# 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.

• You could write down a huge system of equations representing commuting and reduce it. This wouldn't give the intuition, but it would confirm the statement. – Michael Burr Feb 21 '17 at 15:40
• Maybe this analysis could be simplified with Kronecker products. In particular, your matrices are $$G_1 = \pmatrix{1&0\\0&-1} \otimes I\\ G_2 = \pmatrix{0&1\\-1&0} \otimes I\\ G_3 = -i\pmatrix{0&1\\-1&0} \otimes \pmatrix{0&1\\-1&0}\\ G_4 = \pmatrix{0&1\\-1&0} \otimes \pmatrix{1&0\\0&-1}$$ – Omnomnomnom Feb 21 '17 at 16:14

Consider the matrix $$A=\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44} \end{bmatrix}.$$

Then, if $A$ commutes with your first matrix, then $$\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{bmatrix}\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44} \end{bmatrix}= \begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44} \end{bmatrix} \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{bmatrix}$$ In other words, $$\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ -a_{31}&-a_{32}&-a_{33}&-a_{34}\\ -a_{41}&-a_{42}&-a_{43}&-a_{44} \end{bmatrix}= \begin{bmatrix} a_{11}&a_{12}&-a_{13}&-a_{14}\\ a_{21}&a_{22}&-a_{23}&-a_{24}\\ a_{31}&a_{32}&-a_{33}&-a_{34}\\ a_{41}&a_{42}&-a_{43}&-a_{44} \end{bmatrix}$$ This tells you that $a_{31}=0=a_{32}=a_{41}=a_{42}=a_{13}=a_{14}=a_{23}=a_{24}$. Already, $A$ is significantly simplified. The second matrix gives $a_{11}=a_{33}$, $a_{12}=a_{34}$, $a_{21}=a_{43}$, and $a_{22}=a_{44}$. Then, keep going.

• It worked out very well! Thank you! – Jens Roderus Feb 21 '17 at 16:14
• Ah, the Fuerza Bruta method! – Matthew Leingang Feb 22 '17 at 16:59
• @MatthewLeingang waves – Michael Burr Feb 22 '17 at 16:59

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*}

• Best answer so far – YoTengoUnLCD Feb 22 '17 at 4:10
• @YoTengoUnLCD provided you do understand it... For many readers the accepted answer will be the best. – Ruslan Feb 22 '17 at 5:37
• Awesome! How do you get that? Is it a classic result from quantum mechanics? – Taladris Feb 22 '17 at 9:32
• @Taladris I found it by inspection. Not sure if it has any physical significance. – user1551 Feb 22 '17 at 12:54
• @YoTengoUnLCD I agree that this is the best answer (+1) (I'm the author of the accepted answer). – Michael Burr Feb 22 '17 at 20:42

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.

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.

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 $dim M_4(K)=16$.