Gruber P. Convex and Discrete Geometry

360 Geometry of Numbers
This is an immediate consequence of the case i = d of the following proposition:
(3) Let c1,...,cd ∈ L be linearly independent. Then, for i = 1,...,d, we
have the following: There are i linearly independent vectors b1,...,bi ∈ L
such that
and
c1 = u11b1
c2 = u21b1 + u22b2
.........................
ci = ui1b1 +······+uiibi
where u jk ∈ Z, u jj �= 0
lin{b1,...,bi}∩L ={u1b1 +···+uibi : u j ∈ Z}.
Proof by induction: For i = 1, choose, among all points of L on the line lin{c1}, one
with minimum positive distance from o, sayb1. Since L is discrete, this is possible.
Then the assertion in (3), for i = 1, holds with this b1.
Next, let i < d and assume that the assertion in (3) holds for i. Consider the
unbounded parallelotope
P = � α1b1 +···+αibi + αci+1 : 0 ≤ α j < 1, α ∈ R � .
All points of P, which are sufficiently far from o, have arbitrarily large distance
from lin{b1,...,bi}. Since (P ∩ L) \ lin{b1,...,bi} ⊇{ci+1} �= {o} and since L
is discrete, we thus may choose a point bi+1 ∈ P ∩ L which is not contained in
lin{b1,...,bi} and has minimum distance from lin{b1,...,bi}. Then
(4) b1,...,bi, bi+1 ∈ L are linearly independent.
Next, note that, for any point of L in lin{b1,...,bi, bi+1}, we obtain a point of P
by adding a suitable integer linear combination of b1,...,bi . These two points then
have the same distance from lin{b1,...,bi}. Thus bi+1 has minimum distance from
lin{b1,...,bi}, not only among all points of (P ∩L)\ lin{b1,...,bi}, but also among
all points of � lin{b1,...,bi, bi+1}∩L � \ lin{b1,...,bi }. This yields, in particular,
(5) � α1b1 +···+αibi + αi+1bi+1 : 0 ≤ α j < 1 � ∩ L ={o}.
We now show that
(6) lin{b1,...,bi+1}∩L = � u1b1 +···+uibi + ui+1bi+1 : u j ∈ Z � .
Let x ∈ lin{b1,...,bi+1}∩L. Hence x = u1b1 +···+ui+1bi+1 with suitable
ui ∈ R. Then
x −⌊u1⌋b1 −···−⌊ui+1⌋bi+1 ∈ � α1b1 +···+αi+1bi+1 : 0 ≤ α j < 1 � ∩ L ={o}
by (5) and thus x =⌊u1⌋b1 +···+⌊ui+1⌋bi+1. Comparing the two representations
of x and taking into account the fact that b1,...,bi+1 are linearly independent by
(4), it follows that u j =⌊u j⌋ ∈Z, for j = 1,...,i + 1. Thus, the left-hand side in
(6) is contained in the right-hand side. Since the converse is obvious, the proof of (6)
is complete.