Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
eigenvalues of matrices or operators
4
votes
Accepted
Estimates of eigenvalues
Yes, and more may be said.
Assume that $(I+A+B+AB)^{-1}(A+B)z=\lambda z$, then $(I+c(A+B)+AB)z=0$, where $c=1-\lambda^{-1}$.
Denote $Bz=u$, where $u$ may be arbitrary vector such that $(u,z)>0$. Ne …
6
votes
A log inequality for positive definite trace-one matrices
If $\alpha\geqslant \beta\geqslant 0$ are eigenvalues of $S$, $\lambda_1\geqslant \dots \geqslant \lambda_n>0$ are eigenvalues of $T$, we see that there exists non-zero vector $u\in L$ such that $\lambda …
9
votes
Accepted
A log inequality for positive definite trace-one matrices
The proof of the general case, in a strong form suggested in the end of OP.
Denote $X^{1/4} v_i=u_i$, $X^{1/2}=S$, then we have ${\rm tr}\,S^2=1$ and need to prove that
$$
{\rm tr}\,S^2\geqslant
\s …
1
vote
Minimum and maximum eigenvalue
I think, yes. The quadratic form $Q(z) :=z^tMz$ agree with that of $\Lambda$ on the hyperplane $\{y^t\Lambda z=0\}$. Therefore you have a space of codimension 1 on which $Q(z)\geqslant \lambda_1 \|z\| …
6
votes
convergence of 2nd eigenvalue
I claim that the second eigenvalues tends to $\frac{\det(M)(M^{-1})_{33}}{M_{11}}.$ Indeed, the product of eigenvalues equals $n^{2(h_3-h_2)}\det(M(n))$, the largest eigenvalue grows as $n^{2(h_3-h_1)} …
3
votes
Bound on the ratio of top 2 eigenvalues
I claim that $\lambda_2'/\lambda_1'\leqslant 1-\frac{2\tau}{1-(n-2)\tau}$ which is bit stronger than you ask for. This is sharp as the example $D_{11}=D_{22}=1\gg \max(D_{33},\dots,D_{nn})$ shows.
Of …
3
votes
Accepted
Does $\|(I-A)^{-1}\|_{2} \leq 1/(1-\|A\|_{2})$ holds for matrices with spectral radius small...
Since $A$ is positive semidefinite, norm of $A$ is the same as spectral radius of $A$. Thus
$$\|(I-A)^{-1}\|=\|I+A+A^2+\dots\|\leqslant 1+\|A\|+\|A\|^2+\dots=\frac1{1-\|A\|}.$$
4
votes
Non-asympototic version of Gelfand's formula
Yes, this is true for any $c>1$ and large enough $n$.
Let $p(z)=\prod_{i=1}^n (z-\lambda_i)$ be characteristic polynomial of $A$. Denote by $r_k(z)=c_0(k)+c_1(k)z+\dots +c_{n-1}(k)z^{n-1}$ the remai …
2
votes
Accepted
Support of eigenvectors
This looks to be true for $\varepsilon=1/4$.
First of all, the space of symmetric ($u_i=u_{N+1-i}$ for all $i=1,\ldots, N$) vectors is invariant for $M_N$, and so is the space of antisymmetric ($u_i=- …
11
votes
Accepted
Formula expressing symmetric polynomials of eigenvalues as sum of determinants
This is not a reference, but a short proof.
We use the following (probably known, but see later) lemma on representing a symmetric tensor as a linear combination of rank-1 symmetric tensors.
Lemma. Le …