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Results tagged with matrix-analysis
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user 8430
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.
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 …
- 52.3k
1
vote
Accepted
Norm/trace of product inequality involving skew symmetric matrices
Something seems to be missing here, because the inequality is trivially seen to be false. Consider the following randomly picked matrices for instance:
\begin{equation*}
B = \begin{bmatrix}0 & -4 & 4 …
- 28.6k
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 …
- 28.6k
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} …
- 52.3k
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 …
- 28.6k
4
votes
A simple but curious determinantal inequality
EDIT (added some clarifications). The argument below provides a self-contained proof.
Introduce the shorthand $C^{-2}=A^{k+1}$. We need to show that
\begin{equation*}
\det(I+ CBABC) \ge \det(I + CA …
- 28.6k
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}) …
- 28.6k
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 \\ …
- 28.6k
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 …
- 28.6k
6
votes
Accepted
On a trace condition for positive definite $2\times 2$ block matrices
For any unitarily invariant norm it can be shown that
\begin{equation*}
\|X\| = \left\Vert
\begin{bmatrix}
A & C\\
C^* & B
\end{bmatrix}
\right\Vert \le \|A\| + \|B\|.
\end{eq …
- 28.6k
4
votes
Add a multiple of $I$ to a matrix to minimize its operator norm
Not a closed form answer, but this can be solved as a semidefinite program. In particular, we can rewrite the task as
\begin{equation*}
\min_{t\ge 0, s}\ t\quad \text{s.t.}\quad (A-sI)^*(A-sI) \preceq …
- 28.6k
5
votes
Accepted
On proof of the conditionally negative definiteness of a kernel
Here a direct approach. Recall the power-series
\begin{equation*}
\arccos(z) = \frac\pi2 - \sum_{k\ge0}\binom{2k}{k}\frac{z^{2k+1}}{4^k(2k+1)}.
\end{equation*}
From this series it is clear that $\ar …
- 28.6k
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 …
CommunityBot
- 1
5
votes
A curious determinantal inequality
Here is a complementary approach without using majorization. The answer is partial because it has an open "TODO". I am writing it down here already in case someone wishes to complete the argument.
…
- 28.6k
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 $|\lambda(A)|^{\downarrow} \prec_w s(A)$ …
- 28.6k