One simple example with a special matrix, which has somehow "a continuum" as eigenvalue...

Consider some function $ f(x) = K + ax + bx^2 + cx^3 + ... $ having a nonzero radius of convergence. Then think of the infinite matrix of the form
$$ \small \begin{bmatrix}
K & . & . & . & \cdots \\\
a & K & . & . & \cdots \\\
b & a & K & . & \cdots \\\
c & b & a & K & \cdots \\\
\vdots & \vdots & \vdots& \vdots & \ddots
\end{bmatrix} $$
From the properties of finite matrices we would expect, that *K* is an eigenvalue. But consider a type of an infinite vector

$$ V(x) = [1,x,x^2,x^3,x^4,\ldots ] $$ with a scalar parameter $x$ from the range of convergence, then
$$ V(x) \cdot F = f(x) \cdot V(x) $$
This means also: any vector $V(x)$ is an eigenvector of the matrix *F* and corresponds to the eigenvalue $f(x)$. If now $f(x)$ is entire, for instance the exponential function $ f(x)=\exp(x)$, then any value from the complex plane (except $0$ because $\exp(x)$ is never $0$) "is an eigenvalue" of *F* contradicting the "naive" extrapolation from the finite truncation of the matrix ...

Bounded Integral operators on $L^2$ spaces. projecteuclid.org/… – Pietro Majer Oct 14 '12 at 11:21