本文整理汇总了Python中sage.matrix.constructor.Matrix.adjoint方法的典型用法代码示例。如果您正苦于以下问题：Python Matrix.adjoint方法的具体用法？Python Matrix.adjoint怎么用？Python Matrix.adjoint使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类sage.matrix.constructor.Matrix的用法示例。
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示例1: _jordan_2_adic
# 需要导入模块: from sage.matrix.constructor import Matrix [as 别名]
# 或者: from sage.matrix.constructor.Matrix import adjoint [as 别名]
def _jordan_2_adic(G):
r"""
Transform a symmetric matrix over the `2`-adic integers into jordan form.
Note that if the precision is too low, this method fails.
The method is only tested for input over `\ZZ_2` of ``'type=fixed-mod'``.
INPUT:
- ``G`` -- symmetric `n` by `n` matrix in `\ZZ_p`
OUTPUT:
- ``D`` -- the jordan matrix
- ``B`` -- transformation matrix, i.e, ``D = B * G * B^T``
The matrix ``D`` is a block diagonal matrix consisting
of `1` by `1` and `2` by `2` blocks.
The `2` by `2` blocks are matrices of the form
`[[2a, b], [b, 2c]] * 2^k`
with `b` of valuation `0`.
EXAMPLES::
sage: from sage.quadratic_forms.genera.normal_form import _jordan_2_adic
sage: R = Zp(2, prec=3, print_mode='terse', show_prec=False)
sage: A4 = Matrix(R,4,[2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2])
sage: A4
[2 7 0 0]
[7 2 7 0]
[0 7 2 7]
[0 0 7 2]
sage: D, B = _jordan_2_adic(A4)
sage: D
[ 2 7 0 0]
[ 7 2 0 0]
[ 0 0 12 7]
[ 0 0 7 2]
sage: D == B*A4*B.T
True
sage: B.determinant().valuation() == 0
True
"""
R = G.base_ring()
D = copy(G)
n = G.ncols()
# transformation matrix
B = Matrix.identity(R, n)
# indices of the diagonal entrys which are already used
cnt = 0
minval = None
while cnt < n:
pivot = _find_min_p(D, cnt)
piv1 = pivot[1]
piv2 = pivot[2]
minval_last = minval
minval = pivot[0]
# the smallest valuation is on the diagonal
if piv1 == piv2:
# move pivot to position [cnt,cnt]
if piv1 != cnt:
B.swap_rows(cnt, piv1)
D.swap_rows(cnt, piv1)
D.swap_columns(cnt, piv1)
# we are already orthogonal to the part with i < cnt
# now make the rest orthogonal too
for i in range(cnt+1, n):
if D[i, cnt] != 0:
c = D[i, cnt]//D[cnt, cnt]
B[i, :] += -c * B[cnt, :]
D[i, :] += -c * D[cnt, :]
D[:, i] += -c * D[:, cnt]
cnt = cnt + 1
# the smallest valuation is off the diagonal
else:
# move this 2 x 2 block to the top left (starting from cnt)
if piv1 != cnt:
B.swap_rows(cnt, piv1)
D.swap_rows(cnt, piv1)
D.swap_columns(cnt, piv1)
if piv2 != cnt+1:
B.swap_rows(cnt+1, piv2)
D.swap_rows(cnt+1, piv2)
D.swap_columns(cnt+1, piv2)
# we split off a 2 x 2 block
# if it is the last 2 x 2 block, there is nothing to do.
if cnt != n-2:
content = R(2 ** minval)
eqn_mat = D[cnt:cnt+2, cnt:cnt+2].list()
eqn_mat = Matrix(R, 2, 2, [e // content for e in eqn_mat])
# calculate the inverse without using division
inv = eqn_mat.adjoint() * eqn_mat.det().inverse_of_unit()
B1 = B[cnt:cnt+2, :]
B2 = D[cnt+2:, cnt:cnt+2] * inv
for i in range(B2.nrows()):
for j in range(B2.ncols()):
B2[i, j]=B2[i, j] // content
B[cnt+2:, :] -= B2 * B1
#.........这里部分代码省略.........