In the book [Number Theory IV](https://www.springer.com/gp/book/9783540614678) 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



Minoration of linear forms
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.


