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

Question

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.

Answer 1

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: $B=\frac12(B+B^\ast)-\frac i2(iB-iB^\ast)$.

Answer 2

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 $x_3$.