I have encountered the following equality


$\arg\max\_{\text{tr}\left(\boldsymbol{S}^{H}\boldsymbol{S}\right)=1}\text{tr}\left(\boldsymbol{S}^{H}\boldsymbol{A}\boldsymbol{S}\left(\boldsymbol{S}^{H}\boldsymbol{B}\boldsymbol{S}\right)^{-1}\right)=\arg\max\_{\text{tr}\left(\boldsymbol{S}^{H}\boldsymbol{S}\right)=1}\frac{\left|\boldsymbol{S}^{H}\boldsymbol{A}\boldsymbol{S}\right|}{\left|\boldsymbol{S}^{H}\boldsymbol{B}\boldsymbol{S}\right|}$


in the paper "[Trace Ratio vs. Ratio Trace for Dimensionality Reduction](http://cs.yale.edu/homes/wang-huan/WebContent/CVPR07_Wang.pdf)".


However I can't seem to find its proof, how do we prove this equality?



equality between the ratio trace and the determinant ratio
The equality of the [arg max](https://en.wikipedia.org/wiki/Arg_max) of the trace of the ratio and the ratio of determinants follows from the fact that each of these two maximisation problems has the same solution, given by the matrix $S$ composed of column vectors $s\_k$ that solve the generalised eigenvalue problem $As\_k=\tau\_k Bs\_k$, with $\tau\_k$ the $k$-th largest generalised eigenvalue. This solution in terms of the generalised eigenvalues is given in reference 3 of the cited paper.


