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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 $$ c_1 | \xi|^2 \leqslant \sum_{i,j=1}^N a_{ij} (w) \bar\xi_i \xi_j \leqslant c_2 | \xi|^2 ?$$

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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.

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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

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.

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