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The following paper studies relations between $\lambda(BAB^T)$ and $\lambda(A)$: Li, Mathias (1999). The Lidskii-Mirsky-Wielandt theorem – additive and multiplicative versions. Numerische Mathematik. …

Your matrix has entries given by $a_{ij}=\min(i,j)$, where $0\le i,j\le n-1$. Have a look at Section 3 of this paper of mine for a derivation of explicit bounds.

Let the (unique wlog) largest element of $A^{-1}$ be at position $(i,j)$. Since $A^{-1}$ is positive definite, it follows that \begin{equation*} 2e_i^TA^{-1}e_j \le e_i^TA^{-1}e_i + e_j^TA^{-1}e_j. …

The claim is false. In particular, we have $A=kee^T-kI$, so that $\lambda(A)=((n-1)k,-k,\ldots,-k)$, so that $\|A\|_* = 2(n-1)k$. Now generate a random matrix $B$ such that $B_{ii}=0$, $B_{ij}=B_{ji …

Let $\lambda^\downarrow(X)$ denote the vector of eigenvalues of $X$ in decreasing order; define $\lambda^\uparrow(X)$ likewise. … However, when dealing with matrix products, it is more natural to consider singular values rather than eigenvalues. …

This problem is essentially the same as this one. In particular, let $J$ be the anti-diagonal identity matrix, and $P^{-1}$ be the matrix mentioned in the link above. Then, the matrix in the current p …

We show below a slightly more general claim (to simplify notation, I'll write only in terms of matrices). $\newcommand{\reals}{\mathbb{R}}$ Def. We say a kernel $\psi: X \times X \to \reals$ is negat …

Have a look at the references cited in my older answer here, for the results of the kind you are looking for (complexity in there though is shown with a worst case O($\log n$) dependence on the dimens …

This maxQP problem is hard, it includes MaxCUT as a special case---see this paper by M. Charikar on MAXQP. Having $Q$ be positive definite does not really help (take the MAXQP problem in Charikar's pa …

Suppose $x_i > 0$ and $y_j -x_j$, then $c_{ij} = 1/(x_i+x_j)$. These matrices are infinitely divisible, i.e., $[c_{ij}^r]$ is also positive definite for all $r > 0$. Spectral properties of Cauchy-li …

While not a full characterization, the following result on $\sigma_n$ shows that $\sigma_k > \epsilon_n$, since $\sigma_n \ge \epsilon_n$. Following my answer on this MO question here, we see that \ …

As per fedja's comment, the version of the inequality written in the question does not hold. A version that does hold is given below. $\newcommand{\trace}{\mathrm{tr}}$ Define $\trace_n(X) := \fra …