kronecker product的具体含义?

kronecker product的具体含义?
kronecker product的具体含义?

kronecker product的具体含义?
Given an (m\times n) matrix A and a (p\times q) matrix B,their Kronecker product C=(A\otimes B) is an (mp\times nq) matrix with elements defined by
C_{st}=A_{ij} B_{kl},
where s=p(i-1)+k,t=q(j-1)+l.
For example,if
A= A_{11} A_{12}
A_{21} A_{22}
B= B_{11} B_{12}
B_{21} B_{22}
then
C=(A\otimes B)=A_{11}B A_{12}B
A_{21}B A_{22}B

由Kronecker提出的数论符号。（很多符号打不出来）
设d=0或1(mod4),d非平方数且m>0,则克罗内克尔符号D=(d/m)定义为，若p整除d，即p|d,则(d/p)=0,
若d=1(mod8),则(d/2)=1
若d=5(mod8),则(d/2)=-1
若p为奇素数且p不整除d，则D转化为勒让德符号。
对于一般的整数m可以分解成r个素数（可能有重...

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由Kronecker提出的数论符号。（很多符号打不出来）
设d=0或1(mod4),d非平方数且m>0,则克罗内克尔符号D=(d/m)定义为，若p整除d，即p|d,则(d/p)=0,
若d=1(mod8),则(d/2)=1
若d=5(mod8),则(d/2)=-1
若p为奇素数且p不整除d，则D转化为勒让德符号。
对于一般的整数m可以分解成r个素数（可能有重复的，比如8=2x2x2）的乘积，则D=(d/m)=所有这些素因子的克罗内克尔符号的乘积。例如(d/8)=(d/2)x(d/2)x(d/2)
简单的应用（例如量子力学里）就是一个因子D(mn),当m=n时D=1,否则D=0
如果和一个方阵相乘的话则除了对角元其他项都为0.
中文也可以写作克罗内克符号。

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kronecker product(矩阵张量乘)

Kronecker Products Tom Lyche University of Oslo Norway Kronecker Products – p. 1/2 Example: Poisson Problem u=0 u=0 u=f u=0 2u 2u u = 2 + 2 x y m ∈ N, h = 1/(m + 1...

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kronecker product(矩阵张量乘)

Kronecker Products Tom Lyche University of Oslo Norway Kronecker Products – p. 1/2 Example: Poisson Problem u=0 u=0 u=f u=0 2u 2u u = 2 + 2 x y m ∈ N, h = 1/(m + 1) u=0 4h 3h 2h h 0 0 h 2h 3h 4h u(jh,kh) f j,k vj,k ≈ u(jh, kh) 1,3 1,2 1,1 2,3 2,2 2,1 3,3 3,2 3,1 v in grid j,k Kronecker Products – p. 2/2 Discrete Equation g(t h) 2g(t) + g(t + h) g (t) ≈ h2 v j,k+1 vj-1,k vj,k vj+1,k vj,k-1 From differential equation for j, k = 1, 2, . . . , m (vj1,k +2vj,k vj+1,k )+(vj,k1 +2vj,k vj,k+1 ) = h2 fj,k (1) From boundary conditions vj,k = 0 hvis j = 0, m + 1 eller k = 0, m + 1 (2) Kronecker Products – p. 3/2 Matrix Equation JV + V J = h F 2 1 1 2 v11 v12 v11 v12 + v21 v22 v21 v22 1 f11 f12 2 1 = 9 f21 f22 1 2 2 v11 . V := . . vm1 J := Jm = tridiagm (1, 2, 1) ∈ Rm,m f11 f1m v1m . ∈ Rm,m F := . . ∈ Rm,m . . . . . . vmm fm1 fmm m (1), (2) JV + V J = h2 F m JV + V J jk = i=1 Jji vik + i=1 vji Jik = l.h.s in (1) = h2 (F )jk Kronecker Products – p. 4/2 Convert JV + V J = h F to Ax = b 1,3 1,2 1,1 2,3 2,2 2,1 3,3 3,2 3,1 > 7 4 1 8 5 2 9 6 3 2 v in grid j,k x i in grid -v j,k+1 -v j-1,k 4 vj,k -x i+m -x i-1 4x i -xi+1 -v j+1,k --> -vj,k-1 -x i-m 4xi xi1 xi+1 xim xi+m = bi Kronecker Products – p. 5/2 Linear system Ax = b For m = 3 or n = 9 we have the following linear system: 4x1 x2 x4 = h2 f11 x1 4x2 x3 x5 = h2 f21 x2 4x3 x6 = h2 f31 x1 4x4 x5 x7 = h2 f12 x2 x4 4x5 x6 x8 = h2 f22 x3 x5 4x6 x9 = h2 f32 x4 4x7 x8 = h2 f13 x5 x7 4x8 x9 = h2 f23 x6 x8 4x9 = h2 f33 Kronecker Products – p. 6/2 Blockstructure of A A= 0 0 0 4 1 0 1 1 4 1 0 1 0 0 0 1 4 0 0 1 0 1 0 0 4 1 0 1 0 1 0 1 4 1 0 0 0 1 0 1 4 0 0 0 0 1 0 0 4 0 0 0 0 1 0 1 0 0 0 0 0 1 0 J + 2I I 0 A = I J + 2I I 0 I J + 2I 0 0 0 0 0 0 0 0 1 0 0 1 1 0 4 1 1 4 Kronecker Products – p. 7/2 Kronecker Product Denition 1. For any positive integers p, q, r, s we dene the Kronecker product of two matrices A ∈ Rp,q and B ∈ Rr,s as a matrix C ∈ Rpr,qs given in block form as We denote the Kronecker product of A and B by C = A B . Ab1,1 Ab1,2 Ab2,1 Ab2,2 C= . . . . . . Abr,1 Abr,2 Ab1,s Ab2,s . . . . Abr,s Dened for rectangular matrices of any dimension # of rows(columns) = product of # of rows(columns) in A and B Kronecker Products – p. 8/2 Kronecker Product Example 2 1 J= , 1 2 1 0 I= 0 1 0 0 2 1 1 2 0 0 J 0 J I = = 0 0 J 0 2 1 0 0 1 2 2 0 1 0 0 0 1 2 2I I I J = = 1 I 2I 0 2 0 0 1 0 2 Kronecker Products – p. 9/2 Poisson Matrix = J I + I J A= J 2I I 0 + I 2I I = J I + I J J 0 I 2I J Denition 2. Let for positive integers r, s, k , A ∈ Rr,r , B ∈ Rs,s and Ik be the identity matrix of order k . The sum A Is + Ir B is known as the Kronecker sum of A and B . The Poisson matrix is the Kronecker sum of J with itself. Kronecker Products – p. 10/2 Given A ∈ Rr,r , B ∈ Rs,s , F ∈ Rr,s . Find V ∈ Rr,s such that AV B T = F Matrix equation Kronecker For B ∈ Rm,n dene b1 b2 vec(B) := . ∈ Rmn , . . bn Lemma 3. b1j b2j bj = . . . bmj j th column AV B T = F AV + V B T = F (A B)vec(V ) = vec(F ) (3) (A Is + Ir B)vec(V ) = vec(F ). (4) Kronecker Products – p. 11/2 Proof We partition V , F , and B T by columns as V = (v1 , . . . , vs ), F = (f1 , . . . , fs ) and B T = (b1 , . . . , bs ). Then we have (A B)vec(V ) = vec(F ) Ab11 . . . Abs1 A j Ab1s . . . Abss f1 v1 . . . = . . . vs fs bij vj = fi , i = 1, . . . , m AV B T = F. [AV b1 , . . . , AV bs ] = F This proves (3) Kronecker Products – p. 12/2 Proof Continued (A Is + Ir B)vec(V ) = vec(F ) T (AV Is + Ir V B T ) = F AV + V B T = F. This give a slick way to derive the Poisson matrix. Recall AV + V B T = F (A Is + Ir B)vec(V ) = vec(F ) Therefore, JV + V J = h2 F (J I + I J)vec(V ) = h2 vec(F ) Kronecker Products – p. 13/2 Properties of Kronecker Products The usual arithmetic rules hold. Note however that (A B)T = AT B T AB =BA (A B)1 = A1 B 1 if A and B are nonsingular (A B)(C D) = (AC) (BD) (mixed product rule) The mixed product rule is valid for any matrices as long as the products AC and BD are dened. Kronecker Products – p. 14/2 Proof Mixed Product Rule If B ∈ Rr,t and D ∈ Rt,s for some integers r, t then Cd1,1 Cd1,s E1,1 E1,s Ab1,1 Ab1,t . . . . = . . . . . . . . . . . . . . Abr,1 Abr,t Cdt,1 Cdt,s Er,1 Er,s where for all i, j t Ei,j = k=1 bi,k dk,j AC = (AC)(BD)i,j = ((AC) (BD))i,j . where in the last formula i, j refers to the ij -block in the Kronecker product. Kronecker Products – p. 15/2 Eigenvalues and Eigenvectors The Kronecker product of two vectors u ∈ Rp and v ∈ Rr is a pr given by u v = uT v , . . . , uT v T vector u v ∈ R 1 r Suppose now A ∈ Rr,r and B ∈ Rs,s and Aui = λi ui , i = 1, . . . , r, Bvj = j vj , j = 1, . . . , s, then for i = 1, . . . , r, j = 1, . . . , s (5) (6) (A Is + Ir B)(ui vj ) = (λi + j )(ui vj ). (A B)(ui vj ) = λi j (ui vj ), Thus the eigenvalues of a Kronecker product(sum) are the products (sums) of the eigenvalues of the factors. The eigenvectors of a Kronecker product(sum) are the products of the eigenvectors of the factors. Kronecker Products – p. 16/2 Proof of Eigen-formulae This follows directly from the mixed product rule. For (5) (AB)(ui vj ) = (Aui )(Bvj ) = (λi ui )(j vj ) = (λi j )(ui vj ). From (5) (AIs )(ui vj ) = λi (ui vj ), and (Ir B)(ui vj ) = j (ui vj ) The result now follows by summing these relations. Kronecker Products – p. 17/2 Need eigenvalues and eigenvectors of the J matrix. Let h = 1/(m + 1). 1. We have Jsj = λj sj for j = 1, . . . , m, where sj = (sin (jπh), sin (2jπh), . . . , sin (mjπh))T , jπh ). λj = 4 sin ( 2 2 The Poisson Matrix A = J I + I J (7) (8) 2. The eigenvalues are distinct and the eigenvectors are orthogonal sT sk j 1 = δj,k , 2h j, k = 1, . . . , m. (9) Kronecker Products – p. 18/2 Eigenvalues/-vectors A = J I + I J 1. We have Axj,k = λj,k xj,k for j, k = 1, . . . , m, where xj,k = sj sk , λj,k 2 (10) (11) (12) sj = (sin (jπh), sin (2jπh), . . . , sin (mjπh))T , jπh 2 kπh ) + 4 sin ( ). = 4 sin ( 2 2 2. The eigenvectors are orthogonal xT xp,q j,k 1 = 2 δj,p δk,q , 4h j, k, p, q = 1, . . . , m. (13) 3. A is symmetric AT = (J I + I J)T = J T I T + I T J T = A 4. A is positive denite (positive eigenvalues) Kronecker Products – p. 19/2 A = J I + I J for m = 2 s1 = sin( π ) 3 sin( 2π ) 3 √ 3 1 , = 2 1 2 s2 = sin( 2π ) 3 sin( 4π ) 3 2 √ 3 1 = 2 1 π λ1 = 4 sin ( ) = 1, 6 2π λ2 = 4 sin ( ) = 3 6 i, j = 1, 2 Axij = ij xij , ij = λi + λj , xij = si sj 11 = 2, 12 = 21 = 4, 22 = 6. 3 s1 s1 = [1, 1, 1, 1]T , 4 3 s2 s1 = [1, 1, 1, 1]T , 4 xT xp,q j,k 9 = δj,p δk,q , 4 3 s1 s2 = [1, 1, 1, 1]T , 4 3 s2 s2 = [1, 1, 1, 1]T 4 j, k, p, q = 1, 2. Kronecker Products – p. 20/2 Check 2 1 1 0 1 0 2 1 A= + 1 2 0 1 0 1 1 2 1 1 4 1 1 0 1 1 4 0 1 1 = 2 , 1 1 0 4 1 1 1 1 0 1 1 4 1 1 4 1 1 0 1 1 4 0 1 1 = 4 1 1 0 4 1 1 1 1 0 1 1 4 etc. Kronecker Products – p. 21/2 Summary Studied the Poisson matrix A Shown that A can be written as a Kronecker sum studied properties of general Kronecker products and Kronecker sums used this to derive properties of A Kronecker Products – p. 22/2

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