This is because operators, and functionals can be considered as (possibly) infinite dimensional matrices.
Of we solve $Av=\lambda v$, the eigenvalue(s) we seek, are the values that make the mapping $Av-\lambda v$ non injective, so we cannot invert the map.
For a matrix this corresponds to a matrix being singular, I.e uninvertible.
The eigenvectors are the elements in the kernel of this non invertible mapping.