Is $M_0(\mathbb F)$ a valid construct? Or, is there a $\text{dim}\ 0$ matrix ring? I assume you can define $M_0(\mathbb F) := \{[]\}$ and $\mathbf I_0 := []$, but I couldn't find any confirmation on this.
 A: Yes, your definition is right (and useful). See Carl de Boor, An empty exercise.
A: There is no generally accepted consensus on what a "dimension $0$ matrix" would be. Here is what I would say:
I prefer linear transformation instead of matrices.
For $n\geq 1$, the matrix algebra $M_n(\mathbb{F})$ is isomorphic to the algebra of linear operators on $\mathbb{F}^n$, so we might just say that  $M_0(\mathbb{F})$ is the algebra of linear operators on $\mathbb{F}^0$.
But what is $\mathbb{F}^0$?
Well, for $n,m\geq 1$, we have natural isomorphisms $\mathbb{F}^n\times\mathbb{F}^m\to\mathbb{F}^{n+m}$, so we might just try to extend this to allow the same type of isomorphism for $n$ or $m=0$. That is, we should have $\mathbb{F}^n\times\mathbb{F}^0\cong\mathbb{F}^n$. Taking the dimensions on both sides, we obtain $n+\dim(\mathbb{F}^0)=n$, i.e., $\dim(\mathbb{F}^0)=0$. So a natural definition is $\mathbb{F}^0=\left\{0\right\}$, the null space.
This definition behaves quite well with respect to the natural isomorphisms: $\mathbb{F}^n\times\mathbb{F}$ is naturally isomorphic to $\mathbb{F}^{n+m}$ for all $n,m\geq 0$.
Ok, so $M_0(\mathbb{F})$ should be the algebra of linear operators on $\mathbb{F}^0=\left\{0\right\}$. This is just the null algebra: $M_0(\mathbb{F})=\left\{0\right\}$. This is the definition I would adopt, although I make no claims on how useful it is.
