Lemma 12.6.981EECB4E9700

Proof. By the trace formula, the congruence $M\mapsto P^*MP$ by a fixed invertible matrix $P$ is an isometry. Taking $P=A^{-1/2}$, we may assume that the endpoints are $I$ and $C=A^{-1/2}BA^{-1/2}>0$. The curve $\exp (t\log C)$, $0\leq t\leq 1$, has constant speed $\| \log C\| _{\mathrm F}/\sqrt n$, which gives the upper bound for the distance.

For the reverse inequality, diagonalization gives the convergent resolvent formula for the matrix logarithm and its derivative

On a neighborhood of $M$ where the spectrum is bounded below by a positive constant, differentiation under the integral sign is justified by an integrable bound. In a unitary basis diagonalizing $M$, with eigenvalues $d_ i>0$, the derivative has entries $c_{ij}U_{ij}$, where

Writing $d_ i/d_ j=\mathrm e^{2s}$ gives $\sqrt{d_ id_ j}|c_{ij}|=|s/\sinh s|\leq 1$, including the limiting value at $s=0$. Hence

Applying this along a piecewise $C^1$ path of positive matrices from $I$ to $C$ and integrating, we bound its length below by $\| \log C\| _{\mathrm F}/\sqrt n$. This proves the distance formula. In terms of the original endpoints, the length-minimizing curve is

Its determinant is $(\det A)^{1-t}(\det B)^ t$, so for endpoints with equal determinant it stays in the fixed-determinant hypersurface. $\square$