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5

YCor's comment contains the essential idea needed for the proof, but maybe a few more details would be helpful. (If YCor does provide something similar later, feel free to award YCor's answer the bounty.) Consider the mapping $\sigma:\mathrm{GL}(2,\mathbb{H})\to M_2(\mathbb{H})$ given by $$ \sigma(A) = A^* A $$ where $A^*$ is the conjugate transpose of $A$ ...


1

Q1: $dZ/dK$ is a $p\times q$ matrix with elements $$\bigl[dZ/dK\bigr]_{ij}=\sum_{k=1}^n\frac{\partial \mu_k}{\partial K_{ij}}\,\mathbb{E}[f(x)\,\Sigma^{-1}\cdot(x-\mu)]_k.$$ Q2: To evaluate this you will have to specify the function $f(x)$. For some $f$ you will be able to evaluate the expectation value symbolically, but in general a numerical evaluation ...


10

Your question is equivalent to whether the category $\mathcal{E}$ of pairs $(V,f)$ consisting of a finitely generated free (right) $R$-module and an endomorphism $f$ of $V$ is a Krull-Schmidt category, i.e., an additive category where every object decomposes as a direct sum of finitely many indecomposable objects and the decomposition is unique up to ...


2

Let me elaborate on the use of Pólya enumeration for this problem. Let $G\simeq S_m \times S_n$ act on $X=\{x_{i,j}\colon 1\leq i \leq m, 1\leq j \leq n\}$ via $(\sigma,\pi)\cdot x_{i,j} = x_{\sigma(i),\pi(j)}$, and via this action view $G\subseteq S_{mn}$. For $g\in G$ let $c_i(g)$ be the number of $i$-cycles of $g$ (in this embedding into $S_{mn}$), and ...


2

Indeed, the condition $\det M\ne0$ can be expressed as a certain non-independence condition, as follows. For $i$ and $j$ in $[J]:=\{1,\dots,J\}$, let \begin{equation*} p_{i|j}:=P(X=i|Y=j), \end{equation*} so that $M=[p_{i|j}]_{i,j\in[J]}$. Suppose that $\det M=0$. Then for some real $c_1,\dots,c_J$ not all of which are $0$ and for all $i\in[J]$ we have \...


0

$\DeclareMathOperator\diag{diag}\DeclareMathOperator\rank{rank}$Here is how we can construct solutions in $P$. Necessarily, $r\leq d$, $\rank(P)=r$, $\rank(S)=r_1\geq r$. There is $Q$ invertible s.t. $QAQ^T=I_d$, $QSQ^T=\diag((\lambda_i)_{i\leq p_1},(-\mu_j)_{j\leq q_1},0_{s})$, where $\lambda_i,\mu_j>0$ and $p_1+q_1=r_1$, $p_1+q_1+s=d$. Let $p$, $q$ be s....


0

As Federico explained, systems of quadratic are as general as arbitrary systems of polynomial equations of any degrees. So a general method for solving such systems analytically is out of the question. It is only for special systems of quadratic equations where there is additional algebraic structure that one can solve the system explicitly. A notable ...


4

There is certainly nothing special for quadratics, as you suggest in your comments. Indeed, a generic system of polynomial equations can be reduced to a system of quadratic ones by introducing extra variables. For instance, if you have a term $x^3y^2$ in a polynomial system you can introduce the three auxiliary variables $t=y^2, u=x^2, v=tu$, and then you ...


1

When the $(B_i)$ are in general position, there are always $2^s$ complex solutions but we don't have control over the number of real solutions. Let $n=2^{s-1}$. In generic cases, the system can be decomposed as follows: $\{x_i=Q_i(x_s),i<s,P(x_s)=0\}$ where $Q_i \in \mathbb{R}_{2n-1}[x],P \in\mathbb{R}_{2n}[x]$. The complexity is in the resolution of $P(x)...


8

A simple brute force method worked (even though I'm not happy with this). Let $\zeta=\xi+i\eta$ be a non-positive eigenvalue of $M$ and $\left[\begin{matrix} x & y & z \end{matrix}\right]^T$ be a corresponding eigenvector. This gives equations \begin{align} Ax + By \qquad &= \zeta x \\ B^Tx + Cy - B^Tz &= \zeta y\\ -Ax -By + Az &= \zeta ...


5

This post is the solution to the limit case, as suggested by Denis Serre, where the Schur complement $S=C-B^TA^{-1}B$ is zero. This means that $W:=A^{-1/2}BC^{-1/2}$ is an orthogonal matrix. Now put $S=\operatorname{diag}(A^{1/2}, C^{1/2}W^T, A^{1/2})$ and consider $$N:=S^{-1}MS =\left[\begin{matrix} A & D & 0 \\ A & D & -A \\ -A & -D &...


2

Let $\mathcal{A}$ be a finite-dimensional commutative associative unital $\mathbb{R}$-algebra. (This covers split-complex, hyberbolic and a lot of other number "systems".) We want to extend the function $\operatorname{sign}(x)$ to $\mathcal{A}$ in a meaningful way (to be determined) and understand the number of values in its range. If $\mathcal{A}$ ...


2

Since there is not much progress on this question, let me give a partial result and a direct consequence. Denote $S=C-B^TA^{-1}B$ the Schur complement of $A$ in $N$, which is positive definite. Then $M$ factorizes $UL$ with $$U=\begin{pmatrix} A & B & 0 \\ 0 & S & -B^T \\ 0 & 0 & A \end{pmatrix},\qquad L=\begin{pmatrix} I & 0 &...


1

This is only for the case n=1. Clearly the claim is true if B=0. Hence it suffices to show that there is never a zero or purely imaginary eigenvalue. The case of a zero eigenvalue leads to $a(ac-b^2)=0$, but this quantity is positive by assumption. The case of a purely imaginary eigenvalue $iy$ leads to the pair of equations $$A(AC-B^2)-y^2(2A+C)=0,$$ $$-A^2+...


6

The algorithm used in Kwant, mentioned by Carlo uses adaptive sampling to resolve level crossings, and it it fairly involved. For for completeness I'll present a simpler approach that uses eigenvector continuity and works with the simplistic grid sampling. The detailed description is in this blog post. The key idea is to take a set of eigenvectors $|\psi_i(k)...


6

The open source Python package Kwant has built in routines to find the bands for any lattice Hamiltonian (tight-binding representation). It resolves the crossings by ensuring a continuous derivative. One developer of the package, Anton Akhmerov, has posting on this problem, which I found quite illuminating.


2

Following the ideas from my answer to Carlo's question, let $M(z_1, \ldots, z_n)$ be the symmetric matrix with $M(z_1, \ldots, z_n)_{ij} = \Sigma_{ij}$ for $i \neq j$ and $M(z_1, \ldots, z_n)_{ii} = z_i$. Use any sort of numerical optimization algorithm to maximize $$\log \det M(z_1, \ldots, z_n) - \sum_{i=1}^n \Sigma^{-1}_{ii} z_i$$ on the open set of $(z_1,...


13

The conjecture is true. More precisely, here is what I will prove: Theorem Partition $\{ (i,j) : 1 \leq i \leq j \leq n \}$ into two disjoint sets, $A \sqcup B$. Let $X$ and $Y$ be positive definite $n \times n$ matrices and suppose that $X_{ij} = Y_{ij}$ for $(i,j) \in A$ and $(X^{-1})_{ij} = (Y^{-1})_{ij}$ for $(i,j) \in B$. Then $X = Y$. Lemma 1 Consider ...


3

This answer arose by a discussion with @JamieGabe in the comments. One can prove that the map $$\Phi: M_n(A) \to M_n(A): A \mapsto n \operatorname{Diag}(A)-A$$ is completely positive [Paulsen, "Completely bounded maps and operator spaces", exercise 3.6]. In particular, it is positive. Hence, writing $A= (a_{i,j})$ as in the OP, we obtain $$A \le n \...


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