Commutator of $B$, $C$ vanishes if $A$, $B$, $C$, $AB$, $AC$ are Hermitian

Suppose 3 operators $$A$$, $$B$$, $$C$$ are Hermitian operators. Assume $$A$$ has a non-degenerate spectrum, and $$AB$$ and $$AC$$ are also Hermitian. Show that $$[B,C] = 0$$

From the conditions $$A$$, $$B$$, $$C$$, $$AB$$ and $$AC$$ are Hermitian operators, one can derive that $$[A,B]=[A,C]=0$$ How can one proceed to show that $$[B,C]=0$$?

• Note that $(AB)^{\dagger} = B^{\dagger}A^{\dagger}$. Think: how does that look if $A, B$ and $AB$ are Hermitian? what does that tell us? – yu-v Feb 3 at 17:11
• I know that gives us $[A,B]=0$, same argument gives $[A,C]=0$, but how do we relate this to [B,C]? – LY3000 Feb 3 at 17:16
• oh sorry, I didn't read the question carefully enough. You can show that $[A, BC]=0$. This, together with the fact that $A$ is non degenerate, should allow you to show that $BC$ is also hermitian. – yu-v Feb 3 at 17:30
• This is where I get unsure, I tried to write $A=\sum_i a_i|a_i><a_i|$, with $a_i \neq a_j$. But I'm not sure how to proceed – LY3000 Feb 3 at 17:46
• Great, so eigenvectors are orthogonal, $A=\sum_i a_i ~|i\rangle\langle i|$, $B=\sum_i b_i ~|i\rangle\langle i|$, and $C=\sum_i c_i ~|i\rangle\langle i|$. – Cosmas Zachos Feb 3 at 22:25

You are nearly there. If $$A$$ commutes with $$B$$ it means that they can be diagonalized simultaneously. Now use the fact the the eigenvalues of $$A$$ are non-degenerate. This means that also $$B$$ is diagonal in the same basis. Repeat with $$C$$ and you are done.