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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 …
Fedor Petrov's user avatar
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 …
Fedor Petrov's user avatar
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 …
Fedor Petrov's user avatar
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\| …
Fedor Petrov's user avatar
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)} …
Fedor Petrov's user avatar
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 …
Fedor Petrov's user avatar
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 $$\|(I-A)^{-1}\|=\|I+A+A^2+\dots\|\leqslant 1+\|A\|+\|A\|^2+\dots=\frac1{1-\|A\|}.$$
Fedor Petrov's user avatar
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 …
Fedor Petrov's user avatar
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=- …
Fedor Petrov's user avatar
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 …
Fedor Petrov's user avatar