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The answer to your question is positive, in every dimension: Let $A=I_n+T$ be symmetric positive semi-definite, with ${\rm diag}T=(0,\ldots,0)$ and $T$ tridiagonal. Then the largest eigenvalue $\rho( …

Here is the proof. Please let me use lighter notations: $$H=\begin{pmatrix} B & C \\ C^T & D \end{pmatrix},$$ so that $$A=\begin{pmatrix} 0_p & B^{-1}C \\ D^{-1}C^T & 0_q \end{pmatrix}.$$ From Schur c …

These two problems are fundamentally different, because their data are. As far as linear systems are concerned, controlability is a property for an ODE $\dot x=Ax+Bu$, where $u$ is the control, while …

By density, it is enough to prove the property when $A$ is positive definite. Then $$A(I+BA)^{-1}=A^{1/2}(I+A^{1/2}BA^{1/2})^{-1}A^{1/2}$$ is congruent to $(I+A^{1/2}BA^{1/2})^{-1}$, which itself is …

$M$ is congruent to ${\rm diag}(A-BC^{-1}B^T,C)$. Therefore the condition $A-BC^{-1}B^T\le0$ amounts to saying that the maximal dimension of positive subspace is the size $p$ of $C$.If $A-BC^{-1}B^T<0 …

I am amazed that nobody noticed that Iosif asks for the calculation of the transfinite diameter of the interval $[0,1]$. This notion applies to arbitrary compact domains $K$ in ${\mathbb R}^n$ : $$d(K …

This is related to so-called hyperbolic polynomials, studied by L. Gaarding in the fifties. More generally, let $\lambda(\xi)$ be the least eigenvalue of $A(\xi)=\sum_\alpha\xi_\alpha A^\alpha$, where …

Here is a partial answer. But let me first modify the question by replacing the reals by the complex. Hence, $x$ runs over ${\mathbb C}^n$ and $A_1,A_2$ are Hermitian. Then $A_1,A_2$ can be viewed as …

The set of matrices $A$ is a cone, smooth away from the origin. With the Frobenius norm $\sqrt{{\rm Tr}(M^TM)}$, you can use differential calculus. The minimum is achieved at some $A$. If $A\ne0_n$, t …

The first inequality is true. Write $$f=\frac{a}{a+b}f_0+\frac{b}{a+b}f_1,$$ where $f_0$ and $f_1$ correspond to the case $b=0$ and $a=0$, respectively. You know that $f_0\ge2\sqrt a$ and $f_1\ge\frac …

Edited. Conjecture for $d=1$: Define the sequence $v_1,\ldots,v_n$ by $v_1=1$ and $$v_{k+1}=\left(\sqrt{1+\frac1k}-1\right)(v_1+\cdots v_k).$$ Then the $\min\max$ equals $a$ where $$a^2\sum_1^nv_j^2 …