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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 $x$.

For which $a \in \mathbb{R}$ is the function/mapping $L$ invertible?

I'm not sure how to do this. But I tried that:

$$f(x)=\lambda_{1}e^x \sin(ax)+ \lambda_{2}e^x \cos(ax)$$

$$f'(x)= \lambda_{1}e^x \sin(ax)+ \lambda_{1}a \cos(ax)+ \lambda_{2}e^x \cos(ax) - \lambda_{2}e^xa \sin(ax)$$

and as I understood from task this derivative would be $L$.

But what we do now? I have really no idea how to continue :s

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Since $Lf_1$ is the function $$ e^x\sin(ax)+ae^x\cos(ax)=f_1(x)+af_2(x) $$ and $Lf_2$ is the function $$ e^x\cos(ax)-ae^x\sin(ax)=-af_1(x)+f_2(x) $$ the matrix of $L$ with respect to the basis $A$ is $$ \begin{bmatrix} 1 & -a \\ a & 1 \end{bmatrix} $$ which has determinant $1+a^2\ne0$ (since $a$ is real). So $L$ is invertible for every $a$.

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    omg i did and understood it completely wrong then, thank you ! – 2017-01-15
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    could you maybe help me with http://math.stackexchange.com/questions/2100667/what-is-the-representation-of-g-bbl-in-mathbbr2-times-2-of-l-in i would be very grateful because i cannot get the solutions anywhere since this is from a very old exam – 2017-01-16