isLLL  is a basis an LLL basis?
Synopsis

 Usage:
 isLLL m

Inputs:

m, a matrix, over ZZ, whose columns are linearly independent

Optional inputs:

Threshold => ... (missing documentation),

Outputs:

a Boolean value, Whether the columns of the matrix form an LLL basis with respect to the Threshold (which has default 3/4)
Description
This function is provided by the package
LLLBases.
If the matrix is not in LLL reduced form, then the offending conditions are displayed. For example,
i1 : m = matrix {{1, 0}, {1, 1}, {1, 2}, {1, 3}}
o1 =  1 0 
 1 1 
 1 2 
 1 3 
4 2
o1 : Matrix ZZ < ZZ

i2 : isLLL m
LLL size failure 1,0: 1.5
LLL Lovasz failure 1: .833333
o2 = false

i3 : n = LLL m
o3 =  1 1 
 1 0 
 1 1 
 1 2 
4 2
o3 : Matrix ZZ < ZZ

i4 : isLLL n
o4 = true

If the optional argument Threshold is given, the conditions are checked using that value.
i5 : m = matrix {{1, 0}, {1, 1}, {1, 2}, {1, 3}}
o5 =  1 0 
 1 1 
 1 2 
 1 3 
4 2
o5 : Matrix ZZ < ZZ

i6 : isLLL(m, Threshold=>1)
LLL size failure 1,0: 1.5
LLL Lovasz failure 1: 1
o6 = false

Caveat
A GramSchmidt reduction is done over QQ, so this can be computationally intensive for larger matrix sizes. It is usually easier and faster to see if LLL returns the same matrix. This routine was used to debug and test the LLL routines here, and is provided as an alternate check of correctness.The matrix must be defined over the ring ZZ. It should be possible to allow real and rational matrices too, but this is not yet implemented.
See also

LLL  compute an LLL basis

gcdLLL  compute the gcd of integers, and small multipliers

kernelLLL (missing documentation)

hermite (missing documentation)