All Questions
14 questions
53
votes
7
answers
51k
views
Determinant of sum of positive definite matrices
Say $A$ and $B$ are symmetric, positive definite matrices. I've proved that
$$\det(A+B) \ge \det(A) + \det(B)$$
in the case that $A$ and $B$ are two dimensional. Is this true in general for $n$-...
- 541
3
votes
2
answers
840
views
An inequality for the spectral radius of block matrices
Let $d,m$ be positive integers. Suppose that $A_{i,j}$ is a $d\times d$-matrix with real entries whenever $i,j\in\{1,\dots m\}$.
Let $A$ be the $dm\times dm$ matrix that can be written as a block ...
10
votes
2
answers
7k
views
Bounding the trace of a matrix product by the operator norms; generalized Hölder inequality?
$\DeclareMathOperator\Tr{Tr}$Let $A_i$ with $i=1,\dotsc,N$ and $p$ be real $M\times M$ matrices.
Further, let $p$ be positive definite, i.e., $p\succ 0$, with $\Tr(p)=1$. Let $0< a_i<1$ and $\...
- 103
4
votes
1
answer
977
views
Ratio sum comparison on operators
It is known by the Lidskii inequality, that $\sum_{i=1}^n \left|s_i(S)-s_i(T)\right|\le\sum_{i=1}^n s_i(S+T)$,
where $s_i(S)$ is the $i$-th singular value of $S$.
How would one prove that
$$\sum_{i=1}^...
- 41
35
votes
3
answers
4k
views
A curious determinantal inequality
In my study, I come across the following curious inequality, which I do not know a proof yet (so I am asking it here).
Let $A, B$ be $n\times n$ (Hermitian) positive definite matrices. It is very ...
- 1,748
19
votes
1
answer
856
views
A possible extension of a determinant inequality
It is well known that if $A, B$ are positive semidefinite matrices, then $$\det (A+B)\ge \det A+\det B.$$
I am considering a possible extension of this result. Let $\mathbb{M}_m(\mathbb{M}_n)$ ...
- 1,748
17
votes
1
answer
2k
views
Hlawka inequality for determinants of positive definite matrices
It is mentioned here that if $A, B, C\in M_{n}(\mathbb C)$ are positive semidefinite, then $$\det (A+B+C)+\det C\ge \det (A+C)+\det (B+C)$$ (quoted from this article) and the special case ($C=\bf 0$) $...
- 13.4k
13
votes
1
answer
1k
views
An inequality for the spectral radius of matrices used by J. Bochi
I am interested in the history of an inequality for the spectral radius of a $d\times d$ real or complex matrix, which occurs in Jairo Bochi's 2002 article Inequalities for numerical invariants of ...
- 6,206
12
votes
0
answers
218
views
Which ordering of factors is needed to obtain this kind of determinantal inequalities?
Let $A$ and $B$ be $n\times n$ Hermitian positive definite matrices. The curious determinantal inequality given here, which can be stated as $$\det (A^{4}+ ABBA+BAAB+B^{4})\ge\det(A^{4}+ AABB+BBAA+B^{...
- 13.4k
10
votes
1
answer
630
views
Minimum distance of a symmetric matrix to diagonal matrices
Let $A=(a_{ij})$ be an arbitrary $n \times n$ real symmetric matrix and $n\geq 2$. Let $\| \cdot \|$ denote the operator $2$-norm or equivalently the maximum absolute value of eigenvalues for ...
- 4,484
10
votes
1
answer
3k
views
Reverse Minkowski (and related) Determinant Inequalities
For positive semidefinite matrices $A,B,C \in \mathbb{R}^{n\times n}$, the following inequalities are well known:
$$(\det(A+B))^{1/n} \geq (\det A)^{1/n} + (\det B)^{1/n} $$
and
$$\det(A+B+C) + \...
- 716
6
votes
2
answers
2k
views
Tight bound for sum of entries of the inverse of a nonnegative matrix
While playing around with certain non-negative matrices, I got stuck at the following question.
Let $A$ be a strictly positive-definite $n \times n$ matrix ($n \ge 3$), with ones on the diagonal, and ...
- 28.6k
5
votes
1
answer
474
views
An inequality for certain positive-semidefinite matrices
Suppose that $G=(G_{ij})$ is a positive-semidefinite symmetric matrix with the diagonal entries all equal $1$ and all off-diagonal entries $\le0$. Does it then necessarily follow that
$$\sum_{i,j}(G^5)...
- 128k
3
votes
1
answer
80
views
prove spectral equivalence bounds for fractional power of matrices
Let $A, D \in \mathbb{R}^{n \times n}$ be two symmetric,positive definite and tri-diagonal matrices for that we know that they are spectrally equivalent, thus ist holds
$$ c^- x^\top D x \le x^\top A ...
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