Minoration of linear forms

Question

In the book Number Theory IV from Parshin, one can find this statement (precisely at p. 215) with the comment "it is easy to see that...":

Let $a_{ij}$ ($1\le i,j\le m$) be complex numbers with $|a_{i,j}|\le H_i$ ($1\le j\le m$). One assumes that the linear forms $L_i(\underline X)=\sum_{i=1}^ma_ix_i$ are linearly independent. Then, for any $\overline w=(w_1,\cdots,w_m)\in\mathbb C^m$, one has $$\sum_{i=1}^m\frac{|L_i(\overline w)|}{H_i}\ge c_1\frac{|\Delta|}{H_1\cdots H_m}$$ where $\Delta=\det(a_{i,j})$ and $c_1=\frac1{m!}\max_{1\le j\le m}|w_j|$.

Unfortunately, for me it is not easy. Can one have hints to prove that? Thanks in advance

Answer 1

We claim that $$ \sum_{i=1}^m \frac{|L_i(w)|}{H_i} \geqslant \frac{|w|_\infty}{(m-1)!}\frac{|\Delta|}{H_1 \cdots H_m}, $$ where $|w|_\infty = \max_{1 \leqslant j \leqslant m} |w_j|$.

Indeed, first assume that $|A|_\infty = \max_{1\leqslant i,j \leqslant m} |a_{ij}|\leqslant 1$, and $H_i = 1$ for all $i$. Because $|Bw|_\infty \leqslant |B|_\infty |w|_1$ for any matrix $B$ and $w \in \mathbb{C}^m$ (where $|w| = \sum_{j=1}^m |w_j|$), we have $$ |w|_\infty \leqslant |A^{-1}|_\infty|Aw|_1. $$ Now $A^{-1} = \frac{1}{|\Delta|} \mathrm{Com}(A)^\top$ where $\mathrm{Com}(A)$ is the comatrix of $A$. Note that $|\mathrm{Com}(A)|_\infty \leqslant (m-1)!$ by the Leibniz formula for determinants, since $|A|_\infty \leqslant 1$. Thus $$ |A^{-1}|_\infty \leqslant \frac{(m-1)!}{|\Delta|}. $$ Therefore, by definition of $L_i$ $(i = 1, \dots, m)$ we get $$ \sum_{i=1}^m |L_i(w)| = |Aw|_1 \geqslant \frac{|w|_\infty |\Delta|}{(m-1)!}, $$ which is the sought result in the case $|A|_\infty \leqslant 1$ and $H_1 = \cdots = H_m = 1$.

Finally take a general $A$ and assume that $\max_{1 \leqslant j \leqslant m}|a_{ij}| \leqslant H_i$ for $i = 1, \dots, m$. Writing $A = \begin{pmatrix} L_1 \\ \vdots \\ L_m \end{pmatrix}$ and applying the above estimate to the matrix $ \tilde A = \begin{pmatrix} L_1 / H_1 \\ \vdots \\ L_m / H_m \end{pmatrix}, $ we get the claim.