In my tekstbook powervectors are discussed where $(L-\lambda I)^p \textbf{v}=0$ was mentioned. Here is $L$ a linear operator on the vectorspace $V$, $\lambda$ a scalar and $\textbf{v} \in V $.
My question: How is $L-\lambda I$ definied? I'm confused because $L$ is here a function and not a matrix.
On
Robert did fully answer the question, but I'd like to add a different example: The differential equation of the harmonic oscillator is
$$\frac{\mathrm{d}^2 u(t)}{\mathrm{d}t^2}+2\beta \frac{\mathrm{d}u(t)}{\mathrm{d}t}+\omega^2 u(t)=0$$
And it can be rewritten as
$$(D^2+2\beta D+\omega^2 I)u(t)=0$$
Where $D$ is the differential operator: $Du(t)=\frac{\mathrm{d}u(t)}{\mathrm{d}t}$.
Or the Klein-Gordon equation in $1+1$ dimensions:
$$\left(\frac{1}{c^2} \partial_t^2-\partial_x^2+\mu I\right)u(t)=0$$
This kind of notation is very useful, because the differential operator will take it's eigenvalue on it's eigenfunction, so you can easily get the dispersion relation, with just substitutions of the eigenvalues: $\partial_t \to (-i \omega)$ and $\partial_x \to (ik)$ in the case of a plane wave.
$I$ is the identity operator on the vector space: $Ix = x$ for all $x \in V$. So $$(L - \lambda I) x = L x - \lambda I x = L x - \lambda x$$