1
$\begingroup$

Let $A\in\mathbb{R}^{n\times n}$ be a non-negative matrix and let $D\in\mathbb{R}^{n\times n}$ be a signature matrix, i.e. a diagonal matrix having either $+ 1$ or $-1$ elements on its diagonal. Notice that it holds $D=D^{-1}$. Consider the matrix $$ B:=AD. $$ Observe that $B$ is equal to $A$ except for the fact that some of its columns may have opposite sign.

The problem. I'm interested in some estimates of the 2-norm of the matrix exponential of $B$. In particular, I'm wondering whether it's possible to derive upper bounds on $\|\exp (B)\|_2$ of the form $$ \|\exp (B)\|_2\leq f(A,D)\|\exp(A)\|_2, $$ for some function $f$ of $A$ and/or $D$. Any suggestions?

Note 1. Using the subadditivity and submultiplicativity of matrix 2-norm, we can obtain the bound $$ \|\exp (B)\|_2\leq \exp(\|A\|_2). $$ However, this bound is quite loose and not very informative.

Note 2. It can be assumed that $A$ is symmetric, if that helps in deriving a bound.

  • 0
    I don't think you'll be able to do any better than note 1. Since $A$ is symmetric, you can express that 2-norm in terms of the eigenvalues of $A$.2017-02-13
  • 0
    Is your 2-norm the operator norm (the "spectral norm") or the Frobenius norm?2017-02-13
  • 0
    @Omnomnomnom: The operator norm. However, if it's possible to derive a bound for the Frobenius norm, I'm interested in it as well.2017-02-13
  • 0
    I think if you want any kind of improvement to your inequality, you'll have to say something about the eigenvectors of $A$.2017-02-13
  • 0
    Alternatively, if we partition the matrix $B$ in the form $$ \begin{bmatrix}A_{11} &-A_{12} \\ A_{21} &-A_{22} \end{bmatrix},$$ is it possible to derive a bound on $\|\exp(B)\|_2$ using the norm of the exp of the diagonal blocks?2017-02-13
  • 0
    that does give us a *little* more information, but I have a feeling that the answer is no: there's always a worst-case that occurs when one of the eigenvectors of $A$ is a standard basis vector.2017-02-13

0 Answers 0