If a matrix commutes with a set of other matrices, what conclusions can be drawn? 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.
 A: 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*}
A: The first condition for commuting with $diag(1,1,-1,-1)$ determines already $8$coefficients of "the object" to be zero. It is indeed really a matter of computation then to see that the matrix has to be a multiple of the identity. This also shows that generating the full matrix algebra is not the same as forcing the matrix to be scalar. Already commuting with $4$ given matrices here is sufficient, although $dim M_4(K)=16$.
A: For the example you gave, the conclusion follows from Schur's Lemma since the gamma matrices form an irreducible representation of the complexification of the Clifford algebra $Cl_{1,3}(\mathbf{R})_{\mathbf{C}}$. This comes up when considering irreducible representations of the Lorentz group (specifically the spin representation), which is central to discussions of spin in physics.
A: For two matrices to commute, it is necessary that each matrix preserves the eigenspace of the other matrix (that is, it can't map part of an eigenspace onto a different eigenspace). As multiples of the identity matrix do not change eigenspaces and have all vectors in the same eigenspace, they commute with all other matrices.
If $A$ and $B$ commute, and $x$ is an eigenvector of $A$ with eigenvalue $\lambda$, then,
$$
ABx=BAx=B\lambda x=\lambda Bx
$$
and thus $Bx$ must also be an eigenvector of $A$ with the same eigenvalue... or a zero vector.
Suppose that $v=(a,b,c,d)$ is an eigenvector of our matrix. Then we must find, in the same eigenspace, $(a,b,-c,-d)$, $(d,c,-b,-a)$, $(-d,c,b,-a)$, and $(c,-d,-a,b)$ - (I have dropped the $i$ from the third matrix, as it's just a constant multiplier).
Adding and subtracting the middle two together, we can also see that $(0,c,0,-a)$ must be in the eigenspace, as must $(d,0,-b,0)$. At least one of these must be non-zero. Let's assume that $(0,c,0,-a)$ is non-zero.
Then we can also see that $(0,c,0,a)$ is in the eigenspace (from the first matrix), and so both $(0,c,0,0)$ and $(0,0,0,a)$ are in the eigenspace. Let's assume that $(0,0,0,a)$ is non-zero, and thus $(0,0,0,1)$ is in the eigenspace. Now, from the last matrix, $(0,0,1,0)$ is in the eigenspace.  From the third matrix, we can then determine that $(0,1,0,0)$ and $(0,0,0,1)$ are in the eigenspace.
A similar analysis works if you assume $b$, $c$, or $d$ is non-zero.
Therefore, the eigenspace is the set of all 4D vectors, all sharing the same eigenvalue. This tells us that our matrix must be the identity matrix multiplied by that eigenvalue.
