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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 ...

0

Consider the following argument for a slightly changed problem (with $s^2$ in $C(s)$ replaced by $-s^2$) . Not sure if this would be of any help, but writing it anyway. Note that due to concavity of $-s^2$ in C(s) and the rest of the terms being either constant or linear in $s$ (in $C(s)$), we have: $$C(\lambda s_1 + (1-\lambda)s_2) \succeq \lambda C(s_1) +... 1 Kadison in this paper proves that self-adjoint A and B will have a greatest lower bound if and only if they are comparable (meaning A\leq B or B\leq A or A=B). Another way to put this is that the Loewner order is an anti-lattice. Since this is for all self-adjoint then one can flip the order by multiplying by -1 to get that A and B will have ... 2 These matrices are a special case of a much broader class of positive semidefinite (psd) matrices constructed as follows. Let (\alpha_{i,j})_{i,j=1}^n be any psd matrix satisfying |\alpha_{i,j}| < 1 for all i,j \in \{1,\ldots,n\}. Then the matrix ((1-\alpha_{i,j})^{-c})_{i,j=1}^n is positive semidefinite for all c \geq 0. This follows ... 2 This answer has been updated with an explicit example showing more conditions are needed. After the example we give some additional necessary conditions. Let f(x,y) = \sum_{j=0}^n a_j x^{n-j}y^j. One other simple condition not in the question is a_j \leq \binom{n}{j}. But we also need some lower bounds of the coefficients a_j. Consider the nonlinear ... 10 Benjamin Steinberg answered the question, but I wanted to unwind his idea into an explicit formula. Let V be the representation with basis e_1,\dotsc, e_n, where a permutation \sigma acts by sending e_i to e_{\sigma(i)}. Let W be an irreducible representation of S_n. Fix a linear form l on W. We can map W to V^{\otimes n} by sending w \... 13 Turning my comment into an answer, Robert Steinberg proved in Complete sets of representations of algebras that if you have a faithful representation of a finite semigroup S, then every irreducible representation of S appears as a composition factor of a tensor power of that representation. In fact, he shows that the semigroup algebra KS acts ... 6 This matrix is also related to the Nevanlinna-Pick Theorem. Namely, if z_i, \lambda_i \in \mathbb D, 1\leq i\leq n then$$\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$$if and only if there is a holomorphic function \varphi : \mathbb D \rightarrow \overline{\mathbb D} such that \... 2 Of course, without assumptions on the behavior of the eigenvectors, your desired conclusion will not hold. E.g., for t:=\lambda, let$$A(t):=\left( \begin{array}{cc} 2+\cos t & \sin t \\ \sin t & 2-\cos t \\ \end{array} \right),\quad L:=\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right).$$The eigenvalues of A(t) are 3 and 1 for all t... 3 Assuming all the \alpha_j are nonzero, the matrices X are Cauchy-like matrices, since you can rewrite them as$$ X_{ij} = \frac{\alpha_j^{-1}}{\alpha_j^{-1}-\bar{\alpha}_i} $$so there are analogous formulas for their determinant and inverse. In particular, XA^{-1} is a Cauchy matrix, where A = diag(\alpha_i), so these formulas follow directly from ... 7 No, the expression$$ \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),$$where U and V run over all partial isometries (or contractions) of rank (at most) k. The only nontrivial is \le part of the rightmost equality. For the proof of this, we ... 3 To see what you might expect for a relation, consider the case of a 2\times 2 matrix M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}, with eigenvalues \lambda_\pm=\tfrac{1}{2}(a+d)\pm\sqrt{4bc+(a-d)^2}. Knowledge of a and d is a constraint on the sum of the eigenvalues (the trace of M), but there is no other constraint. I'm pretty sure this ... 12 Since rank(A)\geq 2, the matrix A has a 2\times 2 submatrix a with nonzero determinant; the determinant is a continuous function of the matrix elements, so adding a sufficiently small perturbation \alpha B to A will leave \det a\neq 0 and hence the rank of A+\alpha B remains \geq 2. If instead of small \alpha you wish to take large \... 5 Your actual Question has nothing to do with operads. Perhaps it is clarifying to consider the following more general setting: let G be a group, X and Y be right G-sets, and f : X \to Y be a function. If g, h \in G and f commutes with the actions of g and of h then it commutes with the action of gh:$$f(x) \cdot (gh) = (f(x) \cdot g) \cdot ...

2

There is a pretty direct argument here. Note first that $G = {\rm PGL}(3,2) = {\rm SL}(3,2)$ is a finite simple group of order 168. Suppose it were so generated. Note next that $a$ inverts both $ab$ and $ac$, so normalizes $\langle ab, ac \rangle$. Now $bc = (ca)(ba)^{-1} \in \langle ab,ac \rangle$, so that $a$ normalizes $\langle ab,bc,ac \rangle.$ The ...

11

This cannot be done. Let $G_1$ and $G_2$ be the groups $$G_1 = \langle a,b,c | a^2 = b^2 = c^2 = (ab)^4 = (bc)^4 = (ac)^2 \rangle$$ $$G_2 = \langle a,b,c | a^2 = b^2 = c^2 = (ab)^3 = (bc)^3 = (ac)^3 \rangle.$$ You are looking for surjections from the $G_j$ onto $PGL_3(\mathbb{F}_2)$, which is the simple group of order $168$. Now, $G_1$ and $G_2$ are Coxeter ...

2

Neat question! $2^n$ (for diagonal matrices) is indeed an upper bound, but the answer might not be a power of 2. To see the upper bound: start by throwing out, WLOG, any entries from your set that don't change the determinant. Now changing a single entry affects the determinant as some linear and generically non-constant function. Also, if you change two ...

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-...

1

The invariance of the Haar measure implies that the probability to draw the matrix $U$ from the unitary group is unchanged if you replace $U$ by $U_0 U U_0^\dagger$, with $U_0$ an arbitrary unitary matrix. Since this conjugation changes the unitary matrix of eigenvectors from $V$ into $U_0V$, it means that $V$ and $U_0V$ are equally probably in the unitary ...

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.

7

Yes. Let the size of your matrix be $n$. Your condition implies that there is an $n-1\times n-1$ submatrix whose determinant is not identically equal to $0$. Assume without loss of generality that this is the submatrix formed by the first $n-1$ rows and columns. Then we can set $u_n=1$ and find a vector $u(z)$ such that $M(z)u(z)=0$ by solving the system of $... 0 Given any$n$distinct points$\{x_i/x_i\in(0,1)\}$which are pairwise distinct. For any$c_i,i = 1,2,3,...n$, and not all zeros. Using the given expression for$g(x)$we can deduce that $$\sum_{i=1}^n\sum_{j=1}^nc_ic_jg(x_i-x_j) = \sum_{\eta\in\mathbb{Z}} \left(\frac{1}{1+\gamma\eta^2} \left|\sum_{i=1}^n c_i e^{2\pi i \eta x_i}\right|^2 \right)> 0$$ as$\...

2

Re your answer: Statement 1 follows from Weyl's inequalities and $\rho_1(C) = O(\lambda^{-1})$. $2 \implies 3$ seems problematic; what if $C = \frac{1}{\lambda^3}I$, which should be possible given the definition of big-O? Then $\rho_1(M^{-1}) \sim \lambda^2 \not\in O(\lambda)$. Statement 4 seems just the matrix norm inequality $\|M^{-1}e\| \leq \|M^{-1}\|\|e\... 1 My proof (some statements are without proof, for which I need references) Let$X$be any$n\times n$square matrix and Let$\rho_1(X), \rho_2(X), \rho_3(X),...\rho_i(X),...\rho_n(X)$denote the eigenvalues in descending order, of the matrix$X$. Statement : 1* $$\rho_1(A+C) = n+O(1/\lambda)$$ $$\rho_i(A+C) = 0+O(1/\lambda),i = 2,3...n$$ Let$M = A+C+\frac{1}{...

3

Decided to turn my comment into an answer since it seems to be independent of the ambiguity noticed by @LSpice. Take $$A= \begin{pmatrix} 0&1&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{pmatrix},$$ $$... 3 Assuming that you are asking for an isomorphism taking B to B': No. Take A to be the n\times n Jordan block with eigenvalue zero. The centralizer of A is exactly \mathbb C[A]. Hence for any B,B' \in \mathbb C[A], we have \mathbb C[A,B] = \mathbb C[A] = \mathbb C[A,B']. However, B and B' need not be conjugate: take B = A^2, B' = A^3. As ... 0 By combining the useful comments of Rodrigo and Todd, the methodology to solve this system is shown here below. One caveat is that the method is probably not very efficient, since you need to use the decomposition of B two times to find the final solution. 1. Build Equations with Schur Complement By using Schur Complement on the original problem, one can ... 2 I am not sure I get what you mean by "brought to Jordan form", but if you don't consider the structure of D while changing basis for N then it won't work. Example:$$ U = \begin{bmatrix} 1 & 1& 0 & 0\\ 0 & 1& 0 & 0\\ 0 & 0 & 2 & 1\\ 0 & 0 & 0 & 2 \end{bmatrix} $$has$$ N = \begin{bmatrix} 0 &...

4

Sure, let $\mu_n = \frac{1}{|n|+ 1}$, then for each $n$ the vector $\sqrt{\mu_{n-1}}e_n\oplus \pm\sqrt{\mu_n}e_{n-1}$ is an eigenvector with eigenvalue $\pm\sqrt{\mu_n\mu_{n-1}}$.

9

Note that $$T^2 = \begin{pmatrix}lMrM&0\\0&rMlM\end{pmatrix},$$ and hence the eigenvectors of $T^2$ are $$v_j = (e_j, 0) , \qquad w_j = (0, e_j),$$ with corresponding eigenvalues $$\lambda_j = \frac{1}{(1 + |j|) (1 + |j+1|)} \, , \qquad \mu_j = \frac{1}{(1 + |j|) (1 + |j-1|)} \, ,$$ respectively. In particular, the eigenspaces of $T^2$ are four-...

4

What you really need to show is that $$f(a\circ_ib)=(-1)^{(n-1)(m-1)}f(a)\circ_if(b).$$ Here, $n$ is the arity of $a$, $m$ is the arity of $b$, and $\circ_i$ is the infinitesimal composition in $\mathcal{O}$ (once you twist the definition of the infinitesimal composition by your sign, you get the usual equation for operad morphisms). You achieve this with ...

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