Task from an exam:
Given is the real vector space where $a \in \mathbb{R}$ is fixed:
$V = \left\{ f: \mathbb{R} \rightarrow \mathbb{R}: f(x) = \lambda_{1}e^{x}\sin(ax)+ \lambda_{2}e^{x}\cos(ax) \text{ for } x \in \mathbb{R}, \lambda_{1}, \lambda_{2} \in \mathbb{R} \right\}$
with the basis $A=(f_{1},f_{2})$, whereby $\text{ }$ $\text{ }$$f_{1}(x)=e^{x}\sin(ax), \text{ } \text{ }f_{2}(x)= e^{x}\cos(ax)$.
And because every element of this vector space is characterized by an ordered pair $(\lambda_{1}, \lambda_{2}) \in \mathbb{R}^{2}$ of coefficients, with allocation $f=\lambda_{1}f_{1}+\lambda_{2}f_{2} \mapsto (\lambda_{1}, \lambda_{2})$, every linear mapping of $V$ in $V$ can be seen as a $2\times 2$ matrix.
Let $L: V \rightarrow V$ be the linear mapping which assigns every function $f \in V$ its derivative $f' \in V$ to $x$.
Show that $L(f_{1})=f_{1}+af_{2}$ and $L(f_{2})=f_{2}-af_{1}$
So $L(f_{1}) = f_{1}'(x)=e^xsin(ax)+e^xa \cdot cos(ax)$
Then we have: $e^xsin(ax)+e^xa \cdot cos(ax) = e^xsin(ax)+a \cdot e^xcos(ax) $
Thus $L(f_{1}) = f_{1}+af_{2}$
I don't do it for the other because it's same procedure.
But please I need to know if I did it correctly like that?