# Copying a Matrix

I'm trying to take the Singular Value Decomposition of the adjacency matrix of a graph A, but all my "reasonable" attempts at doing so have failed, see the code below.

I believe the problem has something to do with the matrix being immutable, so I've been trying to copy A into a different matrix M and then perform SVD on M. he only thing that I've found to work is to essentially build A from scratch, but there must be a better method.

sage: A=graphs.CycleGraph(3).adjacency_matrix()
#Method 1: Try and get SVD directly from A; gives error at last step.
#sage: A.change_ring(RDF)
#sage: A.SVD()

#Method 2: Try and copy A to a different matrix and then do SVD; gives error at last step.
# sage: M=copy(A)
# sage: M.change_ring(RDF)
# sage: M.SVD()

#Method 3: Build A up again from scratch and then do SVD; works, but is tedious.
sage: M=matrix(RDF,3,3)
sage: for i in [0..2]:
for j in [0..2]:
if(A[i,j]==1):
M[i,j]=1
sage: M.SVD()

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Note that A.change_ring(RDF) does not change A "in place", (so as it is the case with the sort method, applied on a list,) instead, it delivers an other object living over RDF. To see this:

sage: G = graphs.CycleGraph(3)
sage: A.base_ring()
Integer Ring
sage: A.change_ring( RDF )
[0.0 1.0 1.0]
[1.0 0.0 1.0]
[1.0 1.0 0.0]
sage: A
[0 1 1]
[1 0 1]
[1 1 0]
sage: A.base_ring()
Integer Ring


Now the solution is simple:

sage: G = graphs.CycleGraph(3)
sage: a = A.change_ring( RDF )
sage: a.SVD()
(
[   -0.5773502691896255 2.7755575615628914e-16     0.8164965809277258]
[   -0.5773502691896257    -0.7071067811865475    -0.4082482904638628]
[   -0.5773502691896255     0.7071067811865476   -0.40824829046386313],

[               2.0                0.0                0.0]
[               0.0 0.9999999999999999                0.0]
[               0.0                0.0 0.9999999999999999],

[ -0.577350269189626                -0.0 -0.8164965809277259]
[-0.5773502691896256  0.7071067811865475 0.40824829046386313]
[-0.5773502691896256 -0.7071067811865476 0.40824829046386313]
)
sage:

more

Thank you so much for that excellent answer, it was exactly what I was looking for!

( 2018-01-23 19:19:57 -0600 )edit