From: andrew@rentec.com (Andrew Mullhaupt)
Newsgroups: sci.math
Subject: Re: Tensor Diagonalization
Date: 23 Nov 92 18:51:48 GMT
Organization: Renaissance Technologies Corp., Setauket, NY.

In article <1992Nov20.221459.3566@news.media.mit.edu> perry@media.mit.edu (Chris Perry) writes:
>
>        I'm wondering if anyone has seen a tensor of rank greater than two
>factored into its principle components - i.e., given a tensor A of rank n,
>is there a decomposition similiar to
>         -1
>        S   A S  =  L
>
>  where L is a "diagonal" tensor of rank n, and S is a tensor of rank n that
>is built out of the eigentensors (!) of A? Thanks for any advice or
>references on this topic -

I can help, but I look at tensor products from the mathematical,
instead of physicist's approach.


There are some facts about tensor products of vector spaces which are
needed, and I give them here, using @ for the tensor product.

Let U and V be vector spaces, and U@V their tensor product. If {e_i}
is a basis for U and {f_j} is a basis for V then {e_i @ f_j} is a basis
for U@V. The standard basis for U@V obtains when e_i and f_j are the standard
bases for U and V, respectively.

Let A and B be linear transformations on U and V, then the tensor product
A@B is defined by the linear transformation on U@V which satisfies

	(A@B) (u@v) = (Au) @ (Bv)

for all vectors u in U and v in V. This determines that the identity 
transformation on U@V is the tensor product of the identities on U and V.
The inverse of A@B is also determined by this; an inverse can be formed
by A^-1 @ B^-1, 

	(A@B) (A^-1 @ B^-1) = (A A^-1) @ (B B^-1) = 1@1.

Now the inverse is unique, so A@B is invertible iff A and B are, and 
(A@B)^-1 = A^-1 @ B^-1.

Similarity transforms are given by the obvious tensor product. Suppose
S^-1 A S = J and T^-1 B T = K. Then

	(S@T)^-1 (A@B) (S@T) = (S^-1 A S) @ (T^-1 B T) = J @ K.

It follows that a transformation A@B is diagonalized with respect to a basis
{e_i @ f_j} if and only if A is diagonalized in {e_i} and B is diagonalized
in {f_j}, and the similarity is given by the obvious tensor product.

Note that the dual space of U@V is isomorphic (i.e. equal) to U* @ V*, and

if U and V have inner products, then so does U@V and you calculate it by:

	<u_1 @ v_1, u_2 @ v_2> = <u_1, u_2> <v_1, v_2>.

This lets you determine what orthogonal vectors are in U@V: u_1 @ v_1 is
orthogonal to u_2 @ v_2 if and only if either u_1 is orthogonal to u_2 or
v_1 is orthogonal to v_2.

The inner product determines what adjoints of linear transformations are; 
in particular if A+ is the adjoint of A and B+ is the adjoint of B then

	<u_1 @ v_1, (A@B) u_2 @ v_2> = <u_1, A u_2> <v_1, B v_2> =

	= <A+ u_1, u_2> <B+ v_1, v_2> = <(A+ @ B+)u_1 @ v_1, u_2 @ v_2>

so the adjoint of (A@B) must be (A@B)+ = (A+ @ B+).

In the case of real vector spaces, this means that the tensor product of
orthogonal transformations is orthogonal.

Finally, we get to the classic example. If A and B are symmetric, and
S diagonalizes A to J and T diagonalizes B to K, then

	(S@T)^-1 (A@B) (S@T) = (S^-1 A S) @ (T^-1 B T) = J @ K

Note that the diagonalization is orthogonal.

You cannot extend this trivially to Jordan normal form, since the
tensor product of transformations of the form (diagonal + nilpotent) is
not necessarily (diagonal + nilpotent).

Later,
Andrew Mullhaupt