Equality in trace duality

That follows is a case of equality of the relation

$(*)$ $|tr(A^TB)|\leq \sigma_1(B)\sum_i \sigma_i(A)$.

I don't know if there are other such instances (at least, I'm pretty sure there are not when $n=2$).

Let $A,B$ be $n\times n$ symmetric matrices that commute. Then, up to an orthonormal change of basis, we may assume that they are diagonal. Moreover, they have the form

1. $B=diag(\lambda I_k,\lambda_1,\cdots,\lambda_{n-k})$ where, for every $i$, $|\lambda|\geq |\lambda_i|$.

2. $A=diag(\mu_1,\cdots,\mu_k,0_{n-k})$ where, for every $j$, $\mu_j\geq 0$ (or, for every $j$, $\mu_j\leq 0$).

EDIT. $\textbf{Proposition}$. If $(*)$ is an equality, then $A^TB$ is symmetric $\geq 0$ or $\leq 0$.

$\textbf{Proof}$. One always has $|tr(A^TB)|\leq \sum_i\sigma_i(A^TB)\leq \sigma_1(B)\sum_i \sigma_i(A)$.

Then $|tr(A^TB)|= \sum_i\sigma_i(A^TB)$, and consequently, $spectrum(A^TB)\subset [0,+\infty[$ or $]-\infty,0]$, and the SVD of $A^TB$ is a diagonalization via an orthonormal basis. $\square$