Can epsilon be a matrix? Question
In the following expression can $\epsilon$ be a matrix?
$$ (H + \epsilon H_1) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) = (E |m\rangle + \epsilon E|m_1\rangle + \epsilon^2 E_2 |m_2\rangle + \dots) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) $$
Background
So in quantum mechanics we generally have a solution $|m\rangle$ to a Hamiltonian:
$$ H | m\rangle = E |m\rangle $$
Now using perturbation theory:
$$ (H + \epsilon H_1) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) = (E |m\rangle + \epsilon E|m_1\rangle + \epsilon^2 E_2 |m_2\rangle + \dots) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) $$
I was curious and substituted $\epsilon$ as a matrix:
$$ \epsilon =
\left( \begin{array}{cc}
0 & 0 \\
1 & 0 \end{array} \right) $$
where $\epsilon$ now, is the nilpotent matrix, we get:
$$ \left( \begin{array}{cc}
H | m \rangle & 0 \\
H_1 |m_1 \rangle + H | m\rangle & H |m_1 \rangle \end{array} \right) = \left( \begin{array}{cc}
E | m \rangle & 0 \\
E_1 |m_1 \rangle + E | m\rangle & E |m_1 \rangle \end{array} \right)$$
Which is what we'd expect if we compared powers of $\epsilon$'s. All this made me wonder if $\epsilon$ could be a matrix? Say something like $| m_k\rangle \langle m_k |$ ? Say we chose $\epsilon \to \hat I \epsilon$
then there exists a radius of convergence. What is the radius of convergence in a general case of any matrix?

$\epsilon$ is not generally used as a matrix. When matrices are needed, as you will notice from your example, epsilon is usually set beside the matrices of concern as a scalar which indicates that one is infinitesimally close to the origin (or zero) in whichever space you seek to be perturbing. 
