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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.

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.

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. I think the matrix which should go on the numerator is the one with the largest product of $\prod_k(1+\lambda_{ik})$ where $\lambda_{ik}$ are the eigenvalues of $a_iv_iv_i^H$. Or in the case of PD which is equal to largest determinant $\det(a_iv_iv^H_i)$. Is this claim right?

So I proceeded: $$\det \left( I+\frac{a_iv_iv_i^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right)=\det \left( I+\sum_{j} a_jv_jv_j^H \right) \det \left( \left( I+\sum_{j\neq i} a_jv_jv_j^H \right)^{-1}\right).$$ The matrices $a_iv_iv^H_i$ are PSD and have non negative eigenvalue. Could someone please help to complete the proof.

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

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 positive real scalars and $v_i$ are 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. $$\det \left( I+\frac{a_iv_iv_i^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right).$$

I think the matrix which should go on the numerator is the one with the largest product of $\prod_k(1+\lambda_{ik})$ where $\lambda_{ik}$ are the eigenvalues of $a_iv_iv_i^H$. Or in the case of PD which is equal to largest determinant $\det(a_iv_iv^H_i)$. Is this claim right?

So I proceeded: $$\det \left( I+\frac{a_iv_iv_i^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right)=\det \left( I+\sum_{j} a_jv_jv_j^H \right) \det \left( \left( I+\sum_{j\neq i} a_jv_jv_j^H \right)^{-1}\right).$$ The matrices $a_iv_iv^H_i$ are PSD and have non negative eigenvalue. Could someone please help to complete the proof.

P.S.: I believe this question is related to Determinant of sum of positive definite 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. I think the matrix which should go on the numerator is the one with the largest product of $\prod_k(1+\lambda_{ik})$ where $\lambda_{ik}$ are the eigenvalues of $a_iv_iv_i^H$. Or in the case of PD which is equal to largest determinant $\det(a_iv_iv^H_i)$. Is this claim right?

So I proceeded: $$\det \left( I+\frac{a_iv_iv_i^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right)=\det \left( I+\sum_{j} a_jv_jv_j^H \right) \det \left( \left( I+\sum_{j\neq i} a_jv_jv_j^H \right)^{-1}\right).$$ The matrices $a_iv_iv^H_i$ are PSD and have non negative eigenvalue. Could someone please help to complete the proof.

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