Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $A$ is positive definite(uniformly to $w$) $N \times N$ matrix whose components ${a_{ij}} = a_{ij} (w) $, then how can I prove that there exist constants $c_1 = c_1 (M), c_2 = c_2 (M) >0$ such that for any $\xi, w \in \Bbb C^N$, if $|w| \leqslant M $ then $$ c_1 | \xi|^2 \leqslant \sum_{i,j=1}^N a_{ij} (w) \bar\xi_i \xi_j \leqslant c_2 | \xi|^2 ?$$

share|improve this question
add comment

2 Answers

up vote 0 down vote accepted

The closed unit ball of $\Bbb C^N$ is compact, and so is $X:=\{(\omega,\xi),|\omega|\leq M,|\xi|=1\}$. Let $f(\omega,\xi)=\sum_{k,j=1}^Na_{kj}(\omega)\bar{\xi_k}\xi_j$. Assuming the $a_{i,j}$ continuous, and applying the result of this thread to $X$ and this function, we get the wanted result.

share|improve this answer
    
Thank you. But if we use the Rayleigh quotient of $A$, (say $R$), $ \lambda_{min} \leqslant R \leqslant \lambda_{max}$ where $\lambda$ is eigenvalues. So I think $c_1 = \lambda_{min}$ and $c_2 = \lambda_{max}$. Is this wrong? –  Ann Oct 24 '12 at 10:14
    
You are right (and you can try to show it). –  Davide Giraudo Oct 24 '12 at 11:33
add comment

I only work with real numbers. For a real positive matrix $A$, it is obvious $$c_1 (w) \|\xi\|^2\le \xi^T A \xi \le c_2(w) \|\xi\|^2$$ $c_1(w)$ and $c_2(w)$ actually are the minimum and maximum eigenvalues, respectively. Note that an eigenvalue is a continuous function of the elements of $A$ and hence $w$. Since the set of $w$ is compact, $c_1(w)$ and $c_2(w)$ respectively have positive lower and upper bounds by the Weierstrass theorem.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.