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:

$$ \epsilon =
\left( \begin{array}{cc}
0 &amp; 0 \\
1 &amp; 0 \end{array} \right) $$

where $\epsilon$ now, is the nilpotent matrix, we get:

$$ \left( \begin{array}{cc}
H | m \rangle &amp; 0 \\
H_1 |m_1 \rangle + H | m\rangle &amp; H |m_1 \rangle \end{array} \right) = \left( \begin{array}{cc}
E | m \rangle &amp; 0 \\
E_1 |m_1 \rangle + E | m\rangle &amp; E |m_1 \rangle \end{array} \right)$$

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?
I would say there's nothing preventing you from using a matrix as perturbation of a matrix equation as long as you let the limit of the norm converge to $0$.
Yes and No. When I say yes I mean it is possible in several ways. But it does not make sense.

As usual for a physicist you did not specify what the space is in which your states (kets) live and thus not what the $H$ and $H_1$ are. But of course they are meant to be operators which you can consider to be generalizations of matrices to infinite dimensions. So when you assume $\epsilon$ to be a matrix you could as well absorb this into $H_1$. 

Further you put $\epsilon$ to be a constant matrix. Then you can just leave the $\epsilon$ away. The epsilon is there to control the perturbation $H_1$. If you set $\epsilon$ to be a constant you directly solve the problem.