Let $M_n(\mathbb{C})$ denote the algebra of $n\times n$ matrices over the field of complex numbers $\mathbb{C}$. Let $h_1,h_2\in M_n(\mathbb{C})$ be two Hermitian matrices. Suppose that $h_1,h_2$ are "relatively irreducible", i.e. they don't have common invariant proper subspaces. What is the dimension of the subalgebra $\mathcal{A}$ generated by $e,h_1,h_2$, where $e$ is the $n\times n$ identity matrix? Especially, what is the condition on $h_1,h_2$ such that $\mathcal{A}=M_n(\mathbb{C})$? [Note: it is easy to see that if $h_1,h_2$ can be simultaneously block diagonalized, then $e,h_1,h_2$ cannot generate $M_n(C)$.]
For example when $n=2$, the Pauli matrices $\sigma^z,\sigma^x$ are enough to generate $M_2(\mathbb{C})$. When $n=3$, it is easy to check that the Gellmann matrices $$\lambda_1 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},~~\lambda_4 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},$$ along with identity matrix $e$ could generate $M_3(\mathbb{C})$. I'm curious about what can be said in general.
Let $V$ denote the $n$-dimensional vector space where $M_n(\mathbb{C})$ acts on. Then $h_1, h_2$ are "relatively irreducible" is equivalent to that $V$ is an irreducible representation of $\mathcal{A}$. The density theorem of representation theory (Sec.3.2 of the rep-th book) says that if $V$ is an irreducible finite dimensional representation of a unital associative algebra $\mathcal{A}$, then the representation map $\rho: \mathcal{A}\to \mathrm{End}V$ is surjective. But $\mathrm{End}V=M_n(\mathbb{C})$ and $\mathcal{A}$ is by definition a subalgebra of $M_n(\mathbb{C})$, therefore $\mathcal{A}=M_n(\mathbb{C})$.
In conclusion, relative irreducibility of $h_1,h_2$ already guarantees $\mathcal{A}=M_n(\mathbb{C})$--we don't need anymore conditions.