Kronecker1

this file is obs, but keep for a while. It is now part of tensor wedge doc 1.n The Kronecker Product Let S be a linear operator in V, so S is an n x n matrix with matrix elements Sij. Let T be a linear operator in W, so T is an n' x n' matrix with matrix elements Tij. Then: v' = Sv = a vector in V S: V→V v'i = ΣaSiava w' = Tw = a vector in W T:W→W w'i = ΣaTiawa We want create a meaning for ST which is the tensor product of these two matrices S and T. A reasonable way to do this is as follows: (ST)(vw) = (v'w') = (Sv)(Tw) ST : VW → VW Consider the following processing steps, (ST)([αv1 + βv2]w) = (S[αv1 + βv2])(Tw) // definition of action of (ST) = (α Sv1+ βSv2) (Tw) // S:V→V is linear = α (Sv1)(Tw) + β(Sv2)(Tw) // using the first rule for vectors = α (ST)(v1w) + β (ST)(v2w) // definition of action of (ST) This shows that (ST)(vw) is linear in v. A similar argument shows it is also linear in w. Thus, the operator (ST) as defined above is a bilinear operator on VW, and we confirm the essential characteristic of the tensor product, which is its bilinearity. ____________________________________________________________________ Exercise: Compute the action of (ST) on a general element of VW . Applying (ST) to a general element of VW we get (ST)[ ΣijFij eie'j] = ΣijFij (ST)( eie'j) = ΣijFij (Sei)(Te'j) The action of S on a vector v (and T on w) can be written as (Sv) = Σa[Sv]aea = Σa(ΣbSabvb)ea = Σab(Sabvb)ea (Tw) = Σc[Tw]ce'c = Σc(ΣdTcdwd)e'c = Σcd(Tcdwd)e'c Therefore (Sei) = ΣabSab(ei)b ea (Te'j) = ΣcdTcd(e'j)d e'c Then (Sei)(Te'j) = [ ΣabSab(ei)b ea] [ ΣcdTcd(e'j)d e'c] = Σabcd Sab(ei)bTcd(e'j)d (eae'c) and so (ST)[ ΣijFij eie'j] = ΣijFij (Sei)(Te'j) = Σijabcd FijSab(ei)bTcd(e'j)d (eae'c) = Σac { Σijbd FijSab(ei)bTcd(e'j)d } (eae'c) = Σac Gac (eae'c) where Gac = Σijbd FijSab(ei)bTcd(e'j)d In the special case that ei and e'j are the standard unit vector bases for V and W, the result simplifies, Gac = Σijbd FijSabδi,bTcd δj,d = Σij FijSaiTcj = Σij SaiFijTTjc = (SFTT)ac or G = SFTT . This shows explicitly how (ST) acts on a general element of VW to produce another element of VW. The matrix S is n x n, the matrix T is n' x n', while the matrices F and G are n x n'. For example, if n < n' we would have this view of how the matrices "conform" for G = SFTT : _______________________________________________________________ It is useful now to consider the component analysis of the action of ST on a pure element of VW in the sense of our outer product extension of the tensor product of two vectors discussed earlier. Then (ST)(vw) = (Sv)(Tw) // = v'w' [(ST)(vw)]ii' = [(Sv)(Tw)]ii' // = (v'w')ii' (**) The right side is easily processed as already done above, [(Sv)(Tw)]ii' = (Sv)i(Tw)i' = (Σj Sijvj)(Σj'Ti'j'wj') = Σjj' SijTi'j' vj wj' = Σjj' SijTi'j' (vw)jj' (***) It is helpful to visualize the object on the left of * in this manner, [(ST)(vw)]ii' = Σjj' (ST)ii',jj' (vw)jj' so then * becomes (v'w')ii' = Σjj' (ST)ii',jj' (vw)jj' as if we were multiplying a vector (vw) by a matrix (ST), but each usual summation index is replaced by two summation indices. In a multiindex notation one might write the above as [(ST)(vw)]I = ΣJ (ST)I,J (vw)J I = {i,i'} J = {j,j'} Comparing (**) and (***) we find that (ST)ii',jj' = SijTi'j' Note carefully how the indices are arranged. The object (ST)ii',jj' is a rank-4 tensor, since it is the outer product of two rank-2 tensors Sij and Ti'j', and as such it has four indices. One could just write (ST)ii'jj' without the comma, but the comma helped us visualize the multi-index matrix multiplication above. Is there some way to write ST as a standard matrix with two indices instead of four? Go back to our equation (v'w')ii' = Σjj' (ST)ii',jj' (vw)jj' or (v'iw'i') = Σjj' (ST)ii',jj' (vjwj') (ST)ii',jj' = (SijTi'j') We want to write this somehow in a form x'r = Σs Mrs xs . For illustration purposes, assume n = 2 and n' = 3. Then write the components (vjwj') as a single column vector in this obvious manner, where the w component index moves fastest, = If viwj → xr, one can compute r from i,j as follows ( here 3 = n' = dim(W) for this special case ) r = (i-1)3 + j (r-1) = (i-1)3 + (j-1) = (i-1) + int() = i-1 and rem () = j-1 Thus we can compute i and j from r in this way (integer part and remainder) i = 1+int( ) j = 1+rem( ) . Therefore, the desired 6x6 matrix Mrs is given by Mrs = (ST)ii',jj' = SijTi'j' where i = 1+int( ) j = 1+int( ) i' = 1+rem( ) j' = 1+rem( ) It is a bit tedious to compute and display the matrix M by hand, so we let Maple do it for us: For the general case where dim(V) = n and dim(W) = n', the rule for generating this matrix M is Mrs = (ST)ii',jj' = SijTi'j' where i = 1+int( ) j = 1+int( ) i,j = 1,2..n i' = 1+rem( ) j' = 1+rem( ) i',j' = 1,2..n' r,s = 1,2...nn' Symbolically we can write M = (ST), with the meaning shown above. Our general matrix M is a special case of what is known as a Kronecker product of two matrices. In our application, S and T are each square matrices, whereas the general Kronecker products allows for S and T to each be non-square. This matrix is an example of what is known as a Kronecker product. viwj → xr where r = (i-1)n' + j Then we have a matrix equation of the form = By staring at (v'iw'i') = Σjj' (ST)ii',jj' (vjwj') we conclude that M11 = (ST)11,11 M21 = (ST)12,11 M31 = (ST)13,11 M41 = (ST)21,11 M12 = (ST)11,12 M22 = (ST)12,12 M32 = (ST)13,12 M42 = (ST)21,12 M13 = (ST)11,13 M23 = (ST)12,13 M33 = (ST)13,13 M43 = (ST)21,13 M14 = (ST)11,21 M24 = (ST)12,21 M34 = (ST)13,21 M44 = (ST)21,21 M15 = (ST)11,22 M25 = (ST)12,22 M65 = (ST)13,22 M45 = (ST)21,22 M16 = (ST)11,23 M26 = (ST)12,23 M66 = (ST)13,23 M46 = (ST)21,23 where the last two columns of M follow the same pattern. Then using (ST)ii',jj' = SijTi'j' this becomes M11 = S11T11 M21 = S11T21 M31 = S11T31 M41 = S11T31 M12 = S11T12 M22 = S11T22 M32 = S11T32 M42 = S11T32 M13 = S11T13 M23 = S11T23 M33 = S11T33 M43 = S11T33 M14 = S12T11 M24 = S12T21 M34 = S12T31 M44 = S12T31 M15 = S12T12 M25 = S12T22 M35 = S12T32 M45 = S12T32 M16 = S12T13 M26 = S12T23 M36 = S12T33 M46 = S12T33 Notice that: We can then write = Write this in an abbreviated notation xii' = Σjj Aii The trick is to write the n*n components (vw)ab = vawb in a column vector in a certain obvious manner. As a simple case, if n = 2 and n' = 3 we would write (vw)ab → = (v1w1, v1w2, ...v1wn', v2w1, v2w2, ...v2wn', ......vnw1, vnw2, ...vnwn')T If these components are written as (x1, x2, ......xnn')T, then xr = (n'-1)i + j