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Let $A,B$ be Hermitian matrices so that $0 \le A,B < I$ and also $(1-\varepsilon)(I-B)\le I - A \le (1+\varepsilon)(I-B)$.

For every $t \in \mathbb{N}$, consider the matrix $A_{t} = \sum_{i=0}^{t}A^{i}$ and likewise $B_{t} = \sum_{i=0}^{t}B^{i}$. I am interested in how "close" $A_{t}$ and $B_{t}$ are, given that we know $A_{\infty}$ and $B_{\infty}$ are close.

Formally, define $\gamma_{i}$ as the minimal $\gamma \ge 1$ satisfying $A_{i} \le \gamma B_{i}$ (provided it exists). We know that $\gamma_{0} = 1$ and that $\gamma_{t}$ approaches $1/(1-\varepsilon)$.

What about intermediate values? How does the sequence $\gamma_{0},\gamma_{1},\ldots$ behave? For scalars, this sequence is monotone, but I did find (numerically) examples of $4 \times 4$ matrices for which this sequence was not monotone, however the deviation from $\gamma_{\infty}$ was small.

Can we, under some reasonable assumptions, bound $|\gamma_{i}-\gamma_{\infty}|$ (with something smaller than a multiple of $i$)? Or, is there a natural example of $A$ and $B$ that behave badly in that sense?

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  • $\begingroup$ what do you mean by $0<A<I$? Do you mean that all entries are between $0$ and $1$? Do your inequalities hold elementwise $\endgroup$
    – user100927
    Commented Sep 28, 2017 at 13:39
  • $\begingroup$ For a submultiplikative matrix norm we have $$ \|A_t-B_t\|=\left\|\frac{I-A^{t+1}}{I-A}-\frac{I-B^{t+1}}{I-B}\right\|=\|A_\infty(I-A^{t+1})-B_\infty(I-B^{t+1})\| \\ \leq \|A_\infty-B_\infty\|+\|A_\infty\|\|A\|^{t+1}+\|B_\infty\|\|B\|^{t+1} $$ $\endgroup$
    – user100927
    Commented Sep 28, 2017 at 14:25
  • $\begingroup$ There is the simple bound $\|A_\infty-A_t\| \leq \|A \|^{t+1}\|(I-A)^{-1}\|=O(\rho ^ t)$ with $\rho <1$. Therefore $A_i \leq A_\infty \leq (1+O(\rho ^i)) A_i$. Similar estimate for $B$ which give an exponential decay for $|\gamma_i -\gamma_\infty|$. Is it what you are looking for? $\endgroup$
    – RaphaelB4
    Commented Sep 28, 2017 at 18:32
  • $\begingroup$ @user100927, I mean that both $A$ and $I-A$ are positive definite. $\endgroup$
    – Daniel86
    Commented Sep 29, 2017 at 17:36
  • $\begingroup$ @RaphaelB4, I am looking for a multiplicative approximation (and $(1+\epsilon)^t$ is too high), so I don't see how you get the bound on $|\gamma_i - \gamma_\infty|$. $\endgroup$
    – Daniel86
    Commented Sep 29, 2017 at 17:47

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