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1、September 21, 20158:8Matrices: Algebra, Analysis and Applications - 9in x 6inb2108-ch05page 325Chapter 5Elements of Multilinear Algebra5.1Tensor Product of Two Free ModulesLet D be a domain. Recall that N is called a free finite dimensionalmodule over D if N has a finite basis e1,.,en, i.e. dim N =
2、n. Then N?:= Hom (N,D) is a free n-dimensional module. Furthermore we can identify Hom (N?,D) with N. (See Problem 1.)Definition 5.1.1 Let M,N be two free finite dimensional mod- ules over an integral domain D. Then the tensor product M N isidentified with Hom (N?,M). Moreover, for each m M,n N we i
3、dentify m n M DN with the linear transformation m n : N? M given by f ? f(n)m for any f N?.Proposition 5.1.2 Let M,N be free modules over a domain D with bases d1,.,dm,e1,.,en respectively.Then M DN is a free module with the basis di ej,i = 1,.,m,j = 1,.,n. In particulardim M N = dim M dim N.(5.1.1)
4、(See Problem 3.) For an abstract definition of M DN for any two D-modules see Problem 16.325September 21, 20158:8Matrices: Algebra, Analysis and Applications - 9in x 6inb2108-ch05page 326326MatricesIntuitively, one views M N as a linear span of all elements of the form m n, where m M,n N satisfy the
5、 following natural properties:1. a(m n) = (am) n = m (an) for all a D.2. (a1m1+a2m2)n = a1(m1n)+a2(m2n) for all a1,a2 D.(Linearity in the first variable.)3. m (a1n1+ a2n2) = a1(m n1) + a2(u n2) for all a1,a2 D.(Linearity in the second variable.)The element mn is called a decomposable tensor, or deco
6、mposable element (vector), or rank one tensor.Proposition 5.1.3 Let M,N be free modules over a domain D with bases d1,.,dm, e1,.,en respectively. Then any MD N is given by =i=m,j=n?i=j=1aijdi ej,A = aij Dmn.(5.1.2)Let u1,.,um,v1,.,vn be different bases of M,N respectively. Assume that =?m,n i,j=1bij
7、ui vjand let B = bij Dmn. Then B = PAQT, where P and Q are the transition matrices from the bases d1,.,dm to u1,.um and e1,.,en to v1,.,vn. (That is, d1,.,dm = u1,.umP, e1,.,en = v1,.,vnQ.)See Problem 6.Definition 5.1.4 Let M,N be free finite dimensional modules over a domain D. Let M DN be given by
8、 (5.1.2). The rank of , denoted by rank , is the rank of the representation matrix A, i.e. rank = rank A. The tensor rank of , denoted by Rank , is the minimal k such that =?k l=1ml nlfor some ml M,nl N,l = 1,.,k.September 21, 20158:8Matrices: Algebra, Analysis and Applications - 9in x 6inb2108-ch05
9、page 327Elements of Multilinear Algebra327The rank is independent of the choice of bases in M and N. (See Problem 7.) Since M DN has a basis consisting of decomposable tensors it follows thatRank min(dim M,dim N) for any M DN.(5.1.3)See Problem 8.Proposition 5.1.5 Let M,N be free finite dimensional
10、modules over a domain D. Let M DN. Then rank Rank . If D is a Bezout domain then rank = Rank .Proof. Assume that M,N have bases as in Proposition 5.1.3. Suppose that (5.1.2) holds. Let =?k l=1ml nl. Clearly, each ml nl=?m,n i,j=1aij,ldi ej, where Al:= aij,lm,ni,j=1 Dmnis rankone matrix. Then A =?k l
11、=1Al. It is straightforward to show that rank A k. This shows that rank Rank . Assume that D is a Bezout domain. Let P GL(m,D) such that PA = bij Dmnis a Hermite normal form of A. In particular,the first r := rank A rows of B are nonzero rows, and all other rows of B are zero rows.Let u1,.,um := d1,
12、.,dmP1be a basis in M. Proposition 5.1.3 yields that =?m,n i,j=1bijui ej.Define nl=?n j=1bljej,l = 1,.,r. Then =?r l=1ul nl. Hence r Rank , which implies that rank = Rank .?Proposition 5.1.6 Let Mi,Nibe free finite dimensional mod- ules over D.Let Ti: Mi Nibe homomorphisms.Then there exists a unique
13、 homomorphism on T : M1 M2 N1 N2such that T(m1 m2) = (T1m1) (T2m2) for all m1 M1,m2 M2. This homomorphism is denoted by T1 T2. Suppose furthermore that W1,W2are free finite dimensional D- modules, and Pi: Ni Wi,i = 1,2 are homomorphisms.Then (P1 P2)(T1 T2) = (P1T1) (P2T2).See Problem 9. Since each h
14、omomorphism Ti: Mi Ni,i = 1,2 is representedby a matrix, one can reduce the definition of T1 T2to the notionOctober 6, 20159:21Matrices: Algebra, Analysis and Applications - 9in x 6inb2108-ch05page 328328Matricesof tensor product of two matrices A1 Dn1m1,A2 Dn2m2. This tensor product is called the K
15、ronecker product.Definition 5.1.7 Let A = aijm,ni,j=1 Dmn,B = bijp,qi,j=1 Dpq. Then A B Dmpnqis the following block matrix:A B :=a11Ba12B.a1nB a21Ba22B.a2nB . am1Bam12B.amnB.(5.1.4)In the rest of the section, we discuss the symmetric and skew symmetric tensor products of M M.Definition 5.1.8 Let M be a free finite dimensional module over D. Denote M2:= M M. The submodule Sym2M M2, called a 2-symmetric power of M, is spanned by tensors of the form sym2(m,n) := m n + n m for all m,n M. sym2(m,n) =
