Return to Question

Let A and B be two positive definite, real, symmetric matrices. Their eigenvalues are constraint to be $\lambda(A), \lambda(B)$ respectively. I know (from 1) that the eigenvalues of AB must obey $$ \lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec \lambda(AB) \prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B) $$ where $\downarrow$ indicates decreasing order, $\uparrow$ increasing order, $x \cdot y := (x_1y_1,\ldots ,x_ny_n)$ for $x,y \in \mathbb{R}^n$ and $\prec$ is the majorization preorder.

My question is, for a given set of eigenvalues $\lambda(A),\, \lambda(B)$ and $\lambda(AB)$ which satisfies the above, does there necessarily exist an A and B such that they and AB have the desired eigenvalues?

I've found plenty of material on similar inequalities, but nothing that states if every solution of the inequalities can be realised.

Let A and B be two positive definite, real, symmetric matrices. The eigenvalues of A, B and AB, denoted by $\lambda(X)$, obey the relation (from Bhatia): $$ \lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec \lambda(AB) \prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B) $$ where $\downarrow$ indicates decreasing order, $\uparrow$ increasing order, $x \cdot y := (x_1y_1,\ldots ,x_ny_n)$ for $x,y \in \mathbb{R}^n$ and $\prec$ is the majorization preorder.

My question is, for a given set of eigenvalues $\lambda(A),\, \lambda(B)$ and $\lambda(AB)$ which satisfies the above, does there necessarily exist an A and B such that they and AB have the desired eigenvalues?

I've found plenty of material on similar inequalities, but nothing that states if every solution of the inequalities can be realised.

Let A and B be two positive definite, real, symmetric matrices. Their eigenvalues are constraint to be $\lambda(A), \lambda(B)$ respectively. I know (from 1) that the eigenvalues of AB must obey $$ \lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec \lambda(AB) \prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B) $$ where $x \cdot y := (x_1y_1,\ldots ,x_ny_n)$ for $x,y \in \mathbb{R}^n$ and $\prec$ is the majorization preorder.

My question is, for a given set of eigenvalues $\lambda(A),\, \lambda(B)$ and $\lambda(AB)$ which satisfies the above, does there necessarily exist an A and B such that they and AB have the desired eigenvalues?

I've found plenty of material on similar inequalities, but nothing that states if every solution of the inequalities can be realised.

Let A and B be two positive definite, real, symmetric matrices. Their eigenvalues are constraint to be $\lambda(A), \lambda(B)$ respectively. I know (from 1) that the eigenvalues of AB must obey $$ \lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec \lambda(AB) \prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B) $$ where $\downarrow$ indicates decreasing order, $\uparrow$ increasing order, $x \cdot y := (x_1y_1,\ldots ,x_ny_n)$ for $x,y \in \mathbb{R}^n$ and $\prec$ is the majorization preorder.

My question is, for a given set of eigenvalues $\lambda(A),\, \lambda(B)$ and $\lambda(AB)$ which satisfies the above, does there necessarily exist an A and B such that they and AB have the desired eigenvalues?

I've found plenty of material on similar inequalities, but nothing that states if every solution of the inequalities can be realised.

Let A and B be two positive definite, real, symmetric matrices. Their eigenvalues are constraint to be $\lambda(A), \lambda(B)$ respectively. I know (from 1) that the eigenvalues of AB must obey $$ \lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec \lambda(AB) \prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B) $$

My question is, given a set of eigenvalues $\lambda(A),\, \lambda(B)$ and $\lambda(AB)$ which satisfies the above, does there necessarily exist an A and B such that they and AB have the desired eigenvalues?

I've found plenty of material on similar inequalities, but nothing that states if every solution of the inequalities can be realised.

Let A and B be two positive definite, real, symmetric matrices. Their eigenvalues are constraint to be $\lambda(A), \lambda(B)$ respectively. I know (from 1) that the eigenvalues of AB must obey $$ \lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec \lambda(AB) \prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B) $$ where $x \cdot y := (x_1y_1,\ldots ,x_ny_n)$ for $x,y \in \mathbb{R}^n$ and $\prec$ is the majorization preorder.

My question is, for a given set of eigenvalues $\lambda(A),\, \lambda(B)$ and $\lambda(AB)$ which satisfies the above, does there necessarily exist an A and B such that they and AB have the desired eigenvalues?

I've found plenty of material on similar inequalities, but nothing that states if every solution of the inequalities can be realised.