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Questions about the properties of vector spaces and linear transformations, including linear systems in general.
7
votes
1
answer
265
views
A conjectured trace inequality for some products of powers of matrices
Let $B, R\in M_{n}(\mathbb{C})$ hermitian and $B$ positive semidefinite.
Let $s,t \in \mathbb{R}$ and $s,t \ge 0$ .
Does then hold $Tr[B^s (B R^2 B)^t] \ge Tr[B^s (R B^2 R)^t]$ ?
6
votes
0
answers
105
views
A conjectured trace inequality for some products of powers of matrices II
Let $B, R\in M_{n}(\mathbb{C})$ hermitian and $B$ positive semidefinite.
Let $s \in \mathbb{R}$ and $s \ge 0$ .
Does then hold $Tr[B^s (I + B R^2 B)^{-1}] \le Tr[B^s (I + R B^2 R)^{-1}]$ ?
See also …
7
votes
What are the possible eigenvalues of these matrices?
Consider the case where the $8\times 8$ matrix is positive semidefinite and assume that the 5 largest eigenvalues are all equal. Then by the argument of Federico Poloni they equal a. Then it follows $ …
4
votes
Accepted
What are the possible eigenvalues of these matrices?
Let $V$ be the real vector space of the $8\times 8$ matrices of the form given in the question.
Where is an open set in $\mathbb{R}^8$ of possible $8$-tuples of eigenvalues of matrices in $V$.
Proof …
3
votes
Vacuum vector and basis defined by anti-commuting operators
Proof of the existence of $\Omega$ :
It is easy to see that the number operators $N_i = A_i^* A_i$ commute.
Therefore there exists an orthonomal basis of common eigenvectors.
Now let $\Omega$ one of t …
3
votes
Accepted
A simple but curious determinantal inequality
Let $C = A^{-(k+1)/2} B A B A^{-(k+1)/2}$ and $D = A^{-(k-1)/2} B A^{-1} B A^{-(k-1)/2}$ .
We have to show that $det(I+C) \ge det(I+D)$ .
Now my goal is to apply equation (5.21) in Ando, Majorizatio …
5
votes
Accepted
How to prove this determinant is positive-II?
Let $q(x,y) = x^H J y$ for $x,y \in \mathbb{C}^{2n}$ where $J = diag(I_n,-I_n)$ and let $S = \{A \in M_{2n}(\mathbb{R}) : q(Ax, Ax) \ge q(x,x) $ $\forall x \in \mathbb{C}^{2n}\}$.
Obviously $S$ is a s …
6
votes
How to prove this determinant is positive?
The conjecture ist true :
It follows from $$det(I + e^{M}) = det(I + i e^{M/2}) * det(I - i e^{M/2}) = {\left\lvert{det(I + i e^{M/2})}\right\lvert}^2 \ge 0$$ .
I.e. it follows from the fact that ev …
4
votes
Accepted
A singular value-eigenvalue inequality
The conjecture is true.
Lemma 1 : For every matrix $Z$ holds $\lambda_j(Z^* Z + Z^* + Z ) \leq \lambda_j(Z^* Z + 2 (Z^* Z)^{1/2})$ .
For a proof see the proof of $\lambda_j(Z^* + Z ) \leq \lambda_j …
7
votes
Accepted
Well known matrix inequality?
The desired inequality follows from an majorisation argument :
According to Bhatia Matrix Analysis Corollary III.4.6 we have :
$log \lambda(AB) \succ log \lambda^\downarrow(A) + log \lambda^\uparro …