# Show that if $A$ commutes with every matrix of certain form, then $A$ commutes with every matrix in $M_2(\mathbb{C})$

In Exercise 1.16 in Hall's book Lie Group, Lie Algebras, and Representations, we are asked to show that if $$A$$ commutes with every matrix $$X$$ of the form \begin{align} \begin{pmatrix}x_1 & x_2+ix_3 \\ x_2-ix_3 & -x_1\end{pmatrix}, \end{align} (where $$x_1,x_2,x_3\in\mathbb{R}$$) then $$A$$ commutes with every element of $$M_2(\mathbb{C})$$. It is possible to write $$A=\begin{pmatrix}\alpha & \beta \\ \gamma & \delta\end{pmatrix}$$ and use the commutativity to exploit restrictions on $$\alpha,\beta,\gamma,\delta$$. However, is there any other way to prove this without explicitly investigating $$\alpha,\beta,\gamma,\delta$$?

Thanks in advance for any comment, hint and answer.

• My guess is that $x_1,x_2,x_3\in\mathbb R$. Am I right? – José Carlos Santos Oct 1 '18 at 10:23
• @JoséCarlosSantos Yes you are right. – Hopf eccentric Oct 1 '18 at 10:58

## 2 Answers

The condition literally says that $$A$$ commutes with every traceless Hermitian matrix. Since every Hermitian matrix $$H$$ is the sum of a traceless Hermitian part and a scalar part (i.e. $$H=\left(H-\frac{\operatorname{tr}(H)}2I\right)+\frac{\operatorname{tr}(H)}2I$$), $$A$$ must also commute with every Hermitian matrix. It follows that $$A$$ commutes with every complex matrix $$B$$, because $$B$$ can always be written as a complex linear combination of Hermitian matrices: $$B=\frac12(B+B^\ast)-\frac i2(iB-iB^\ast)$$.

Recall: Commuting endomorphisms respect each others' (generalized) eigenspace decomposition.

So immediately from commuting with diag(1,-1) we must have $$A$$ diagonal. Then commuting with $$\begin{pmatrix}0&1\\1&0\end{pmatrix}$$ gives the diagonal and antidiagonal subspaces are also invariant under $$A$$, hence $$A$$ is a multiple of the identity matrix. Note there is no need to use $$x_3$$.