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Drew Brady
Drew Brady
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Let $z$ denote a unit vector. Fix a finite collection of positive semidefinite matrices $\mathcal{P}$. Define the function and set $$ f_{\mathcal{P}}(X) = \sum_{P \in \mathcal{P}} z^T X(PX + I)^{-1} z, \quad \mbox{and} \quad \mathcal{X} = \{X \succeq 0,~\mathrm{trace}(X) = 1\}. $$ Above, $X \succeq 0$ means that $X$ is a real, symmetric, and positive semidefinite matrix.

I am interested in the following quantities, $$ f^\star_{\mathcal{P}} = \sup_{X \in \mathcal{X}} f_{\mathcal{P}}(X) \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \mathrm{argmax}_{X \in \mathcal{X}} f_{\mathcal{P}}(X). $$ By continuity and compactness, the supremum is attained and both quantities are well-defined.

In the case $|\mathcal{P}| = 1$, I am able to directly compute this. I obtain $$ f^\star_{\mathcal{P}} = z^T(I + P)^{-1} z \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \frac{(I+P)^{-1} zz^T (I + P)^{-1}}{sqrt{z^T (I+P)^{-2} z}. $$$$ f^\star_{\mathcal{P}} = z^T(I + P)^{-1} z \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \frac{(I+P)^{-1} zz^T (I + P)^{-1}}{z^T (I+P)^{-2} z}. $$ Above, $\mathcal{P} = \{P\}$. However, my proof does not generalize to $|\mathcal{P}| > 1$. I am wondering if there is a more clever way to obtain this result.

Let $z$ denote a unit vector. Fix a finite collection of positive semidefinite matrices $\mathcal{P}$. Define the function and set $$ f_{\mathcal{P}}(X) = \sum_{P \in \mathcal{P}} z^T X(PX + I)^{-1} z, \quad \mbox{and} \quad \mathcal{X} = \{X \succeq 0,~\mathrm{trace}(X) = 1\}. $$ Above, $X \succeq 0$ means that $X$ is a real, symmetric, and positive semidefinite matrix.

I am interested in the following quantities, $$ f^\star_{\mathcal{P}} = \sup_{X \in \mathcal{X}} f_{\mathcal{P}}(X) \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \mathrm{argmax}_{X \in \mathcal{X}} f_{\mathcal{P}}(X). $$ By continuity and compactness, the supremum is attained and both quantities are well-defined.

In the case $|\mathcal{P}| = 1$, I am able to directly compute this. I obtain $$ f^\star_{\mathcal{P}} = z^T(I + P)^{-1} z \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \frac{(I+P)^{-1} zz^T (I + P)^{-1}}{sqrt{z^T (I+P)^{-2} z}. $$ Above, $\mathcal{P} = \{P\}$. However, my proof does not generalize to $|\mathcal{P}| > 1$. I am wondering if there is a more clever way to obtain this result.

Let $z$ denote a unit vector. Fix a finite collection of positive semidefinite matrices $\mathcal{P}$. Define the function and set $$ f_{\mathcal{P}}(X) = \sum_{P \in \mathcal{P}} z^T X(PX + I)^{-1} z, \quad \mbox{and} \quad \mathcal{X} = \{X \succeq 0,~\mathrm{trace}(X) = 1\}. $$ Above, $X \succeq 0$ means that $X$ is a real, symmetric, and positive semidefinite matrix.

I am interested in the following quantities, $$ f^\star_{\mathcal{P}} = \sup_{X \in \mathcal{X}} f_{\mathcal{P}}(X) \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \mathrm{argmax}_{X \in \mathcal{X}} f_{\mathcal{P}}(X). $$ By continuity and compactness, the supremum is attained and both quantities are well-defined.

In the case $|\mathcal{P}| = 1$, I am able to directly compute this. I obtain $$ f^\star_{\mathcal{P}} = z^T(I + P)^{-1} z \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \frac{(I+P)^{-1} zz^T (I + P)^{-1}}{z^T (I+P)^{-2} z}. $$ Above, $\mathcal{P} = \{P\}$. However, my proof does not generalize to $|\mathcal{P}| > 1$. I am wondering if there is a more clever way to obtain this result.

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Drew Brady
Drew Brady
  • 460
  • 4
  • 16

Let $z$ denote a unit vector. Fix a finite collection of positive semidefinite matrices $\mathcal{P}$. Define the function and set $$ f_{\mathcal{P}}(X) = \sum_{P \in \mathcal{P}} z^T X(PX + I)^{-1} z, \quad \mbox{and} \quad \mathcal{X} = \{X \succeq 0,~\mathrm{trace}(X) = 1\}. $$ Above, $X \succeq 0$ means that $X$ is a real, symmetric, and positive semidefinite matrix.

I am interested in the following quantities, $$ f^\star_{\mathcal{P}} = \sup_{X \in \mathcal{X}} f_{\mathcal{P}}(X) \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \mathrm{argmax}_{X \in \mathcal{X}} f_{\mathcal{P}}(X). $$ By continuity and compactness, the supremum is attained and both quantities are well-defined.

In the case $|\mathcal{P}| = 1$, I am able to directly compute this. I obtain $$ f^\star_{\mathcal{P}} = z^T(I + P)^{-1} z \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \frac{(I+P)^{-1} zz^T (I + P)^{-1}}{\sqrt{z^T (I+P)^{-2} z}}. $$$$ f^\star_{\mathcal{P}} = z^T(I + P)^{-1} z \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \frac{(I+P)^{-1} zz^T (I + P)^{-1}}{sqrt{z^T (I+P)^{-2} z}. $$ Above, $\mathcal{P} = \{P\}$. However, my proof does not generalize to $|\mathcal{P}| > 1$. I am wondering if there is a more clever way to obtain this result.

Let $z$ denote a unit vector. Fix a finite collection of positive semidefinite matrices $\mathcal{P}$. Define the function and set $$ f_{\mathcal{P}}(X) = \sum_{P \in \mathcal{P}} z^T X(PX + I)^{-1} z, \quad \mbox{and} \quad \mathcal{X} = \{X \succeq 0,~\mathrm{trace}(X) = 1\}. $$ Above, $X \succeq 0$ means that $X$ is a real, symmetric, and positive semidefinite matrix.

I am interested in the following quantities, $$ f^\star_{\mathcal{P}} = \sup_{X \in \mathcal{X}} f_{\mathcal{P}}(X) \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \mathrm{argmax}_{X \in \mathcal{X}} f_{\mathcal{P}}(X). $$ By continuity and compactness, the supremum is attained and both quantities are well-defined.

In the case $|\mathcal{P}| = 1$, I am able to directly compute this. I obtain $$ f^\star_{\mathcal{P}} = z^T(I + P)^{-1} z \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \frac{(I+P)^{-1} zz^T (I + P)^{-1}}{\sqrt{z^T (I+P)^{-2} z}}. $$ Above, $\mathcal{P} = \{P\}$. However, my proof does not generalize to $|\mathcal{P}| > 1$. I am wondering if there is a more clever way to obtain this result.

Let $z$ denote a unit vector. Fix a finite collection of positive semidefinite matrices $\mathcal{P}$. Define the function and set $$ f_{\mathcal{P}}(X) = \sum_{P \in \mathcal{P}} z^T X(PX + I)^{-1} z, \quad \mbox{and} \quad \mathcal{X} = \{X \succeq 0,~\mathrm{trace}(X) = 1\}. $$ Above, $X \succeq 0$ means that $X$ is a real, symmetric, and positive semidefinite matrix.

I am interested in the following quantities, $$ f^\star_{\mathcal{P}} = \sup_{X \in \mathcal{X}} f_{\mathcal{P}}(X) \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \mathrm{argmax}_{X \in \mathcal{X}} f_{\mathcal{P}}(X). $$ By continuity and compactness, the supremum is attained and both quantities are well-defined.

In the case $|\mathcal{P}| = 1$, I am able to directly compute this. I obtain $$ f^\star_{\mathcal{P}} = z^T(I + P)^{-1} z \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \frac{(I+P)^{-1} zz^T (I + P)^{-1}}{sqrt{z^T (I+P)^{-2} z}. $$ Above, $\mathcal{P} = \{P\}$. However, my proof does not generalize to $|\mathcal{P}| > 1$. I am wondering if there is a more clever way to obtain this result.

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Drew Brady
Drew Brady
  • 460
  • 4
  • 16

Maximum of $\sum_{n=1}^N z^T X(P_n X + I)^{-1}z$ over unit trace, positive semidefinite matrices?

Let $z$ denote a unit vector. Fix a finite collection of positive semidefinite matrices $\mathcal{P}$. Define the function and set $$ f_{\mathcal{P}}(X) = \sum_{P \in \mathcal{P}} z^T X(PX + I)^{-1} z, \quad \mbox{and} \quad \mathcal{X} = \{X \succeq 0,~\mathrm{trace}(X) = 1\}. $$ Above, $X \succeq 0$ means that $X$ is a real, symmetric, and positive semidefinite matrix.

I am interested in the following quantities, $$ f^\star_{\mathcal{P}} = \sup_{X \in \mathcal{X}} f_{\mathcal{P}}(X) \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \mathrm{argmax}_{X \in \mathcal{X}} f_{\mathcal{P}}(X). $$ By continuity and compactness, the supremum is attained and both quantities are well-defined.

In the case $|\mathcal{P}| = 1$, I am able to directly compute this. I obtain $$ f^\star_{\mathcal{P}} = z^T(I + P)^{-1} z \quad \mbox{and} \quad X^\star_{\mathcal{P}} = \frac{(I+P)^{-1} zz^T (I + P)^{-1}}{\sqrt{z^T (I+P)^{-2} z}}. $$ Above, $\mathcal{P} = \{P\}$. However, my proof does not generalize to $|\mathcal{P}| > 1$. I am wondering if there is a more clever way to obtain this result.