Let $V$ be the vector space over the complex numbers of all functions from $\textbf{R}$ into $\textbf{C}$, i.e., the space of all complex-valued functions on the real line. Let $f_{1}(x) = 1$, $f_{2}(x) = e^{ix}$ and $f_{3}(x) = e^{-ix}$.
(a) Prove that $f_{1}$, $f_{2}$ and $f_{3}$ are linear independent.
(b) Let $g_{1}(x) = 1$, $g_{2}(x) = \cos(x)$ and $g_{3}(x) = \sin(x)$. Find a $3\times 3$ matrix $P$ such that \begin{align*} g_{j} = \sum_{i=1}^{3}P_{ij}f_{i} \end{align*}
MY ATTEMPT
(a) The given functions are indeed linear independent. In fact, we have \begin{align*} & af_{1} + bf_{2} + cf_{3} = a + b\cos(x) + c\cos(x) + i(b\sin(x) - c\sin(x)) = 0 \end{align*}
If $x = 0$, then $a + b + c = 0$. If $x = \pi/2$, then $b = c$ and $a = 0$. Consequently, $a = b = c = 0$, from whence we conclude that $f_{1}$, $f_{2}$ and $f_{3}$ are linear independent.
(b) Such sought matrix $P$ is the change of basis matrix, that is to say \begin{align*} P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2i} \\ 0 & \frac{1}{2} & -\frac{1}{2i} \end{bmatrix} \end{align*}
Could someone double-check if the matrix $P$ is expressed correctly?
(a) Your reasoning is not sound. Scalars $a,b,c$ are complex, so $a = -1 - i$, $b = 1$, $c = i$ satisfy your two equations also. A hint is that you are on the right track—just go one step further.
(b) I got the same $P$.