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8

An equivalent trick : Let $J:={\rm diag}(1,i,-1,-i)$. Then $J^*AJ=iB$ where $B$ is real and skew-symmetric. Hence the spectrum of $iB$ (thus that of $A$) comes by pairs $\pm\lambda$.


20

For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with a matrix $X$ that squares to the identity: $$X=\left( \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ \end{array} \right),\;\;XA+AX=0,\;\;X^2=I.$$ Hence the spectrum ...


2

Using (say) the Jordan form, approximate $A$ by a nonsingular matrix $C_t$ so that $C_t\to A$ as $t\to0$. Then $$C_t^\top (C_t X^{-1} C_t^\top + B)^{-1} C_t\le C_t^\top (C_t X^{-1} C_t^\top)^{-1} C_t=X.$$ Letting now $t\to0$, we get $$A^\top (A X^{-1} A^\top + B)^{-1} A \le X,$$ as desired.


4

Let us denote the matrices in question by $\rho_{nm}$ and, first, consider the use of the measure generated by the Hilbert-Schmidt metric \begin{equation} \mbox{d} s^2_{HS}=\frac{1}{2} \mbox{Tr}[(\mbox{d} \rho_{nm})^2]. \end{equation} Then, A. Lovas and A. Andai LovasAndaiPaper MR3673324 have formally answered the question for the case of symmetric $\rho_{...


3

I answer the version where you assume that the diagonal entries of $Df$ are constant (all $1$, in your problem), and the off-diagonal entries of each $Df(x)$ are all the same value $k(x)$. I assume that, by linear, you mean affine linear. Then this is clear if $n = 1$. Otherwise, fix $i$, let $f_i$ be the appropriate component function of $f$, and choose ...


2

These matrices are a special case of a much broader class of positive semidefinite (psd) matrices constructed as follows. Let $(\alpha_{i,j})_{i,j=1}^n$ be any psd matrix satisfying $|\alpha_{i,j}| < 1$ for all $i,j \in \{1,\ldots,n\}$. Then the matrix $((1-\alpha_{i,j})^{-c})_{i,j=1}^n$ is positive semidefinite for all $c \geq 0$. This follows ...


6

This matrix is also related to the Nevanlinna-Pick Theorem. Namely, if $z_i, \lambda_i \in \mathbb D, 1\leq i\leq n$ then $$\left[\begin{matrix} \frac{1- \overline{z_j}z_i}{1-\overline{\lambda_j}\lambda_i}\end{matrix}\right]_{i,j=1}^n \geq 0$$ if and only if there is a holomorphic function $\varphi : \mathbb D \rightarrow \overline{\mathbb D}$ such that $\...


3

Assuming all the $\alpha_j$ are nonzero, the matrices $X$ are Cauchy-like matrices, since you can rewrite them as $$ X_{ij} = \frac{\alpha_j^{-1}}{\alpha_j^{-1}-\bar{\alpha}_i} $$ so there are analogous formulas for their determinant and inverse. In particular, $XA^{-1}$ is a Cauchy matrix, where $A = diag(\alpha_i)$, so these formulas follow directly from ...


0

The following is an attempt to validate the conclusion proposed by @vidyarthi. Theorem: Every doubly-stochastic partial-isometric matrix is the product of a permutation matrix and a doubly-stochastic projection. Proof: Given a doubly-stochastic partial-isometric matrix $A$, one has that $A^tA$ and $AA^t$ are doubly-stochastic projections, so by Theorem 2 ...


-1

From here, we have that a square matrix is a partial isometry if and only if it is of the form $A=UD=EU$, where $D, E$ are idempotent and $U$ is unitary. Translating this to our case, we have that a doubly stochatic matrix is a partial isometry when it is a product of an orthogonal matrix (scaled by a scalar) with an idempotent matrix (again scaled by ...


7

No, the expression $$ \left\|\left[\begin{matrix}A \\ B\end{matrix}\right]\right\|_p = \| |A|^p + |B|^p\|^{1/p}. $$ is not a norm for any $2<p<\infty$ (so it is a norm if and only if $p=2,\infty$). I will justify that by proving that the triangle inequality for this expression would imply that the map $t\mapsto t^{p/2}$ is operator monotone, which is ...


2

We prove that $$\sum_{i=1}^k \sigma^\downarrow_i(AB) = \sup_{U}|\mathrm{Tr}(UAB)| \le \sup_{U,V}|\mathrm{Tr}(UAV^*B)| =\sum_{i=1}^k \sigma^\downarrow_i(A)\sigma^\downarrow_i(B),$$ where $U$ and $V$ run over all partial isometries (or contractions) of rank (at most) $k$. The only nontrivial is $\le$ part of the rightmost equality. For the proof of this, we ...


3

To see what you might expect for a relation, consider the case of a $2\times 2$ matrix $M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$, with eigenvalues $\lambda_\pm=\tfrac{1}{2}(a+d)\pm\sqrt{4bc+(a-d)^2}$. Knowledge of $a$ and $d$ is a constraint on the sum of the eigenvalues (the trace of $M$), but there is no other constraint. I'm pretty sure this ...


12

Since rank$(A)\geq 2$, the matrix $A$ has a $2\times 2$ submatrix $a$ with nonzero determinant; the determinant is a continuous function of the matrix elements, so adding a sufficiently small perturbation $\alpha B$ to $A$ will leave $\det a\neq 0$ and hence the rank of $A+\alpha B$ remains $\geq 2$. If instead of small $\alpha$ you wish to take large $\...


0

This does not seem a simple matrix equation to solve. Computationally, my first attempt would be with Newton's method, even if it takes $O(k^6)$ per iteration, where $k$ is the size of the matrices. The Jacobian of the map in the LHS is $$ L_B f[H] = \sum_{i=1}^n (A_i+B)^{-1}H(A_i+B)^{-1}, $$ and to solve the equation $L_Bf[H] = Y$ for $H$ given $Y$ you need ...


0

Inspired by @Mateusz Wasilewski I find another method. \begin{eqnarray*} \langle x,Ax\rangle & = & \langle Lx,Lx\rangle\\ & = & \sum_{i=1}^{n}u_{i}^{2} \end{eqnarray*} where $u_{i}=\sum_{j=i}^{n}\frac{1}{j}x_{j}$. \begin{eqnarray*} \sum_{i=1}^{n}u_{i}^{2} & = & \sum_{i=1}^{n}(\sum_{k=i}^{n}b_{k})^{2}\quad(\text{where} \ b_{k}=\frac{1}...


0

Just for those who might want to know, I think the min-max theorem mentioned by @Michael Renardy is Courant-Fischer: Supporse a real symmetric matrix A's eigen values are $\lambda_1 \leq \lambda_2 \leq ... \leq \lambda_n $. then k-th eigen value: $$ \lambda_k = \underset{V_k}{\text{min}} \space max\{x^TAx | x \in V_k, ||x||_2 = 1\}$$ $V_k$ is any k-...


3

First, by transposing and interchanging two rows $S({\bf x},{\bf y}) = -\left|\begin{matrix} x_0 & y_0 & 2y_2 \\ x_1 & y_1 & y_0 \\ x_2 & y_2 & y_1 \end{matrix} \right|$. Notice also that if $T = \left[\begin{matrix}0&0&2 \\ 1&0 &0 \\ 0& 1& 0\end{matrix} \right]$ then $T{\bf y} = \left[\begin{matrix} 2y_2 \\ ...


1

In this answer I show that the largest eigenvalue is bounded by $5< 3 + 2\sqrt{2}$. I will first use the interpretation of this matrix as the covariance matrix of the Brownian motion at times $(\frac{1}{n},\dots, 1)$ (I reversed the order so that the sequence of times is increasing, which is more natural for me). We have $A_{ij} = \mathbb{E} (B_{t_{i}} B_{...


3

$$B=\left[\matrix{1&1&1&1\\1&1&-1&-1\\-1&-1&-1&1\\-1&1&-1&1}\right]$$ and probably many other solutions. I'm also voting to close because you didn't pose a research-level problem. If you have an interesting general case, pose that.


14

$B^{-1/2}XAX^TB^{-1/2}=I$, so $B^{-1/2}XA^{1/2}=Q$ must be orthogonal. On the other hand, for any orthogonal $Q$, it is simple to verify that $X = B^{1/2}QA^{-1/2}$ solves the equation, so this is a complete parametrization of the solutions. Here $A^{1/2}$ is the symmetric square root of $A$ (if you prefer you can work with the Cholesky factor and obtain ...


0

Regarding the inequality for $\Vert AB \Vert_2 $, although it does not hold in the form you wrote it as Denis pointed out, replacing the 2-norm by the 1-norm leads to Holder's inequality for Schatten norms: \begin{equation} \Vert AB \Vert_1 \leq \Vert A \Vert _p \Vert B \Vert _q \end{equation}


3

Decided to turn my comment into an answer since it seems to be independent of the ambiguity noticed by @LSpice. Take $$ A= \begin{pmatrix} 0&1&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{pmatrix}, $$ $$...


3

Assuming that you are asking for an isomorphism taking $B$ to $B$': No. Take $A$ to be the $n\times n$ Jordan block with eigenvalue zero. The centralizer of $A$ is exactly $\mathbb C[A]$. Hence for any $B,B' \in \mathbb C[A]$, we have $\mathbb C[A,B] = \mathbb C[A] = \mathbb C[A,B']$. However, $B$ and $B'$ need not be conjugate: take $B = A^2, B' = A^3$. As ...


2

Let $r_i := p_i - q_i$. $${\bf A} (x) := \begin{bmatrix} p_1 (x) & q_1 (x) & \ldots & q_1 (x)\\ q_2 (x) & p_2 (x) & \ldots & q_2 (x)\\ \vdots & \vdots & \ddots & \vdots\\ q_n (x) & q_n (x) & \ldots & p_n (x)\end{bmatrix} = \mbox{diag} \left( {\bf r} (x) \right) + {\bf q} (x) {\Bbb 1}_n^\top$$ Using the matrix ...


3

That equation is called a (continuous-time) algebraic Riccati equation, and there is ample literature on when they are solvable; just look for this search term. For instance, the book Algebraic Riccati equations by Lancaster and Rodman, or Numerical solution of AREs by Bini, Iannazzo, Meini. In the generic case there is a finite number of solutions that ...


0

By combining the useful comments of Rodrigo and Todd, the methodology to solve this system is shown here below. One caveat is that the method is probably not very efficient, since you need to use the decomposition of $B$ two times to find the final solution. 1. Build Equations with Schur Complement By using Schur Complement on the original problem, one can ...


1

The nuclear norm $\|\cdot\|_*$ is a norm and hence a convex function. On the other hand, the set $$S:=\{X\in\mathbb R^{m\times n}\colon X_{ij}\ge0\text{ and }\sum_{j=1}^n X_{ij}=1\ \forall i,j\}$$ is convex and compact. So, the maximum of the nuclear norm $\|\cdot\|_*$ on the set $S$ is attained at an extreme point of $S$, which is clearly a matrix $X\in S$ ...


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