Let $ V=\{ f\in\text{func} (\mathbb{R}, \mathbb{C}) : f(t) =\alpha \cos (t) +\beta \sin(t), \alpha, \beta \in \mathbb{C} \} $.
(a) show that cos$(t) $, sin$(t) $, and exp(-it), exp(it) both form a basis for $ V$.
(b) Find the change of basis matrix.
(c) Find the matrix representation of $ D:V\to V $ with respect to both bases and check that the change of basis matrix gives the correct relationship between these two matrices. Where $D$ is the derivative function.
I am trying to get insight into this problem from linear algebra. It appears to have some intersection with Fourier analysis, which I don't know about. Any help is appreciated.
Since the vectors are given as "$\alpha cos(x)+ \beta sin(x)$" all that you need to do is show that cos(x) and sin(x) are independent. That is, that A cos(x)+ B sin(x)= 0 (for all x) only A= B= 0. And that's pretty close to "trivial". Take x= 0 and $x= \pi/2$ and see what happens.
To see that $e^{ix}$ and $e{-ix}$ also form a basis use the fact that $cos(x)= \frac{e^{ix}+ e^{ix}}{2}$ and $sin(x)= \frac{e^{ix}- e^{-ix}}{2i}$.
For the "change of basis matrix" you want a matrix that multiplied by a column matrix with coefficients from the first basis gives the column matrix with coefficients from the second basis. The first "basis vector", cos(x)= 1(cos(x))+ 0(sin(x)), would, as I said above, be written as $\frac{e^{ix}+ e^{-ix}}{2}= \frac{1}{2}e^{ix}+ \frac{1}{2}e^{-ix}$. So we want our matrix to map $\begin{pmatrix}1 \\ 0 \end{pmatrix}$ to $\begin{pmatrix}\frac{1}{2} \\ \frac{1}{2}\end{pmatrix}$. Similarly with sin(x) to $\frac{e^{ix}- e^{-ix}}{2i}$ the matrix must map $\begin{pmatrix}0 \\ 1 \end{pmatrix}$ to $\begin{pmatrix} \frac{1}{2i} \\ -\frac{1}{2i} \end{pmatrix}$. Such a matrix is $\begin{pmatrix}\frac{1}{2} & \frac{1}{2i} \\ \frac{1}{2} & -\frac{1}{2i} \end{pmatrix}$.