I have a very specific example from a book on quantum mechanics by Schwabl, in which he states that an object which commutes with all four gamma matrices,
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1\\
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
0 & -1 & 0 & 0\\
-1 & 0 & 0 & 0\\
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 & -i\\
0 & 0 & i & 0\\
0 & i & 0 & 0\\
-i & 0 & 0 & 0\\
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1\\
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
\end{pmatrix},
$$
must be a multiple times the unit matrix. These matrices don't seem to span all $4 \times 4$ matrices so why would this be the case? I have asked around but no one seems to know the answer.
Best Answer
Call your four matrices $A,B,C,D$ respectively. While they indeed don't span $M_4(\mathbb C)$, the point is that the algebra they generate is the whole matrix space. So, any matrix that commutes with $A,B,C,D$ must in turn commute with all members of $M_4(\mathbb C)$. In fact, if we put $X=\frac{B\,(AC-C)\,A}{2i}$ and $Y=\frac{B\,(AC+C)\,A}{2i}$, the canonical basis of $M_4(\mathbb C)$ can be obtained as polynomials in $A,B,C,D$: \begin{align*} E_{11}&=\frac12(X^2+X),&E_{14}&=E_{11}B,&E_{13}&=E_{11}D,\\ E_{22}&=\frac12(X^2-X),&E_{23}&=E_{22}B,&E_{24}&=-E_{22}D,\\ E_{33}&=\frac12(Y^2+Y),&E_{32}&=-E_{33}B,&E_{31}&=-E_{33}D,\\ E_{44}&=\frac12(Y^2-Y),&E_{41}&=-E_{44}B,&E_{42}&=E_{44}D,\\ E_{12}&=E_{13}E_{32},\\ E_{21}&=E_{24}E_{41},\\ E_{34}&=E_{31}E_{14},\\ E_{43}&=E_{42}E_{23}. \end{align*}