# Commuting hermitian matrices

Let $A$, $B$ be $n\times n$ hermitian matrices. Denote by $(\alpha_i)$, $(\beta_i)$, $(\gamma_i)$ the eigenvalues of $A$, $B$, and $A+B$. Assume that there exists permutations $\sigma$, $\tau \in \frak S_n$ so that $$\gamma_i = \alpha_{\sigma(i)} + \beta_{\tau(i)}$$ for all $1\le i \le n$. Does this imply that $A$, $B$ commute?

• No necessarily, Let A and B be two big size nilpotent matrices which are strictly upper triangle – Ali Taghavi Oct 4 '17 at 15:02
• I am sorry I did not pay attention to "hermitian" – Ali Taghavi Oct 4 '17 at 15:04
• may be a related post mathoverflow.net/questions/34252/… – Ali Taghavi Oct 4 '17 at 15:22
• @Dennis Serre: I don't know the exact statement of that theorem. Besides, I guess it describes the possibilities for the spectrum of the sum. Not sure it tells when some ( extreme?) cases occur. – orangeskid Oct 4 '17 at 15:45
• @DenisSerre You might wish to correct the spelling of Allen Knutson's last name :) – Yemon Choi Oct 4 '17 at 20:31

The answer is negative. Let $C$ and $D$ be two noncommuting $n\times n$ hermitian matrices, and define the block matrices $$A=C\oplus (C+D)\oplus 0_n,\ \ \ B=D\oplus(-C)\oplus 0_n,$$ where $0_n$ is the $n\times n$ matrix of all 0's.

This question was the topic of Commutativity and spectra of Hermitian matrices, by Wasin So (1994).

The statement is negative in general: Richard Stanley's counterexample has size $n\geq 6$, a counterexample for $n=3$ is $$A=\begin{pmatrix}0&0&0\\ 0&6-\sqrt 2&2\\ 0&2&4+2\sqrt 2 \end{pmatrix},\;\;B=\begin{pmatrix} 4&0&0\\ 0&-4&0\\ 0&0&0 \end{pmatrix},$$

There are special cases when the statement holds:

If $\gamma_i = \alpha_{\sigma(i)} + \beta_{\tau(i)}$ for a permutation $\sigma(i)$ and $\tau(i)$ of the eigenvalues $\alpha_i,\beta_i,\gamma_i$ in non-increasing order, then the $n\times n$ Hermitian matrices $A,B$ commute if either

• $n=2$
• rank $A=1$
• $\sigma(i)=i$ for all $i$ (identity permutation)
• $\sigma(i)=n+1-i$ for all $i$ (reverse permutation)