2it}$
(1) Show that the set $S = \{v_{1}, v_{2}, v_{3}, v_{4}\}$ is linearly independent
(2) Find a basis for the span of S.
(1) Now, for a set to be linearly independent, we require that for $a_{1},..,a_{4} \in F$, for
$a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3}+a_{4}v_{4} = 0 \Rightarrow a_{1}=...=a_{4}=0$ 
So we consider, $a_{1}e^{2it}+a_{2}\cos^2 t+a_{3}\sin 2t+a_{4}(1+e^{-2it}) = 0$
Now the problem is I'm not too sure where to go from here since I am use to having a matrix and row reducing to see if all the scalars are zero. Any hints on how can we show $a_{1}=...a_{4}=0$?
I wonder if I can equate real and imaginary parts to $0+0i$?
(2) I understand the definition of a basis and verifying if a set of vectors forms a basis. We check for linear independence and whether it spans the our vector space. But I'm not too sure how I would go about actually finding one.
Any help would be appreciated.
 A: Try $t\in \{ \pm {\pi \over 2}, 0 \}$. This will give a complex matrix $A$ such that
$A (a_1,a_2,a_4)^T = 0$. The matrix $A$ is invertible, hence $(a_1,a_2,a_4)^T = 0$. From this it follows that $a_3 = 0$.
While it doesn't work in this instance,
another approach would be to repeatedly differentiate and evaluate at $t=0$.
Since the $v_k$ are linearly independent, they must form a basis for $S$.
A: From the question, the vector space is a "function" space - so just to clarify, the $0$-vector in this case would mean the function $f_0$ mapping all $t \in \mathbb{R}$ to $0 \in \mathbb{C}$. So to check for linear independence, if the composite function $a_{1}e^{2it}+a_{2}\cos^2 t+a_{3}\sin 2t+a_{4}(1+e^{-2it})$ has a nonzero value for some $t \in \mathbb{R}$ then it is not $f_0$. My suggestion would be to write the exponential function in polar form and then simplify...
If you have proved the first part, you have essentially proved the second part, since a linearly independent set spanning a space is a basis.  
