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The study of the properties of real and complex matrices that are more close to analysis and operator theory. For instance: the properties of positive definite matrices, matrix inequalities, perturbation analysis, matrix functions, inequalities between eigenvectors and singular values, majorization.
23
votes
Accepted
Does the matrix exponential preserve the positive-semi-definite ordering?
To supplement Robert's counterexample, let me mention below some interesting facts about the matrix exponential, along with what may be regarded as the "correct" way of obtaining matrix exponential li …
17
votes
Matrix trace & norm
Expanding my comment into an answer, which offers a more general result.
Theorem (von Neumann). Let $A$ and $B$ be arbitrary $n\times n$ complex matrices. Then, $$|\text{trace}(AB)| \le \sum_{i=1} …
15
votes
Accepted
Trace of non-commutable matrices
Your conjecture is a special case of the following result which essentially follows from the Lieb-Thirring inequality.
Let $A$ and $B$ be Hermitian matrices. Then, for every positive integer $p$ we h …
14
votes
Accepted
On the positive definiteness of a linear combination of matrices
The following recent paper: "An exact duality theory for semidefinite programming based on sums of squares" by I. Klep, and M. Schweighofer (both are on MO I think) addresses exactly your question: Wh …
12
votes
Accepted
Optimizing the condition number
Update: This recent paper on this topic may also be of interest; it's quite short and claims to have a fully constructive approach.
I think the closest to answering your question is the following pap …
11
votes
Accepted
Hlawka inequality for determinants of positive definite matrices
My previous answer mis-attributed a claim to the paper of Paksoy, Turmen, and Zhang (also cited in the OP). Their claim is indeed strictly weaker than the inequality conjectured by Wolfgang. The detai …
11
votes
Inverse of a small submatrix
One way to go about this is as follows:
For $i,j \in \mathcal{I}$ Compute $e_i^TA^{-1}e_j$ by using the approach based on Gaussian quadrature; see for instance, a precise algorithm and analysis in ou …
11
votes
Do singular values dominate eigenvalues?
Let $\lambda(A)$ denote the vector of eigenvalues and $s(A)$ the vector of singular values (arranged in decreasing order). The claim of the question is whether $|\lambda(A)|^{\downarrow} \prec_w s(A)$ …
10
votes
Accepted
Solving $\text{trace}\left[\left(I + pY\right)^{-1} \left(I - p^{2}Y\right)\right] = 0$ for ...
Write $Y=SDS^{-1}$, where $D$ is diagonal (since $X_1$ and $X_2$ are psd, $Y$ is diagonalizable). Then, observe that
\begin{equation*}
f(p) = \text{tr}(S^{-1}(I+pSDS^{-1})^{-1}SS^{-1}(I-p^2SDS^{-1}) …
10
votes
Accepted
Concavity of the trace of a matrix power
Unfortunately, the conjectured function is not concave. Here is a simple simpler counterexample.
\begin{equation*}
B = \begin{bmatrix} 1 & 2 \\ 3 & 4\end{bmatrix},\quad
A = \begin{bmatrix} 2 & 0 \\ …
9
votes
Accepted
Matrix-convexity of inverse of the cofactor matrix
Not just $3\times 3$, but in general, the map $A \mapsto \det(A^{-1})A$ is operator convex on positive definite matrices.
Proof sketch.
$\newcommand{\pfrac}[2]{\left(\tfrac{#1}{#2}\right)}$
If suffic …
7
votes
Accepted
Reverse Minkowski (and related) Determinant Inequalities
Inequality ($\star\star$) essentially follows from the original Minkowski plus an implication of Lidkskii's inequality (Fiedler's inequality, noted below).
$\newcommand{\da}{\downarrow} \newcommand{\u …
7
votes
Accepted
On a determinant inequality of positive definite matrices
A quick counterexample to your conjecture is
\begin{equation*}
A = \begin{pmatrix}
13 & 3 & -13 & -5\\
3 & 4 & -3 & 4\\
-13 & -3 & 13 & 5\\
-5 & 4 & 5 & 10\\ …
6
votes
The singular values of the Hilbert matrix
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
\ …
6
votes
Accepted
Optimization version of the Sylvester equation
First recall two basic ideas.
Lemma. Let $A$, $B$, $C$ be arbitrary; then, $\text{vec}(ABC) = (C^T \otimes A)\text{vec}(B)$, where $\otimes$ denotes the Kronecker product and $\text{vec}(\cdot)$ d …