# A determinant problem with symmetric PSD matrices

Suppose we have a a set of matrices in the complex field of the form $a_iv_iv_i^H$ for $i=\{1,\dots,n\}$ where $a_i$ are constant positive real scalars and $v_i$ are constant complex valued finite dimensional matrices all of same dimension. So that makes $v_iv_i^H$ positive semidefinite (PSD). $I$ is the identity matrix.

Now we want to maximize the following determinant over $a_iv_iv_i^H$ for $i=\{1,\dots,n\}$ $$\mathrm{maximize}_{\{1,\dots,n\}}\:\det \left( I+\frac{a_iv_iv_i^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right).$$

Essentially we pick one matrix for the numerator and all the rest go in the denominator. Which matrix should go on the numerator?

P.S.: I believe this question is related to Determinant of sum of positive definite matrices.

• I think that $\det(a_iv_iv_i^H)=0$ for each $i$, so the sorting would not work. The maximum depends on the directions of the vectors. Jan 21, 2014 at 18:03
• @AlexDegtyarev If $a_iv_iv_i^H$ is PD then all eigenvalues are possitive, so det is non zero. But I will change the claim to reflect eigenvalues. Thanks a lot
– MLT
Jan 21, 2014 at 18:40
• $v_i v_i^H$ is of rank 1, i.e. all its eigenvalues except one are 0. Jan 21, 2014 at 19:37
• The product you mention equals $1+a_i\|v_i\|^2$. Still the claim is almost obviously wrong: if, say, all $a_i\|v_i\|^2=1$, the maximum depends dramatically on the directions of the vectors. Jan 21, 2014 at 20:19
• What I mean is that two collinear vectors may overpower an orthogonal vector of larger norm. I don't think there is an easy answer to your question. At least, not without a very bright idea :) Jan 21, 2014 at 20:21

Write the objective function as $$\det \left( \frac{I+\sum_j a_jv_jv_j^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right).$$The numerator is constant and can be ignored, so you are really minimizing
$$\det({I+\sum_{j\neq i} a_jv_jv_j^H}).$$