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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.

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  • $\begingroup$ My guess is that $x_1,x_2,x_3\in\mathbb R$. Am I right? $\endgroup$ – José Carlos Santos Oct 1 '18 at 10:23
  • $\begingroup$ @JoséCarlosSantos Yes you are right. $\endgroup$ – Hopf eccentric Oct 1 '18 at 10:58
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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: $B=\frac12(B+B^\ast)-\frac i2(iB-iB^\ast)$.

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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 $x_3$.

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