Maximizing the determinant $\det (I+SS^T)$

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Let

$$S=\alpha X + (1-\alpha)Y$$

where $X$ and $Y$ are $m \times n$ matrices with $m > n$ whose singular values are not larger than 1, i.e., $\sigma_{\max}(X) \leq 1$ and $\sigma_{\max}(Y) \leq 1$, where $\sigma_{\max}(\cdot)$ is the largest singular value of the argument. Assume

$$\det(I+XX^T) > \det(I+YY^T)$$

I have observed that $$\arg \max_{0 \leq \alpha \leq 1} \det \left(I + SS^T \right) = 1$$

However, although it looks simple, I fail to prove this. Any ideas?