Quote:
Originally Posted by davidrh
Aha, I think I found the ideal place to try and get an answer to my question. Doing multilinear algebra with rank-2 tensors is all fine, but I am unable to get a straight (and simple example) for rank-3 (and hence a generalisation to rank-n).
I have a tensor A_{ijk} where i = 1..10, j=1..5 and k=1..15
and tensor B_{ijk}
Taking a direct step from rank-2 matrices, I want to compute the inner product of Z=AB'
how? and what is the size of Z is it 10x15x10? or...
Actually being even more precise of what I am trying to write out simply is I have A_{ijk} = x_i y_j h_k
and B = A;
So how do I write the dot product Z in terms of x,y and h?
Much appreciated!
David
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I'm afraid the formula you seek doesn't exist. For tensors greater than rank 2 involves messy calculations of each coefficient.
First of all matrix algebra can only be used on tensors up to rank 2.
Vectors or tensors of rank 1 can be defined as a row or column matrix.
Tensors of rank 2 are defined by a MxN matrix. If M=N this can define the dimension of space, while each component in the matrix is defined by a pair of indices (ij for example) which defines the rank.
Secondly the dot product cannot be used on tensors which are not of rank 1.
The generic operation for tensors is the inner product. This involves the outer product of each tensor followed by a dot product with a vector.
For example suppose
C=
ABC and
D=
EFG.
C and
D are tensors of rank 3,
ABC and
EFG are the respective vector (tensor 1) factors.
The outer product
CD=
ABCEFG is a tensor of rank 6.
The inner product with a vector
U is applied to either
A,
B,
C,
D,
E, F or
G. Since the dot product will be different in each case there is obviously no hard and fast rules for calculations.
The only consistant point is that since the dot product of 2 vectors is a scalar the rank of the tensor is reduced from 6 to 4.
In general if the rank of the tensor is n, the dot product reduces the rank to n-2. This is known as contraction.
Hope this is of help.
Steven