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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".

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

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The equality of the 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.

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