I'm working on the following linear algebra problem: Let $V$ be the space of all differentiable functions $f: R \longrightarrow R$ over $R$ (where $R$ denotes the real numbers) and let $D: V \longrightarrow V$ be the linear transformation such that $D(f) = f'$. Show that $2$ is an eigenvalue of $D$ and find an eigenvector corresponding to $2$.
Here is my attempt at a solution:
Suppose $\lambda$ is an eigenvalue of the operator $D$. Let $f(x) \in V$, with $f(x) \neq 0$. Then $Df(x) = \lambda f(x)$
$\Rightarrow$ $f'(x) = \lambda f(x)$.
Let $y = f(x)$. Then finding a solution to the above equation amounts to solving the differential equation $\frac{dy}{dx} = \lambda y$ :
$\frac{dy}{y} = \lambda dx$
$\Rightarrow$ $ln(|y|) = \lambda x + C$
$\Rightarrow$ $Ce^{\lambda x} = |y|$.
Thus, to show $\lambda = 2$ is an eigenvalue of $D$, we will show that letting $\lambda = 2$ in the equation $f'(x) = \lambda f(x)$ gives an appropriate eigenvector $f(x) \in V$:
$f'(x) = 2f(x)$
$\Rightarrow$ $f(x) = Ce^{2x}$, where $C = f(0)$. Thus, $2$ is an eigenvalue of $D$ with corresponding eigenvector $f(x) = Ce^{2x}$, where $C = f(0) \neq 0$ (if $f(0) = 0$, we get $f(x) = 0$, which can't be an eigenvector by definition).
Is the above solution correct? I have my worries - for example, I don't know that I can force the condition that $f(0) \neq 0$ , and if there's some reason that it's impossible that $f(0) = 0$ that I need to deduce.
Thanks!
~Mo
Yes, the solution is correct. In order to show that $2$ is an eigenvalue, all you need to do is find one example of eigenvector $f(x)$ so that $Df(x)=2f(x)$. So for instance choosing $f(x)=Ce^{2x}$ for any nonzero constant $C$ will suffice. It doesn't really matter that when $C=0$, this is not an eigenvector, since all we need is a single example of an eigenvector.