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Science Forum Index » Math - Numerical Analysis Forum » Schmidt decomposition for operators?...
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| Javier Almeida... |
Posted: Fri May 09, 2008 4:43 am |
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Hi everybody,
let O be a matrix with dimension n x n. I would like to write this
operator as a sum of kronecker products, i.e,
O= Sum_{i=1}^{i=k} lambda_i*kron(A_i,B_i)
with A_i and B_i operators of size (ma x ma) and (mb x mb), holding
that ma*mb=n. In particular I would like to obtain such a
decomposition that requires the minimum number of sums, that is, a
decomposition such that k is as low as possible. Does anybody know a
systematic way to obtain one possible set operators A_i and B_i and
the coefficients lambda_i for a given operator O?
Thanks,
Javi.
P.S: the kronecker product of two matrices is the usual tensorial
product of two matrices. |
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| Robert Israel... |
Posted: Fri May 09, 2008 12:52 pm |
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Javier Almeida <javialmeida at (no spam) gmail.com> writes:
Quote: Hi everybody,
let O be a matrix with dimension n x n. I would like to write this
operator as a sum of kronecker products, i.e,
O= Sum_{i=1}^{i=k} lambda_i*kron(A_i,B_i)
with A_i and B_i operators of size (ma x ma) and (mb x mb), holding
that ma*mb=n. In particular I would like to obtain such a
decomposition that requires the minimum number of sums, that is, a
decomposition such that k is as low as possible. Does anybody know a
systematic way to obtain one possible set operators A_i and B_i and
the coefficients lambda_i for a given operator O?
Thanks,
Javi.
P.S: the kronecker product of two matrices is the usual tensorial
product of two matrices.
I suppose ma and mb are given, so you don't allow the trivial solution
k=1 with ma = 1, mb = n, A_1 = (1) and B_1 = O.
The matrix structure is a distraction: what you really are looking at
is a vector V of N = n^2 components, which you want to write as a sum
of tensor products of vectors with Ma = ma^2 and Mb = mb^2 components:
V = sum_{i=1}^k A_i \tensor B_i.
Let V_j, j=1..Mb, be the Ma-component vectors corresponding to the
individual components of the B vector, so that
V_j = sum_{i=1}^k (B_i)_j A_i.
Thus the V_j are all in the span of A_1, ..., A_k. In particular, k
must be at least the dimension of the linear span of V_1, ..., V_{Mb}.
Moreover, we can attain this k by letting the A_i form a basis of
this linear span.
--
Robert Israel israel at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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