Second of the sequel

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Week 2: Way To Orthogonalization
Masum Billal
Dhaka University
May 11, 2014
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Span & Fundamental Parallelepiped
The span of a lattice L is the vector space spanned by its
vectors. In writing,
span(L(B)) = span(B) = {BY : Y ∈ Rn
}
The fundamental parallelepiped of a lattice L is defined as
P(B) = {B · Y : 0 ≤ yi < 1}
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Is A Set Of Vectors A Basis?
Let B = {b1, ..., bn} be a set of vectors. We need to know when
this forms a basis of a lattice L.
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Is A Set Of Vectors A Basis?
Let B = {b1, ..., bn} be a set of vectors. We need to know when
this forms a basis of a lattice L.
Theorem
Let B = {b1, ..., bn} be a set of linearly independent vectors and
L be a lattice. Then B is a basis of L iff
P(B) ∩ L = {(0)}
where (0) indicates the origin i.e. the vector with all coordinates
set to 0.
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Proof
Proof.
Let’s assume first that, B is a basis of L. By definition, L is the
set of their integer combinations. Since, P is the integer
combination in range [0, 1), we have P(B) ∩ L = {(0)}.
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Proof
Proof.
Let’s assume first that, B is a basis of L. By definition, L is the
set of their integer combinations. Since, P is the integer
combination in range [0, 1), we have P(B) ∩ L = {(0)}.
Now, assume that, P(B) ∩ L = {(0)}. We can write a vector
x ∈ L as
∑
bixi for xi ∈ R. And by definition, L is closed under
vector addition. So, x′ =
∑
(xi − ⌊xi⌋)bi also belongs to L.
From our assumption, x′ = (0). Thus, xi − ⌊xi⌋ = 0 i.e. xi is an
integer. Hence, B a basis of L.
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Determinant
Definition
Let Λ = L(B) be a lattice of rank n. Then the determinant of Λ
is the n-dimensional volume of its parallelepiped, and is
denoted as det(Λ). If B is a full rank basis, then
det(Λ) = | det B|. Otherwise, det(Λ) =
√
det(BT B).
The determinant of a lattice is inverse proportional to its
density: the smaller the determinant, the denser the lattice is.
In more precise terms, if one takes a large ball K (in the span of
Λ) then the number of lattice points inside K approaches
vol(K)/det(Λ) as the size of K goes to infinity.
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Gram-Schmidt Orthogonalization
Gram-Schmidt orthogonalization takes n linearly independent
vectors and creates n orthogonal vectors.
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Gram-Schmidt Orthogonalization
Gram-Schmidt orthogonalization takes n linearly independent
vectors and creates n orthogonal vectors.
Definition
Given n linearly independent vectors b1, b2, ..., bn. Their
Gram-Schmidt Orthogonalization is the sequence of
vectors ¯b1, ¯b2, ..., ¯bn defined as
¯bi = bi −
i−1∑
j=0
µij
¯bj
where µij =
⟨bi, ¯bj⟩
⟨ ¯bj, ¯bj⟩
.
In words, ¯bi is the component of bi orthogonal to ¯b1, ..., ¯bi−1.
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Volume In Terms Of Orthogonal Vector
If we consider the orthonormal basis of B given by
¯b1
| ¯b1|
,
¯b2
| ¯b2|
, . . . ,
¯bn
| ¯bn|
we find that, the volume of the parallelepiped generated by Λ is
n∏
i=1
|¯bi|. Here, |a| denotes the absolute value of the vector a. Or
equivalently,
det(Λ) = det B = det(P(B)) =
n∏
i=1
|¯bi|
Masum Billal Week 2: Way To Orthogonalization

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Parallelepiped Condition For A Basis? Determinant Orthogonalization Volume
Orthogonal Vectors
Theorem
Let B is a basis of rank n, and ¯B be its orthogonalized basis.
Then,
λ1(B) ≥ min
i=1,...,n
|¯bi| > 0
This theorem is quite useful for finding a short vector in a
lattice.
Corollary:
Let Λ be a lattice. Then there is a constant ε so that, for any
two non-equal points x, y ∈ Λ, |x − y| > ε.
Masum Billal Week 2: Way To Orthogonalization