# 克洛内克积

In mathematics, the Kronecker product, sometimes denoted by ⊗,[1] is an operation on two matrices of arbitrary size resulting in a block matrix. It is a generalization of the outer product (which is denoted by the same symbol) from vectors to matrices, and gives the matrix of the tensor product with respect to a standard choice of basis. The Kronecker product is to be distinguished from the usual matrix multiplication, which is an entirely different operation. The Kronecker product is also sometimes called matrix direction product.[2]

In mathematics, the Kronecker product, sometimes denoted by ⊗, is an operation on two matrices of arbitrary size resulting in a block matrix. It is a generalization of the outer product (which is denoted by the same symbol) from vectors to matrices, and gives the matrix of the tensor product with respect to a standard choice of basis. The Kronecker product is to be distinguished from the usual matrix multiplication, which is an entirely different operation. The Kronecker product is also sometimes called matrix direction product.

The Kronecker product is named after the German mathematician Leopold Kronecker (1823-1891), even though there is little evidence that he was the first to define and use it. The Kronecker product has also been called the Zehfuss matrix, after Johann Georg Zehfuss, who in 1858 described this matrix operation, but Kronecker product is currently the most widely used.[3]

\end{bmatrix}, [/itex]

## Definition定义

more explicitly:

If A is an m × n matrix and B is a p × q matrix, then the Kronecker product AB is the pm × qn block matrix:

$\displaystyle{ {\mathbf{A}\otimes\mathbf{B}} = \begin{bmatrix} 如果你不喜欢这个，你可以试试看（？） :\lt math\gt \mathbf{A}\otimes\mathbf{B} = \begin{bmatrix} a_{11} b_{11} & a_{11} b_{12} & \cdots & a_{11} b_{1q} & 11} b {11} & a {11} b {12} & cdots & a {11} b {1q } & a_{11} \mathbf{B} & \cdots & a_{1n}\mathbf{B} \\ \cdots & \cdots & a_{1n} b_{11} & a_{1n} b_{12} & \cdots & a_{1n} b_{1q} \\ Cdots & cdots & a _ {1n } b _ {11} & a _ {1n } b _ {12} & cdots & a _ {1n } b _ {1q } \vdots & \ddots & \vdots \\ a_{11} b_{21} & a_{11} b_{22} & \cdots & a_{11} b_{2q} & 11} b {21} & a {11} b {22} & cdots & a {11} b {2q } & a_{m1} \mathbf{B} & \cdots & a_{mn} \mathbf{B} \cdots & \cdots & a_{1n} b_{21} & a_{1n} b_{22} & \cdots & a_{1n} b_{2q} \\ Cdots & cdots & a _ {1n } b _ {21} & a _ {1n } b _ {22} & cdots & a _ {1n } b _ {2q } \end{bmatrix}, }$

  \vdots & \vdots & \ddots & \vdots & & & \vdots & \vdots & \ddots & \vdots \\


Vdots & vdots & vdots & vdots & vdots & ddots & vdots & vdots

  a_{11} b_{p1} & a_{11} b_{p2} & \cdots & a_{11} b_{pq} &


A {11} b { p1} & a {11} b { p2} & cdots & a {11} b { pq } &

more explicitly:

                  \cdots & \cdots & a_{1n} b_{p1} & a_{1n} b_{p2} & \cdots & a_{1n} b_{pq} \\


Cdots & cdots & a _ {1n } b _ { p1} & a _ {1n } b _ { p2} & cdots & a _ {1n } b _ { pq }

  \vdots & \vdots & & \vdots & \ddots & & \vdots & \vdots & & \vdots \\


Vdots & vdots & vdots & vdots & vdots & vdots

$\displaystyle{ {\mathbf{A}\otimes\mathbf{B}} = \begin{bmatrix} \vdots & \vdots & & \vdots & & \ddots & \vdots & \vdots & & \vdots \\ Vdots & vdots & vdots & ddots & vdots & vdots a_{11} b_{11} & a_{11} b_{12} & \cdots & a_{11} b_{1q} & a_{m1} b_{11} & a_{m1} b_{12} & \cdots & a_{m1} b_{1q} & 11} & a _ { m1} b _ {12} & cdots & a _ { m1} b _ {1q } & \cdots & \cdots & a_{1n} b_{11} & a_{1n} b_{12} & \cdots & a_{1n} b_{1q} \\ \cdots & \cdots & a_{mn} b_{11} & a_{mn} b_{12} & \cdots & a_{mn} b_{1q} \\ Cdots & cdots & a _ { mn } b _ {11} & a _ { mn } b _ {12} & cdots & a _ { mn } b _ {1q } a_{11} b_{21} & a_{11} b_{22} & \cdots & a_{11} b_{2q} & a_{m1} b_{21} & a_{m1} b_{22} & \cdots & a_{m1} b_{2q} & 1} b {21} & a _ { m1} b {22} & cdots & a _ { m1} b {2q } & \cdots & \cdots & a_{1n} b_{21} & a_{1n} b_{22} & \cdots & a_{1n} b_{2q} \\ \cdots & \cdots & a_{mn} b_{21} & a_{mn} b_{22} & \cdots & a_{mn} b_{2q} \\ Cdots & cdots & a { mn } b {21} & a { mn } b {22} & cdots & a { mn } b {2q } \vdots & \vdots & \ddots & \vdots & & & \vdots & \vdots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \vdots & & & \vdots & \vdots & \ddots & \vdots \\ Vdots & vdots & vdots & vdots & vdots & ddots & vdots & vdots a_{11} b_{p1} & a_{11} b_{p2} & \cdots & a_{11} b_{pq} & a_{m1} b_{p1} & a_{m1} b_{p2} & \cdots & a_{m1} b_{pq} & A _ { m1} b _ { p1} & a _ { m1} b _ { p2} & cdots & a _ { m1} b _ { pq } & \cdots & \cdots & a_{1n} b_{p1} & a_{1n} b_{p2} & \cdots & a_{1n} b_{pq} \\ \cdots & \cdots & a_{mn} b_{p1} & a_{mn} b_{p2} & \cdots & a_{mn} b_{pq} Cdots & cdots & a _ { mn } b _ { p1} & a _ { mn } b _ { p2} & cdots & a _ { mn } b _ { pq } \vdots & \vdots & & \vdots & \ddots & & \vdots & \vdots & & \vdots \\ \end{bmatrix}. }$

  \vdots & \vdots & & \vdots & & \ddots & \vdots & \vdots & & \vdots \\

  a_{m1} b_{11} & a_{m1} b_{12} & \cdots & a_{m1} b_{1q} &


More compactly, we have

                  \cdots & \cdots & a_{mn} b_{11} & a_{mn} b_{12} & \cdots & a_{mn} b_{1q} \\


$\displaystyle{ a_{m1} b_{21} & a_{m1} b_{22} & \cdots & a_{m1} b_{2q} & (A\otimes B)_{p(r-1)+v, q(s-1)+w} = a_{rs} b_{vw} (a) _ { p (r-1) + v，q (s-1) + w } = a _ { rs } b _ { vw } \cdots & \cdots & a_{mn} b_{21} & a_{mn} b_{22} & \cdots & a_{mn} b_{2q} \\ }$

  \vdots & \vdots & \ddots & \vdots & & & \vdots & \vdots & \ddots & \vdots \\

  a_{m1} b_{p1} & a_{m1} b_{p2} & \cdots & a_{m1} b_{pq} &


Similarly

                  \cdots & \cdots & a_{mn} b_{p1} & a_{mn} b_{p2} & \cdots & a_{mn} b_{pq}


$\displaystyle{ \end{bmatrix}. }$

(A\otimes B)_{i, j} = a_{\lfloor (i-1)/p \rfloor +1,\lfloor (j-1)/q \rfloor +1} b_{i - \lfloor (i-1)/p \rfloor p, j - \lfloor (j-1)/q \rfloor q}.

(a) _ { i，j } = a _ { lfloor (i-1)/p rfloor + 1，lfloor (j-1)/q rfloor + 1} b _ { i-lfloor (i-1)/p rfloor p，j-lfloor (j-1)/q rfloor q }.

[/itex]

More compactly, we have

Using the identity $\displaystyle{ i\%p = i - \lfloor i/p \rfloor p }$, where $\displaystyle{ i \% p }$ denotes the remainder of $\displaystyle{ i/p }$, this may be written in a more symmetric form

$\displaystyle{ \lt math\gt (A\otimes B)_{p(r-1)+v, q(s-1)+w} = a_{rs} b_{vw} (A\otimes B)_{i, j} = a_{\lfloor (i-1)/p \rfloor +1,\lfloor (j-1)/q \rfloor +1} b_{(i-1)\%p +1, (j-1)\%q + 1}. (a) _ { i，j } = a _ { lfloor (i-1)/p rfloor + 1，lfloor (j-1)/q rfloor + 1} b _ (i-1)% p + 1，(j-1)% q + 1}. }$

[/itex]

Similarly

If A and B represent linear transformations and , respectively, then represents the tensor product of the two maps, .

$\displaystyle{ (A\otimes B)_{i, j} = a_{\lfloor (i-1)/p \rfloor +1,\lfloor (j-1)/q \rfloor +1} b_{i - \lfloor (i-1)/p \rfloor p, j - \lfloor (j-1)/q \rfloor q}. }$

$\displaystyle{ Using the identity \lt math\gt i\%p = i - \lfloor i/p \rfloor p }$, where $\displaystyle{ i \% p }$ denotes the remainder of $\displaystyle{ i/p }$, this may be written in a more symmetric form

 \begin{bmatrix}


$\displaystyle{ 1 & 2 \\ 1 & 2 \\ (A\otimes B)_{i, j} = a_{\lfloor (i-1)/p \rfloor +1,\lfloor (j-1)/q \rfloor +1} b_{(i-1)\%p +1, (j-1)\%q + 1}. 3 & 4 \\ 3 & 4 \\ }$
 \end{bmatrix} \otimes


 \begin{bmatrix}


If A and B represent linear transformations V1W1 and V2W2, respectively, then AB represents the tensor product of the two maps, V1V2W1W2.

   0 & 5 \\

   0 & 5 \\


   6 & 7 \\

   6 & 7 \\


### Examples示例

 \end{bmatrix} =


$\displaystyle{ \begin{bmatrix} 开始{ bmatrix } \begin{bmatrix} 1 \begin{bmatrix} 1 begin { bmatrix } 1 & 2 \\ 0 & 5 \\ 0 & 5 \\ 3 & 4 \\ 6 & 7 \\ 6 & 7 \\ \end{bmatrix} \otimes \end{bmatrix} & 结束{ bmatrix } & \begin{bmatrix} 2 \begin{bmatrix} 2 begin { bmatrix } 0 & 5 \\ 0 & 5 \\ 0 & 5 \\ 6 & 7 \\ 6 & 7 \\ 6 & 7 \\ \end{bmatrix} = \end{bmatrix} \\ 结束{ matrix }\\ \begin{bmatrix} 1 \begin{bmatrix} 3 \begin{bmatrix} 3 begin { bmatrix } 0 & 5 \\ 0 & 5 \\ 0 & 5 \\ 6 & 7 \\ 6 & 7 \\ 6 & 7 \\ \end{bmatrix} & \end{bmatrix} & 结束{ bmatrix } & 2 \begin{bmatrix} 4 \begin{bmatrix} 4 begin { bmatrix } 0 & 5 \\ 0 & 5 \\ 0 & 5 \\ 6 & 7 \\ 6 & 7 \\ 6 & 7 \\ \end{bmatrix} \\ \end{bmatrix} \\ \end{bmatrix} = 3 \begin{bmatrix} 0 & 5 \\ \begin{bmatrix} 6 & 7 \\ 1\times 0 & 1\times 5 & 2\times 0 & 2\times 5 \\ 1 * 0 & 1 * 5 & 2 * 0 & 2 * 5 \end{bmatrix} & 1\times 6 & 1\times 7 & 2\times 6 & 2\times 7 \\ 1 * 6 & 1 * 7 & 2 * 6 & 2 * 7 4 \begin{bmatrix} 3\times 0 & 3\times 5 & 4\times 0 & 4\times 5 \\ 0 & 5 \\ 3\times 6 & 3\times 7 & 4\times 6 & 4\times 7 \\ 6 & 7 \\ \end{bmatrix} = \end{bmatrix} \\ \end{bmatrix} = \begin{bmatrix} 0 & 5 & 0 & 10 \\ 0 & 5 & 0 & 10 \\ \begin{bmatrix} 6 & 7 & 12 & 14 \\ 6 & 7 & 12 & 14 \\ 1\times 0 & 1\times 5 & 2\times 0 & 2\times 5 \\ 0 & 15 & 0 & 20 \\ 0 & 15 & 0 & 20 \\ 1\times 6 & 1\times 7 & 2\times 6 & 2\times 7 \\ 18 & 21 & 24 & 28 18 & 21 & 24 & 28 3\times 0 & 3\times 5 & 4\times 0 & 4\times 5 \\ \end{bmatrix}. 3\times 6 & 3\times 7 & 4\times 6 & 4\times 7 \\ }$

 \end{bmatrix} =


Similarly:

 \begin{bmatrix}

    0 &  5 &  0 & 10 \\


$\displaystyle{ 6 & 7 & 12 & 14 \\ \begin{bmatrix} 0 & 15 & 0 & 20 \\ 1 & -4 & 7 \\ 1 & -4 & 7 \\ 18 & 21 & 24 & 28 -2 & 3 & 3 -2 & 3 & 3 \end{bmatrix}. \end{bmatrix} \otimes }$

\begin{bmatrix}

8 & -9 & -6 & 5 \\

8 & -9 & -6 & 5 \\

Similarly:

1 & -3 & -4 & 7 \\

1 & -3 & -4 & 7 \\

2 & 8 & -8 & -3 \\

2 & 8 & -8 & -3 \\

$\displaystyle{ 1 & 2 & -5 & -1 1 & 2 & -5 & -1 \begin{bmatrix} \end{bmatrix} = 1 & -4 & 7 \\ \begin{bmatrix} -2 & 3 & 3 8 & -9 & -6 & 5 & -32 & 36 & 24 & -20 & 56 & -63 & -42 & 35 \\ 8 & -9 & -6 & 5 & -32 & 36 & 24 & -20 & 56 & -63 & -42 & 35 \\ \end{bmatrix} \otimes 1 & -3 & -4 & 7 & -4 & 12 & 16 & -28 & 7 & -21 & -28 & 49 \\ 1 & -3 & -4 & 7 & -4 & 12 & 16 & -28 & 7 & -21 & -28 & 49 \\ \begin{bmatrix} 2 & 8 & -8 & -3 & -8 & -32 & 32 & 12 & 14 & 56 & -56 & -21 \\ 2 & 8 & -8 & -3 & -8 & -32 & 32 & 12 & 14 & 56 & -56 & -21 \\ 8 & -9 & -6 & 5 \\ 1 & 2 & -5 & -1 & -4 & -8 & 20 & 4 & 7 & 14 & -35 & -7 \\ 1 & 2 & -5 & -1 & -4 & -8 & 20 & 4 & 7 & 14 & -35 & -7 \\ 1 & -3 & -4 & 7 \\ -16 & 18 & 12 & -10 & 24 & -27 & -18 & 15 & 24 & -27 & -18 & 15 \\ -16 & 18 & 12 & -10 & 24 & -27 & -18 & 15 & 24 & -27 & -18 & 15 \\ 2 & 8 & -8 & -3 \\ -2 & 6 & 8 & -14 & 3 & -9 & -12 & 21 & 3 & -9 & -12 & 21 \\ -2 & 6 & 8 & -14 & 3 & -9 & -12 & 21 & 3 & -9 & -12 & 21 \\ 1 & 2 & -5 & -1 -4 & -16 & 16 & 6 & 6 & 24 & -24 & -9 & 6 & 24 & -24 & -9 \\ -4 & -16 & 16 & 6 & 6 & 24 & -24 & -9 & 6 & 24 & -24 & -9 \\ \end{bmatrix} = -2 & -4 & 10 & 2 & 3 & 6 & -15 & -3 & 3 & 6 & -15 & -3 -2 & -4 & 10 & 2 & 3 & 6 & -15 & -3 & 3 & 6 & -15 & -3 \begin{bmatrix} \end{bmatrix} 8 & -9 & -6 & 5 & -32 & 36 & 24 & -20 & 56 & -63 & -42 & 35 \\ }$

1 & -3 & -4 & 7 & -4 & 12 & 16 & -28 & 7 & -21 & -28 & 49 \\

2 & 8 & -8 & -3 & -8 & -32 & 32 & 12 & 14 & 56 & -56 & -21 \\

1 & 2 & -5 & -1 & -4 & -8 & 20 & 4 & 7 & 14 & -35 & -7 \\

-16 & 18 & 12 & -10 & 24 & -27 & -18 & 15 & 24 & -27 & -18 & 15 \\

-2 & 6 & 8 & -14 & 3 & -9 & -12 & 21 & 3 & -9 & -12 & 21 \\

{{ordered list

{有序列表

-4 & -16 & 16 & 6 & 6 & 24 & -24 & -9 & 6 & 24 & -24 & -9 \\

|1= Bilinearity and associativity:

| 1 = 双线性和结合性:

-2 & -4 & 10 & 2 & 3 & 6 & -15 & -3 & 3 & 6 & -15 & -3

\end{bmatrix}

The Kronecker product is a special case of the tensor product, so it is bilinear and associative:

[/itex]

\displaystyle{ \begin{align} 1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3 == Properties性质 == \mathbf{A} \otimes (\mathbf{B} + \mathbf{C}) &= \mathbf{A} \otimes \mathbf{B} + \mathbf{A} \otimes \mathbf{C}, \\ 数学时间(mathbf { b } + mathbf { c }) & = mathbf { a }/mathbf { b }/mathbf { a }/mathbf { c } , (\mathbf{B} + \mathbf{C}) \otimes \mathbf{A} &= \mathbf{B} \otimes \mathbf{A} + \mathbf{C} \otimes \mathbf{A}, \\ (mathbf { b } + mathbf { c }) otimes mathbf { a } & = mathbf { b } otimes mathbf { a } + mathbf { c } otimes mathbf { a } , === Relations to other matrix operations === (k\mathbf{A}) \otimes \mathbf{B} &= \mathbf{A} \otimes (k\mathbf{B}) = k(\mathbf{A} \otimes \mathbf{B}), \\ (k mathbf { a }) otimes mathbf { b } & = mathbf { a } otimes (k mathbf { b }) = k (mathbf { a } times mathbf { b }) , {{ordered list (\mathbf{A} \otimes \mathbf{B}) \otimes \mathbf{C} &= \mathbf{A} \otimes (\mathbf{B} \otimes \mathbf{C}), \\ (mathbf { a } otimes mathbf { b }) otimes mathbf { c } & = mathbf { a } otimes (mathbf { b } otimes mathbf { c }) , |1= '''[[Bilinearity]] and [[associativity]]:'''{{paragraph}} \mathbf{A} \otimes \mathbf{0} &= \mathbf{0} \otimes \mathbf{A} = \mathbf{0}, 如果你是一个数学家，那么你就是一个数学家, \end{align} }

The Kronecker product is a special case of the tensor product, so it is bilinear and associative:

where A, B and C are matrices, 0 is a zero matrix, and k is a scalar.

\displaystyle{ \begin{align} \mathbf{A} \otimes (\mathbf{B} + \mathbf{C}) &= \mathbf{A} \otimes \mathbf{B} + \mathbf{A} \otimes \mathbf{C}, \\ |2= Non-commutative: | 2 = 非交换性: (\mathbf{B} + \mathbf{C}) \otimes \mathbf{A} &= \mathbf{B} \otimes \mathbf{A} + \mathbf{C} \otimes \mathbf{A}, \\ (k\mathbf{A}) \otimes \mathbf{B} &= \mathbf{A} \otimes (k\mathbf{B}) = k(\mathbf{A} \otimes \mathbf{B}), \\ In general, and are different matrices. However, and are permutation equivalent, meaning that there exist permutation matrices P and Q such that 一般来说，\lt font color="#ff8000"\gt 和\lt /font\gt 是不同的矩阵。然而，\lt font color="#ff8000"\gt 和\lt /font\gt 是置换等价的，这意味着存在置换矩阵P和Q，使得 (\mathbf{A} \otimes \mathbf{B}) \otimes \mathbf{C} &= \mathbf{A} \otimes (\mathbf{B} \otimes \mathbf{C}), \\ \lt math\gt \mathbf{B} \otimes \mathbf{A} = \mathbf{P} \, (\mathbf{A} \otimes \mathbf{B}) \, \mathbf{Q}. }

 \mathbf{A} \otimes \mathbf{0} &= \mathbf{0} \otimes \mathbf{A} = \mathbf{0},


If A and B are square matrices, then and are even permutation similar, meaning that we can take .

\end{align}[/itex]

The matrices and are perfect shuffle matrices. The perfect shuffle matrix Sp,q can be constructed by taking slices of the Ir identity matrix, where $\displaystyle{ r=pq }$.

where A, B and C are matrices, 0 is a zero matrix, and k is a scalar.

$\displaystyle{ \mathbf{S}_{p,q} = \begin{bmatrix} [数学][数学][ s }{ p，q } = begin { bmatrix } |2= '''Non-[[commutative operation|commutative]]:'''{{paragraph}} \mathbf{I}_r(1:q:r,:) \\ 1: q: r，:) \mathbf{I}_r(2:q:r,:) \\ 2: q: r，:) In general, {{nowrap|'''A''' ⊗ '''B'''}} and {{nowrap|'''B''' ⊗ '''A'''}} are different matrices. However, {{nowrap|'''A''' ⊗ '''B'''}} and {{nowrap|'''B''' ⊗ '''A'''}} are permutation equivalent, meaning that there exist [[permutation matrix|permutation matrices]] '''P''' and '''Q''' such that\lt ref\gt {{Cite journal 一般而言，{nowrap| | | | | | | | | |}}}}和{nowrap| | | | |}}}}和{nowrap|。然而，{nowrap|「「A」、「otimes」、「B」、「B」及{nowrap|「「B」、「otimes」、「A」、「A」均为置换等价物，即存在[[置换矩阵|置换矩阵]、「P」及「Q」，因此\lt ref\gt {{Cite journal \vdots \\ 斑点 |author1=H. V. Henderson |author2=S. R. Searle | year = 1980 \mathbf{I}_r(q:q:r,:) (q: q: r，:) | title = The vec-permutation matrix, the vec operator and Kronecker products: A review \end{bmatrix} }$

| journal = Linear and Multilinear Algebra

| volume = 9


MATLAB colon notation is used here to indicate submatrices, and Ir is the identity matrix. If $\displaystyle{ \mathbf{A} \in \mathbb{R}^{m_1 \times n_1} }$ and $\displaystyle{ \mathbf{B} \in \mathbb{R}^{m_2 \times n_2} }$, then

| pages = 271–288

| number = 4


$\displaystyle{ \mathbf{B} \otimes \mathbf{A} = \mathbf{S}_{m_1,m_2} (\mathbf{A} \otimes \mathbf{B}) \mathbf{S}^\textsf{T}_{n_1,n_2} }$

| doi = 10.1080/03081088108817379


|hdl=1813/32747 | url = https://ecommons.cornell.edu/bitstream/1813/32747/1/BU-645-M.pdf

|3= The mixed-product property:

| 3 = 混合产品属性:

}}</ref>

$\displaystyle{ \mathbf{B} \otimes \mathbf{A} = \mathbf{P} \, (\mathbf{A} \otimes \mathbf{B}) \, \mathbf{Q}. }$

If A, B, C and D are matrices of such size that one can form the matrix products AC and BD, then

If A and B are square matrices, then AB and BA are even permutation similar, meaning that we can take P = QT.

$\displaystyle{ (\mathbf{A} \otimes \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{AC}) \otimes (\mathbf{BD}). }$

(mathbf { a } otimes mathbf { b })(mathbf { c } otimes mathbf { d }) = (mathbf { AC }) otimes (mathbf { BD }).数学

The matrices P and Q are perfect shuffle matrices.引用错误：没有找到与</ref>对应的<ref>标签 The perfect shuffle matrix Sp,q can be constructed by taking slices of the Ir identity matrix, where $\displaystyle{ r=pq }$.

}}</ref>  完全洗牌矩阵Perfect shuffle matrixSp,q可以通过对Ir的切片来构造，其中$\displaystyle{ r=pq }$。


| 4 = Hadamard 乘积(元素级乘法) :

$\displaystyle{ \mathbf{S}_{p,q} = \begin{bmatrix} The mixed-product property also works for the element-wise product. If A and C are matrices of the same size, B and D are matrices of the same size, then '''\lt font color="#ff8000"\gt 混合积性质Mixed-product property\lt /font\gt '''也适用于元素级产品。如果A和C是相同大小的矩阵，B和D是相同大小的矩阵，则 \mathbf{I}_r(1:q:r,:) \\ \lt math\gt (\mathbf{A} \otimes \mathbf{B}) \circ (\mathbf{C} \otimes \mathbf{D}) = (\mathbf{A} \circ \mathbf{C}) \otimes (\mathbf{B} \circ \mathbf{D}). }$

 \mathbf{I}_r(2:q:r,:) \\

 \vdots \\


|5= The inverse of a Kronecker product:

|5. 克洛内克积的倒数:

 \mathbf{I}_r(q:q:r,:)


\end{bmatrix}[/itex]

It follows that is invertible if and only if both A and B are invertible, in which case the inverse is given by

$\displaystyle{ (\mathbf{A} \otimes \mathbf{B})^{-1} = \mathbf{A}^{-1} \otimes \mathbf{B}^{-1}. }$

< math > (mathbf { a } otimes mathbf { b }) ^ {-1} = mathbf { a } ^ {-1} otimes mathbf { b } ^ {-1}.数学

MATLAB colon notation is used here to indicate submatrices, and Ir is the r × r identity matrix. If $\displaystyle{ \mathbf{A} \in \mathbb{R}^{m_1 \times n_1} }$ and $\displaystyle{ \mathbf{B} \in \mathbb{R}^{m_2 \times n_2} }$, then

MATLAB此处使用冒号表示子矩阵，而Irr × r标识矩阵。如果$\displaystyle{ \mathbf{A} \in \mathbb{R}^{m_1 \times n_1} }$$\displaystyle{ \mathbf{B} \in \mathbb{R}^{m_2 \times n_2} }$，则

The invertible product property holds for the Moore–Penrose pseudoinverse as well, that is

$\displaystyle{ \mathbf{B} \otimes \mathbf{A} = \mathbf{S}_{m_1,m_2} (\mathbf{A} \otimes \mathbf{B}) \mathbf{S}^\textsf{T}_{n_1,n_2} }$

$\displaystyle{ (\mathbf{A} \otimes \mathbf{B})^{+} = \mathbf{A}^{+} \otimes \mathbf{B}^{+}. }$

< math > (mathbf { a } otimes mathbf { b }) ^ { + } = mathbf { a } ^ { + } otimes { b } ^ { + }.数学

|3= The mixed-product property:模板:Paragraph

In the language of Category theory, the mixed-product property of the Kronecker product (and more general tensor product) shows that the category MatF of matrices over a field F, is in fact a monoidal category, with objects natural numbers n, morphisms are n-by-m matrices with entries in F, composition is given by matrix multiplication, identity arrows are simply identity matrices In, and the tensor product is given by the Kronecker product.

If A, B, C and D are matrices of such size that one can form the matrix products AC and BD, then

MatF is a concrete skeleton category for the equivalent category FinVectF of finite dimensional vector spaces over F, whose objects are such finite dimensional vector spaces V, arrows are F-linear maps , and identity arrows are the identity maps of the spaces. The equivalence of categories amounts to simultaneously choosing a basis in ever finite-dimensional vector space V over F; matrices' elements represent these mappings with respect to the chosen bases; and likewise the Kronecker product is the representation of the tensor product in the chosen bases.

MatF╱sub>是F上有限维空间的等价类FinVectF╱sub>的具体骨架类，其对象为有限维空间V，箭头为F-线性映射，而恒等箭头为空间的恒等映射。类别的等价性相当于同时在有限维空间V除以F选择一个基；矩阵的元素表示有关所选基的这些映射；同样地， 克洛内克积是所选 张量积的表示。

$\displaystyle{ (\mathbf{A} \otimes \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{AC}) \otimes (\mathbf{BD}). }$

|6= Transpose:

6 = 转置:

This is called the mixed-product property, because it mixes the ordinary matrix product and the Kronecker product.

Transposition and conjugate transposition are distributive over the Kronecker product:

As immediate consequence,

$\displaystyle{ (\mathbf{A}\otimes \mathbf{B})^\textsf{T} = \mathbf{A}^\textsf{T} \otimes \mathbf{B}^\textsf{T} }$ and $\displaystyle{ (\mathbf{A}\otimes \mathbf{B})^* = \mathbf{A}^* \otimes \mathbf{B}^*. }$

$\displaystyle{ \mathbf{A} \otimes \mathbf{B} = (\mathbf{I_2} \otimes \mathbf{B} )(\mathbf{A} \otimes \mathbf{I_1}) = (\mathbf{A} \otimes \mathbf{I_1} )(\mathbf{I_2} \otimes \mathbf{B}) }$.

|7= Determinant:

7 = 行列式:

In particular, using the transpose property from below, this means that if

Let A be an matrix and let B be an matrix. Then

$\displaystyle{ \mathbf{A} = \mathbf{Q} \otimes \mathbf{U} }$

$\displaystyle{ \left| \mathbf{A} \otimes \mathbf{B} \right| = \left| \mathbf{A} \right| ^m \left| \mathbf{B} \right| ^n . }$

and Q and U are orthogonal (or unitary), then A is also orthogonal (resp., unitary).

「Q」与「U」为[[正交矩阵|正交]（或[[酉矩阵|酉]），则「A」亦为正交（分别为酉）。

The exponent in is the order of B and the exponent in is the order of A.

|8= Kronecker sum and exponentiation:

|8 = 克洛内克和与幂:

|4「（阿达玛积（矩阵）」（元素相乘）：」模板:段

If A is , B is and Ik denotes the identity matrix then we can define what is sometimes called the Kronecker sum, ⊕, by

The mixed-product property also works for the element-wise product. If A and C are matrices of the same size, B and D are matrices of the same size, then

$\displaystyle{ \mathbf{A} \oplus \mathbf{B} = \mathbf{A} \otimes \mathbf{I}_m + \mathbf{I}_n \otimes \mathbf{B} . }$

$\displaystyle{ (\mathbf{A} \otimes \mathbf{B}) \circ (\mathbf{C} \otimes \mathbf{D}) = (\mathbf{A} \circ \mathbf{C}) \otimes (\mathbf{B} \circ \mathbf{D}). }$

This is different from the direct sum of two matrices. This operation is related to the tensor product on Lie algebras.

|5= The inverse of a Kronecker product:模板:Paragraph

|5「」一Kronecker积的逆：模板:段

We have the following formula for the matrix exponential, which is useful in some numerical evaluations.

It follows that AB is invertible if and only if both A and B are invertible, in which case the inverse is given by

$\displaystyle{ (\mathbf{A} \otimes \mathbf{B})^{-1} = \mathbf{A}^{-1} \otimes \mathbf{B}^{-1}. }$

$\displaystyle{ \exp({\mathbf{N} \oplus \mathbf{M}}) = \exp(\mathbf{N}) \otimes \exp(\mathbf{M}) }$

The invertible product property holds for the Moore–Penrose pseudoinverse as well,[4] that is

Kronecker sums appear naturally in physics when considering ensembles of non-interacting systems. Let Hi be the Hamiltonian of the ith such system. Then the total Hamiltonian of the ensemble is

$\displaystyle{ (\mathbf{A} \otimes \mathbf{B})^{+} = \mathbf{A}^{+} \otimes \mathbf{B}^{+}. }$

$\displaystyle{ H_{\mathrm{Tot}}=\bigoplus_{i}H^{i} }$.

[数学][数学][数学]。

}}

}}

In the language of Category theory, the mixed-product property of the Kronecker product (and more general tensor product) shows that the category MatF of matrices over a field F, is in fact a monoidal category, with objects natural numbers n, morphisms nm are n-by-m matrices with entries in F, composition is given by matrix multiplication, identity arrows are simply n × n identity matrices In, and the tensor product is given by the Kronecker product.[6]

MatF is a concrete skeleton category for the equivalent category FinVectF of finite dimensional vector spaces over F, whose objects are such finite dimensional vector spaces V, arrows are F-linear maps L : VW, and identity arrows are the identity maps of the spaces. The equivalence of categories amounts to simultaneously choosing a basis in ever finite-dimensional vector space V over F; matrices' elements represent these mappings with respect to the chosen bases; and likewise the Kronecker product is the representation of the tensor product in the chosen bases.

「Mat」F是「F」上有限维空间的等价范畴的具体[[骨架（范畴理论）[骨架类别]、「FinVect」F ，其对象为该等有限维向量空间「V」，箭头为L : VW的「F」—线性映射，而标准箭头为空间的标准映射。类别的等价性相当于在有限维向量空间「V」除以「F」上同时存在[[选择公理|选择一个基]；矩阵元素表示关于所选基的这些映射；同样，克洛内克积是所选基中张量积的表示。

{{ordered list

{有序列表

|1= Spectrum:

1 = Spectrum:

Suppose that A and B are square matrices of size n and m respectively. Let λ1, ..., λn be the eigenvalues of A and μ1, ..., μm be those of B (listed according to multiplicity). Then the eigenvalues of are

Transposition and conjugate transposition are distributive over the Kronecker product:

$\displaystyle{ \lambda_i \mu_j, \qquad i=1,\ldots,n ,\, j=1,\ldots,m. }$

1，ldots，n，，j = 1，ldots，m.数学

$\displaystyle{ (\mathbf{A}\otimes \mathbf{B})^\textsf{T} = \mathbf{A}^\textsf{T} \otimes \mathbf{B}^\textsf{T} }$ and $\displaystyle{ (\mathbf{A}\otimes \mathbf{B})^* = \mathbf{A}^* \otimes \mathbf{B}^*. }$

It follows that the trace and determinant of a Kronecker product are given by

$\displaystyle{ \operatorname{tr}(\mathbf{A} \otimes \mathbf{B}) = \operatorname{tr} \mathbf{A} \, \operatorname{tr} \mathbf{B} \quad\text{and}\quad \det(\mathbf{A} \otimes \mathbf{B}) = (\det \mathbf{A})^m (\det \mathbf{B})^n. }$

Let A be an n × n matrix and let B be an m × m matrix. Then

|2= Singular values:

|2 = 单数值:

$\displaystyle{ \left| \mathbf{A} \otimes \mathbf{B} \right| = \left| \mathbf{A} \right| ^m \left| \mathbf{B} \right| ^n . }$

The exponent in 模板:Abs is the order of B and the exponent in 模板:Abs is the order of A.

模板:Abs 中的指数为「B」的顺序， 模板:Abs 中的指数为「A」的顺序。


If A and B are rectangular matrices, then one can consider their singular values. Suppose that A has rA nonzero singular values, namely

$\displaystyle{ \sigma_{\mathbf{A},i}, \qquad i = 1, \ldots, r_\mathbf{A}. }$

1，ldots，r _ mathbf { a }.数学

|8= Kronecker sum and exponentiation:模板:Anchor模板:Paragraph

Similarly, denote the nonzero singular values of B by

If A is n × n, B is m × m and Ik denotes the k × k identity matrix then we can define what is sometimes called the Kronecker sum, ⊕, by

$\displaystyle{ \sigma_{\mathbf{B},i}, \qquad i = 1, \ldots, r_\mathbf{B}. }$

1，ldots，r _ mathbf { b }.数学

$\displaystyle{ \mathbf{A} \oplus \mathbf{B} = \mathbf{A} \otimes \mathbf{I}_m + \mathbf{I}_n \otimes \mathbf{B} . }$

Then the Kronecker product has rArB nonzero singular values, namely

This is different from the direct sum of two matrices. This operation is related to the tensor product on Lie algebras.

$\displaystyle{ \sigma_{\mathbf{A},i} \sigma_{\mathbf{B},j}, \qquad i=1,\ldots,r_\mathbf{A} ,\, j=1,\ldots,r_\mathbf{B}. }$

1，qquad i = 1，ldots，r _ mathbf { a } ，，j = 1，ldots，r _ mathbf { b }.数学

We have the following formula for the matrix exponential, which is useful in some numerical evaluations.引用错误：没有找到与</ref>对应的<ref>标签

|4= Relation to products of graphs:

| 4 = 与图的乘积的关系:

$\displaystyle{ \exp({\mathbf{N} \oplus \mathbf{M}}) = \exp(\mathbf{N}) \otimes \exp(\mathbf{M}) }$

The Kronecker product of the adjacency matrices of two graphs is the adjacency matrix of the tensor product graph. The Kronecker sum of the adjacency matrices of two graphs is the adjacency matrix of the Cartesian product graph.

}}

}}

Kronecker sums appear naturally in physics when considering ensembles of non-interacting systems.[citation needed] Let Hi be the Hamiltonian of the ith such system. Then the total Hamiltonian of the ensemble is

$\displaystyle{ H_{\mathrm{Tot}}=\bigoplus_{i}H^{i} }$.

}}

The Kronecker product can be used to get a convenient representation for some matrix equations. Consider for instance the equation , where A, B and C are given matrices and the matrix X is the unknown.

We can use the "vec trick" to rewrite this equation as

### Abstract properties 理论性质

$\displaystyle{ 《数学》 {{ordered list {{有序列表 \left(\mathbf{B}^\textsf{T} \otimes \mathbf{A}\right) \, \operatorname{vec}(\mathbf{X}) 左(mathbf { b } ^ textsf { t } otimes mathbf { a } right) ，操作符名称{ vec }(mathbf { x }) |1= '''[[Spectrum (functional analysis)|Spectrum]]:'''{{paragraph}} = \operatorname{vec}(\mathbf{AXB}) = \operatorname{vec}(\mathbf{C}) = operatorname { vec }(mathbf { AXB }) = operatorname { vec }(mathbf { c }) . }$

. math

Suppose that A and B are square matrices of size n and m respectively. Let λ1, ..., λn be the eigenvalues of A and μ1, ..., μm be those of B (listed according to multiplicity). Then the eigenvalues of AB are

$\displaystyle{ \lambda_i \mu_j, \qquad i=1,\ldots,n ,\, j=1,\ldots,m. }$

Here, vec(X) denotes the vectorization of the matrix X, formed by stacking the columns of X into a single column vector.

It follows that the trace and determinant of a Kronecker product are given by

It now follows from the properties of the Kronecker product that the equation has a unique solution, if and only if A and B are nonsingular .

$\displaystyle{ \operatorname{tr}(\mathbf{A} \otimes \mathbf{B}) = \operatorname{tr} \mathbf{A} \, \operatorname{tr} \mathbf{B} \quad\text{and}\quad \det(\mathbf{A} \otimes \mathbf{B}) = (\det \mathbf{A})^m (\det \mathbf{B})^n. }$

If X and AXB are row-ordered into the column vectors u and v, respectively, then

$\displaystyle{ 《数学》 \mathbf{v} = 2009年10月11日 If '''A''' and '''B''' are rectangular matrices, then one can consider their [[singular value decomposition|singular values]]. Suppose that '''A''' has ''r''\lt sub\gt '''A'''\lt /sub\gt nonzero singular values, namely 如果“A”和“B”是矩形矩阵，则一可以考虑它们的[[奇异值分解|奇异值]）。假设“A”有''r''\lt sub\gt '''A'''\lt /sub\gt 非零奇异值，即 \left(\mathbf{A} \otimes \mathbf{B}^\textsf{T}\right)\mathbf{u} 左(mathbf { a } otimes mathbf { b } ^ textsf { t } right) mathbf { u } :\lt math\gt \sigma_{\mathbf{A},i}, \qquad i = 1, \ldots, r_\mathbf{A}. }$

.[/itex]

. math

The reason is that

Similarly, denote the nonzero singular values of B by

$\displaystyle{ :\lt math\gt \sigma_{\mathbf{B},i}, \qquad i = 1, \ldots, r_\mathbf{B}. }$

\mathbf{v} =

2009年10月11日

\operatorname{vec}((\mathbf{AXB})^\textsf{T}) =

Then the Kronecker product AB has rArB nonzero singular values, namely

\operatorname{vec}(\mathbf{B}^\textsf{T}\mathbf{X}^\textsf{T}\mathbf{A}^\textsf{T}) =

$\displaystyle{ \sigma_{\mathbf{A},i} \sigma_{\mathbf{B},j}, \qquad i=1,\ldots,r_\mathbf{A} ,\, j=1,\ldots,r_\mathbf{B}. }$

\left(\mathbf{A} \otimes \mathbf{B}^\textsf{T}\right)\operatorname{vec}(\mathbf{X^\textsf{T}}) =

\left(\mathbf{A} \otimes \mathbf{B}^\textsf{T}\right)\mathbf{u}

Since the rank of a matrix equals the number of nonzero singular values, we find that

.[/itex]

. math

$\displaystyle{ \operatorname{rank}(\mathbf{A} \otimes \mathbf{B}) = \operatorname{rank} \mathbf{A} \, \operatorname{rank} \mathbf{B}. }$

|3= Relation to the abstract tensor product:模板:Paragraph

For an example of the application of this formula, see the article on the Lyapunov equation.

This formula also comes in handy in showing that the matrix normal distribution is a special case of the multivariate normal distribution.

The Kronecker product of matrices corresponds to the abstract tensor product of linear maps. Specifically, if the vector spaces V, W, X, and Y have bases {v1, ..., vm}, {w1, ..., wn}, {x1, ..., xd}, and {y1, ..., ye}, respectively, and if the matrices A and B represent the linear transformations S : VX and T : WY, respectively in the appropriate bases, then the matrix AB represents the tensor product of the two maps, ST : VWXY with respect to the basis {v1w1, v1w2, ..., v2w1, ..., vmwn}模板:Void of VW and the similarly defined basis of XY with the property that AB(viwj) = (Avi) ⊗ (Bwj), where i and j are integers in the proper range.[8]

$\displaystyle{ \mathbf{A} \circ \mathbf{B} = \left(\mathbf{A}_{ij} \circ \mathbf{B}\right)_{ij} = \left(\left(\mathbf{A}_{ij} \otimes \mathbf{B}_{kl}\right)_{kl}\right)_{ij} }$

When V and W are Lie algebras, and S : VV and T : WW are Lie algebra homomorphisms, the Kronecker sum of A and B represents the induced Lie algebra homomorphisms VWVW.

which means that the (ij)-th subblock of the product is the matrix , of which the (k)-th subblock equals the }} matrix }}. Essentially the Tracy–Singh product is the pairwise Kronecker product for each pair of partitions in the two matrices.

|4= Relation to products of graphs:模板:Paragraph

|4与[[图论|图]的[[矩阵相乘|积]的关系：

For example, if A and B both are partitioned matrices e.g.:

$\displaystyle{ \mathbf{A} = 2009年10月11日 The Kronecker product of the [[adjacency matrix|adjacency matrices]] of two [[Graph (discrete mathematics)|graphs]] is the adjacency matrix of the [[tensor product of graphs|tensor product graph]]. The [[Kronecker product#Kronecker sum and exponentiation|Kronecker sum]] of the adjacency matrices of two [[Graph (discrete mathematics)|graphs]] is the adjacency matrix of the [[cartesian product of graphs|Cartesian product graph]].\lt ref name="TAOCP0a"\gt See answer to Exercise 96, D. E. Knuth: [http://www-cs-faculty.stanford.edu/~knuth/fasc0a.ps.gz "Pre-Fascicle 0a: Introduction to Combinatorial Algorithms"], zeroth printing (revision 2), to appear as part of D.E. Knuth: ''[[The Art of Computer Programming]] Vol. 4A''\lt /ref\gt 二个[[图（离散数学）[图]的[[邻接矩阵╱邻接矩阵]的克洛内克积是图的[[张量积╱张量积图]的邻接矩阵。二个[[图（离散数学）[图]的邻接矩阵的[[Kronecker积| Kronecker和和及指数化| Kronecker和]]是[[图的cartesian积| cartesian积图]的邻接矩阵。\lt ref name="TAOCP0a"\gt See answer to Exercise 96, D. E. Knuth: [http://www-cs-faculty.stanford.edu/~knuth/fasc0a.ps.gz "Pre-Fascicle 0a: Introduction to Combinatorial Algorithms"], zeroth printing (revision 2), to appear as part of D.E. Knuth: ''[[The Art of Computer Programming]] Vol. 4A''\lt /ref\gt \left[ 左边[ }} \begin{array} {c | c} 开始{ array }{ c | c } \mathbf{A}_{11} & \mathbf{A}_{12} \\ 11} & mathbf { a }{12} == Matrix equations 矩阵方程式== \hline \hline The Kronecker product can be used to get a convenient representation for some matrix equations. Consider for instance the equation {{nowrap|1='''AXB''' = '''C'''}}, where '''A''', '''B''' and '''C''' are given matrices and the matrix '''X''' is the unknown. kroncker积可用于某些矩阵方程式的方便表示。例如，考虑公式{nowrap|1=「AXB」=「C」}，其中，「A」、「B」及「C」为给定矩阵，而矩阵「X」为未知。 \mathbf{A}_{21} & \mathbf{A}_{22} 21} & mathbf { a }{22} We can use the "vec trick" to rewrite this equation as 我们可以使用“vec技巧”将此公式重新编写为 \end{array} 结束{数组} :\lt math\gt \right] [右] \left(\mathbf{B}^\textsf{T} \otimes \mathbf{A}\right) \, \operatorname{vec}(\mathbf{X}) = = = \operatorname{vec}(\mathbf{AXB}) = \operatorname{vec}(\mathbf{C}) \left[ 左边[ . }$

\begin{array} {c c | c}

1 & 2 & 3 \\

1 & 2 & 3 \\

Here, vec(X) denotes the vectorization of the matrix X, formed by stacking the columns of X into a single column vector.

4 & 5 & 6 \\

4 & 5 & 6 \\

\hline

\hline

It now follows from the properties of the Kronecker product that the equation AXB = C has a unique solution, if and only if A and B are nonsingular 模板:Harv.

7 & 8 & 9

7 & 8 & 9

\end{array}

If X and AXB are row-ordered into the column vectors u and v, respectively, then 模板:Harv

\right]

[右]

$\displaystyle{ ,\quad ，四 \mathbf{v} = \mathbf{B} = 2 = = \left(\mathbf{A} \otimes \mathbf{B}^\textsf{T}\right)\mathbf{u} \left[ 左边[ . }$

\begin{array} {c | c}

The reason is that

\mathbf{B}_{11} & \mathbf{B}_{12} \\

11} & mathbf { b }{12}

$\displaystyle{ \hline \hline \mathbf{v} = \mathbf{B}_{21} & \mathbf{B}_{22} 21} & mathbf { b }{22} \operatorname{vec}((\mathbf{AXB})^\textsf{T}) = \end{array} 结束{数组} \operatorname{vec}(\mathbf{B}^\textsf{T}\mathbf{X}^\textsf{T}\mathbf{A}^\textsf{T}) = \right] [右] \left(\mathbf{A} \otimes \mathbf{B}^\textsf{T}\right)\operatorname{vec}(\mathbf{X^\textsf{T}}) = = = \left(\mathbf{A} \otimes \mathbf{B}^\textsf{T}\right)\mathbf{u} \left[ 左边[ . }$

\begin{array} {c | c c}

1 & 4 & 7 \\

1 & 4 & 7 \\

### Applications应用

\hline

\hline

For an example of the application of this formula, see the article on the Lyapunov equation.

2 & 5 & 8 \\

2 & 5 & 8 \\

This formula also comes in handy in showing that the matrix normal distribution is a special case of the multivariate normal distribution.

3 & 6 & 9

3 & 6 & 9

This formula is also useful for representing 2D image processing operations in matrix-vector form.

\end{array}

\right]

[右]

Another example is when a matrix can be factored as a Hadamard product, then matrix multiplication can be performed faster by using the above formula. This can be applied recursively, as done in the radix-2 FFT and the Fast Walsh–Hadamard transform. Splitting a known matrix into the Hadamard product of two smaller matrices is known as the "nearest Kronecker Product" problem, and can be solved exactly[9] by using the SVD. To split a matrix into the Hadamard product of more than two matrices, in an optimal fashion, is a difficult problem and the subject of ongoing research; some authors cast it as a tensor decomposition problem.[10][11]

,

[/itex]

In conjunction with the least squares method, the Kronecker product can be used as an accurate solution to the hand eye calibration problem.[14]

we get:

## Related matrix operations模板:Anchor相关矩阵运算模板:Anchor

\displaystyle{ \begin{align} 1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3 Two related matrix operations are the '''Tracy–Singh''' and '''[[Khatri–Rao product]]s''', which operate on [[Block matrix|partitioned matrices]]. Let the {{nowrap|''m'' × ''n''}} matrix '''A''' be partitioned into the {{nowrap|''m''\lt sub\gt ''i''\lt /sub\gt × ''n''\lt sub\gt ''j''\lt /sub\gt }} blocks '''A'''\lt sub\gt ''ij''\lt /sub\gt and {{nowrap|''p'' × ''q''}} matrix '''B''' into the {{nowrap|''p\lt sub\gt k\lt /sub\gt '' × ''q\lt sub\gt {{ell}}\lt /sub\gt ''}} blocks '''B'''\lt sub\gt ''kl''\lt /sub\gt , with of course {{nowrap|1=Σ''\lt sub\gt i\lt /sub\gt m\lt sub\gt i\lt /sub\gt '' = ''m''}}, {{nowrap|1=Σ''\lt sub\gt j\lt /sub\gt n\lt sub\gt j\lt /sub\gt '' = ''n''}}, {{nowrap|1=Σ''\lt sub\gt k\lt /sub\gt p\lt sub\gt k\lt /sub\gt '' = ''p''}} and {{nowrap|1=Σ''\lt sub\gt {{ell}}\lt /sub\gt q\lt sub\gt {{ell}}\lt /sub\gt '' = ''q''}}. 两个相关的矩阵运算是“特雷西-辛格”''Tracy–Singh''和''[[Khatri–Rao product]]s''，它们对[[块矩阵|分区矩阵]]进行操作。将 {{nowrap|''m'' × ''n''}} 矩阵''A''划分成 {{nowrap|''m''\lt sub\gt ''i''\lt /sub\gt × ''n''\lt sub\gt ''j''\lt /sub\gt }} 块'''A'''\lt sub\gt ''ij''\lt /sub\gt ，{{nowrap|''p'' × ''q''}} 矩阵 '''B''' 划分为{{nowrap|''p\lt sub\gt k\lt /sub\gt '' × ''q\lt sub\gt {{ell}}\lt /sub\gt ''}} 块 '''B'''\lt sub\gt ''kl''\lt /sub\gt ，自然有{{nowrap|1=Σ''\lt sub\gt i\lt /sub\gt m\lt sub\gt i\lt /sub\gt '' = ''m''}}，{{nowrap|1=Σ''\lt sub\gt j\lt /sub\gt n\lt sub\gt j\lt /sub\gt '' = ''n''}}，{{nowrap|1=Σ''\lt sub\gt k\lt /sub\gt p\lt sub\gt k\lt /sub\gt '' = ''p''}} 和 {{nowrap|1=Σ''\lt sub\gt {{ell}}\lt /sub\gt q\lt sub\gt {{ell}}\lt /sub\gt '' = ''q''}}。 \mathbf{A} \circ \mathbf{B} 1.1.2.2.2.2.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3 = \left[\begin{array} {c | c} = left [ begin { array }{ c | c } === '''\lt font color="#ff8000"\gt 特雷西-辛格积Tracy–Singh product\lt /font\gt '''=== \mathbf{A}_{11} \circ \mathbf{B} & \mathbf{A}_{12} \circ \mathbf{B} \\ 11} circ mathbf { b } & mathbf { a }12} circ mathbf { b } The '''Tracy–Singh product''' is defined as\lt ref\gt {{cite journal |last=Tracy |first=D. S. |last2=Singh |first2=R. P. |year=1972 |title=A New Matrix Product and Its Applications in Matrix Differentiation |journal=Statistica Neerlandica |volume=26 |issue=4 |pages=143–157 |doi=10.1111/j.1467-9574.1972.tb00199.x }}\lt /ref\gt \lt ref\gt {{cite journal |last=Liu |first=S. |year=1999 |title=Matrix Results on the Khatri–Rao and Tracy–Singh Products |journal=Linear Algebra and Its Applications |volume=289 |issue=1–3 |pages=267–277 |doi=10.1016/S0024-3795(98)10209-4 |doi-access=free }}\lt /ref\gt '''\lt font color="#ff8000"\gt 特雷西-辛格积Tracy–Singh product\lt /font\gt '''定义为\lt ref\gt {{cite journal |last=Tracy |first=D. S. |last2=Singh |first2=R. P. |year=1972 |title=A New Matrix Product and Its Applications in Matrix Differentiation |journal=Statistica Neerlandica |volume=26 |issue=4 |pages=143–157 |doi=10.1111/j.1467-9574.1972.tb00199.x }}\lt /ref\gt \lt ref\gt {{cite journal |last=Liu |first=S. |year=1999 |title=Matrix Results on the Khatri–Rao and Tracy–Singh Products |journal=Linear Algebra and Its Applications |volume=289 |issue=1–3 |pages=267–277 |doi=10.1016/S0024-3795(98)10209-4 |doi-access=free }}\lt /ref\gt \hline \hline \mathbf{A}_{21} \circ \mathbf{B} & \mathbf{A}_{22} \circ \mathbf{B} 21} circ mathbf { b } & mathbf { a }{22} circ mathbf { b } :\lt math\gt \mathbf{A} \circ \mathbf{B} = \left(\mathbf{A}_{ij} \circ \mathbf{B}\right)_{ij} = \left(\left(\mathbf{A}_{ij} \otimes \mathbf{B}_{kl}\right)_{kl}\right)_{ij} }

   \end{array}\right]


 ={} &\left[\begin{array} {c | c | c | c}


{} & left [ begin { array }{ c | c | c | c }

which means that the (ij)-th subblock of the mp × nq product A $\displaystyle{ \circ }$ B is the mi p × nj q matrix Aij $\displaystyle{ \circ }$ B, of which the (k模板:Ell)-th subblock equals the mi pk × nj q模板:Ell matrix AijBk模板:Ell. Essentially the Tracy–Singh product is the pairwise Kronecker product for each pair of partitions in the two matrices.

        \mathbf{A}_{11} \otimes \mathbf{B}_{11} & \mathbf{A}_{11} \otimes \mathbf{B}_{12} & \mathbf{A}_{12} \otimes \mathbf{B}_{11} & \mathbf{A}_{12} \otimes \mathbf{B}_{12} \\


        \hline

        \hline


For example, if A and B both are 2 × 2 partitioned matrices e.g.: 例如，如果“A”和“B”都是{nowrap | 2×2}}分块矩阵，如下：

        \mathbf{A}_{11} \otimes \mathbf{B}_{21} & \mathbf{A}_{11} \otimes \mathbf{B}_{22} & \mathbf{A}_{12} \otimes \mathbf{B}_{21} & \mathbf{A}_{12} \otimes \mathbf{B}_{22} \\


\displaystyle{ \mathbf{A} = \hline \hline \left[ \mathbf{A}_{21} \otimes \mathbf{B}_{11} & \mathbf{A}_{21} \otimes \mathbf{B}_{12} & \mathbf{A}_{22} \otimes \mathbf{B}_{11} & \mathbf{A}_{22} \otimes \mathbf{B}_{12} \\ \begin{array} {c | c} \hline \hline \mathbf{A}_{11} & \mathbf{A}_{12} \\ \mathbf{A}_{21} \otimes \mathbf{B}_{21} & \mathbf{A}_{21} \otimes \mathbf{B}_{22} & \mathbf{A}_{22} \otimes \mathbf{B}_{21} & \mathbf{A}_{22} \otimes \mathbf{B}_{22} \hline \end{array}\right] \\ 结束{数组}右] \mathbf{A}_{21} & \mathbf{A}_{22} ={} &\left[\begin{array} {c c | c c c c | c | c c} {} & left [ begin { array }{ c | c c | c c c | c } \end{array} 1 & 2 & 4 & 7 & 8 & 14 & 3 & 12 & 21 \\ 1 & 2 & 4 & 7 & 8 & 14 & 3 & 12 & 21 \\ \right] 4 & 5 & 16 & 28 & 20 & 35 & 6 & 24 & 42 \\ 4 & 5 & 16 & 28 & 20 & 35 & 6 & 24 & 42 \\ = \hline \hline \left[ 2 & 4 & 5 & 8 & 10 & 16 & 6 & 15 & 24 \\ 2 & 4 & 5 & 8 & 10 & 16 & 6 & 15 & 24 \\ \begin{array} {c c | c} 3 & 6 & 6 & 9 & 12 & 18 & 9 & 18 & 27 \\ 3 & 6 & 6 & 9 & 12 & 18 & 9 & 18 & 27 \\ 1 & 2 & 3 \\ 8 & 10 & 20 & 32 & 25 & 40 & 12 & 30 & 48 \\ 8 & 10 & 20 & 32 & 25 & 40 & 12 & 30 & 48 \\ 4 & 5 & 6 \\ 12 & 15 & 24 & 36 & 30 & 45 & 18 & 36 & 54 \\ 12 & 15 & 24 & 36 & 30 & 45 & 18 & 36 & 54 \\ \hline \hline \hline 7 & 8 & 9 7 & 8 & 28 & 49 & 32 & 56 & 9 & 36 & 63 \\ 7 & 8 & 28 & 49 & 32 & 56 & 9 & 36 & 63 \\ \end{array} \hline \hline \right] 14 & 16 & 35 & 56 & 40 & 64 & 18 & 45 & 72 \\ 14 & 16 & 35 & 56 & 40 & 64 & 18 & 45 & 72 \\ ,\quad 21 & 24 & 42 & 63 & 48 & 72 & 27 & 54 & 81 21 & 24 & 42 & 63 & 48 & 72 & 27 & 54 & 81 \mathbf{B} = \end{array}\right]. 结束{数组}右]。 \left[ \end{align} }

\begin{array} {c | c}

\mathbf{B}_{11} & \mathbf{B}_{12} \\

\hline

\mathbf{B}_{21} & \mathbf{B}_{22}

\end{array}

\right]

=

\left[

\begin{array} {c | c c}

Mixed-products properties

1 & 4 & 7 \\

$\displaystyle{ \mathbf{A} \otimes (\mathbf{B}\bull \mathbf{C}) = (\mathbf{A}\otimes \mathbf{B}) \bull \mathbf{C} }$, where $\displaystyle{ \bull }$ denotes the Face-splitting product

\hline

2 & 5 & 8 \\

$\displaystyle{ (\mathbf{A} \bull \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{A}\mathbf{C}) \bull (\mathbf{B} \mathbf{D}) }$,

3 & 6 & 9

\end{array}

Similarly:

\right]

$\displaystyle{ (\mathbf{A} \bull \mathbf{L})(\mathbf{B} \otimes \mathbf{M}) . . . (\mathbf{C} \otimes \mathbf{S}) = (\mathbf{A}\mathbf{B}...\mathbf{C}) \bull (\mathbf{L}\mathbf{M}...\mathbf{S}) }$,

(数学)(数学)(数学)。..(mathbf { c } otimes mathbf { s }) = (mathbf { a } mathbf { b } ... mathbf { c }) bull (mathbf { l } mathbf { m } ... mathbf { s }) </math > ,

,

[/itex]

$\displaystyle{ c^\textsf{T} \bull d^\textsf{T} = c^\textsf{T} \otimes d^\textsf{T} }$, where $\displaystyle{ c }$ and $\displaystyle{ d }$ are vectors,

$\displaystyle{ c^\textsf{T} \bull d^\textsf{T} = c^\textsf{T} \otimes d^\textsf{T} }$，这里 $\displaystyle{ c }$$\displaystyle{ d }$ 都是向量，< br/>

we get: 我们得到：

$\displaystyle{ (\mathbf{A} \bull \mathbf{B})(c \otimes d) = (\mathbf{A}c) \circ (\mathbf{B}d) }$, where $\displaystyle{ c }$ and $\displaystyle{ d }$ are vectors, $\displaystyle{ \circ }$ denotes the Hadamard product

\displaystyle{ \begin{align} Similarly: 类似地: \mathbf{A} \circ \mathbf{B} \lt math\gt (\mathbf{A} \bull \mathbf{B})(\mathbf{M}\mathbf{N}c \otimes \mathbf{Q}\mathbf{P}d) = (\mathbf{A}\mathbf{M}\mathbf{N}c) \circ (\mathbf{B}\mathbf{Q}\mathbf{P}d), }

(数学)(mathbf { a } bull mathbf { b })(mathbf { n } mathbf { n } mathbf { p } d) = (mathbf { a } mathbf { m } mathbf { n } c) circ (mathbf { b } mathbf { q } mathbf { p } d) ，</math >

 = \left[\begin{array} {c | c}

     \mathbf{A}_{11} \circ \mathbf{B} & \mathbf{A}_{12} \circ \mathbf{B} \\


$\displaystyle{ \mathcal F(C^{(1)}x \star C^{(2)}y) =(\mathcal F C^{(1)} \bull \mathcal F C^{(2)})(x \otimes y)= \mathcal F C^{(1)}x \circ \mathcal F C^{(2)}y }$,

1) = (mathcal f ^ {(1)} x star c ^ {(2)} y) = (mathcal f c ^ {(1)} bull mathcal f c ^ {(2)})(xotimes y) = mathf c ^ {(1)} x 圈 f c ^ {(2)} y </math > ，< br/>

     \hline


where $\displaystyle{ \star }$ is vector convolution and $\displaystyle{ \mathcal F }$ is the Fourier transform matrix (this result is an evolving of count sketch properties ),

     \mathbf{A}_{21} \circ \mathbf{B} & \mathbf{A}_{22} \circ \mathbf{B}

   \end{array}\right]


$\displaystyle{ (\mathbf{A} \bull \mathbf{L})(\mathbf{B} \otimes \mathbf{M}) . . . (\mathbf{C} \otimes \mathbf{S})(\mathbf{K} \ast \mathbf{T}) = (\mathbf{A}\mathbf{B}...\mathbf{C}\mathbf{K}) \circ (\mathbf{L}\mathbf{M}...\mathbf{S}\mathbf{T}) }$,

 ={} &\left[\begin{array} {c | c | c | c}


where $\displaystyle{ \ast }$ denotes the Column-wise Khatri–Rao product

        \mathbf{A}_{11} \otimes \mathbf{B}_{11} & \mathbf{A}_{11} \otimes \mathbf{B}_{12} & \mathbf{A}_{12} \otimes \mathbf{B}_{11} & \mathbf{A}_{12} \otimes \mathbf{B}_{12} \\

        \hline


Similarly:

        \mathbf{A}_{11} \otimes \mathbf{B}_{21} & \mathbf{A}_{11} \otimes \mathbf{B}_{22} & \mathbf{A}_{12} \otimes \mathbf{B}_{21} & \mathbf{A}_{12} \otimes \mathbf{B}_{22} \\


$\displaystyle{ (\mathbf{A} \bull \mathbf{L})(\mathbf{B} \otimes \mathbf{M}) . . . (\mathbf{C} \otimes \mathbf{S})(c \otimes d ) = (\mathbf{A}\mathbf{B}...\mathbf{C}c) \circ (\mathbf{L}\mathbf{M}...\mathbf{S}d) }$,

        \hline

        \mathbf{A}_{21} \otimes \mathbf{B}_{11} & \mathbf{A}_{21} \otimes \mathbf{B}_{12} & \mathbf{A}_{22}          \otimes \mathbf{B}_{11} & \mathbf{A}_{22} \otimes \mathbf{B}_{12} \\


$\displaystyle{ (\mathbf{A} \bull \mathbf{L})(\mathbf{B} \otimes \mathbf{M}) . . . (\mathbf{C} \otimes \mathbf{S})(\mathbf{P}c \otimes \mathbf{Q}d ) = (\mathbf{A}\mathbf{B}...\mathbf{C}\mathbf{P}c) \circ (\mathbf{L}\mathbf{M}...\mathbf{S}\mathbf{Q}d) }$, where $\displaystyle{ c }$ and $\displaystyle{ d }$ are vectors

        \hline

        \mathbf{A}_{21} \otimes \mathbf{B}_{21} & \mathbf{A}_{21} \otimes \mathbf{B}_{22} & \mathbf{A}_{22} \otimes \mathbf{B}_{21} & \mathbf{A}_{22} \otimes \mathbf{B}_{22}

 \end{array}\right] \\

 ={} &\left[\begin{array} {c c | c c c c | c | c c}

         1 &  2 &  4 &  7 &  8 & 14 &  3 & 12 & 21 \\

         4 &  5 & 16 & 28 & 20 & 35 &  6 & 24 & 42 \\

         \hline

         2 &  4 &  5 &  8 & 10 & 16 &  6 & 15 & 24 \\

         3 &  6 &  6 &  9 & 12 & 18 &  9 & 18 & 27 \\

         8 & 10 & 20 & 32 & 25 & 40 & 12 & 30 & 48 \\

        12 & 15 & 24 & 36 & 30 & 45 & 18 & 36 & 54 \\

        \hline

         7 &  8 & 28 & 49 & 32 & 56 &  9 & 36 & 63 \\

        \hline

        14 & 16 & 35 & 56 & 40 & 64 & 18 & 45 & 72 \\

        21 & 24 & 42 & 63 & 48 & 72 & 27 & 54 & 81

      \end{array}\right].


\end{align}[/itex]

### Khatri–Rao product 卡特里—拉奥积

• Block Kronecker product
• 块克洛内克积
• Column-wise Khatri–Rao product
• 列式KaTri-RAO乘积

### 分面产品Face-splitting product

Category:Matrix theory

This page was moved from wikipedia:en:Kronecker product. Its edit history can be viewed at 克洛内克积/edithistory

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2. Weisstein, Eric W. "Kronecker Product". mathworld.wolfram.com (in English). Retrieved 2020-09-06.
3. {{Citation 克洛内克积是以德国数学家利奥波德·克洛内克的名字命名的(1823-1891) ，尽管很少有证据表明他是第一个定义和使用它的人。1858年，Johann Georg Zehfuss 描述了这个矩阵运算， 克洛内克积 也被称为 泽弗斯Zehfuss 矩阵，但是克洛内克积是目前使用最广泛的。 | author = G. Zehfuss | year = 1858 | title = Ueber eine gewisse Determinante If A is an matrix and B is a matrix, then the Kronecker product is the block matrix: 如果 A 是一个矩阵，B 是一个矩阵，那么 克洛内克积矩阵就是块矩阵: | journal = Zeitschrift für Mathematik und Physik | volume = 3 [itex]\mathbf{A}\otimes\mathbf{B} = \begin{bmatrix} | pages = 298–301 a_{11} \mathbf{B} & \cdots & a_{1n}\mathbf{B} \\ 11} mathbf { b } & cdots & a _ {1n } mathbf { b } | url = http://resolver.sub.uni-goettingen.de/purl?PPN599415665_0003 \vdots & \ddots & \vdots \\ Vdots & ddots & vdots | postscript = . a_{m1} \mathbf{B} & \cdots & a_{mn} \mathbf{B} 1} mathbf { b } & cdots & a _ { mn } mathbf { b } }}
4. Langville, Amy N.; Stewart, William J. (June 1, 2004). "The Kronecker product and stochastic automata networks". Journal of Computational and Applied Mathematics. 167 (2): 429–447. Bibcode:2004JCoAM.167..429L. doi:10.1016/j.cam.2003.10.010.
5. Langville, Amy N.; Stewart, William J. (June 1, 2004). "The Kronecker product and stochastic automata networks". Journal of Computational and Applied Mathematics. 167 (2): 429–447. Bibcode:2004JCoAM.167..429L. doi:10.1016/j.cam.2003.10.010.
6. MacEdo, Hugo Daniel; Oliveira, José Nuno (2013). "Typing linear algebra: A biproduct-oriented approach". Science of Computer Programming. 78 (11): 2160–2191. arXiv:1312.4818. Bibcode:2013arXiv1312.4818M. CiteSeerX 10.1.1.747.2083. doi:10.1016/j.scico.2012.07.012.
7. MacEdo, Hugo Daniel; Oliveira, José Nuno (2013). "Typing linear algebra: A biproduct-oriented approach". Science of Computer Programming. 78 (11): 2160–2191. arXiv:1312.4818. Bibcode:2013arXiv1312.4818M. CiteSeerX 10.1.1.747.2083. doi:10.1016/j.scico.2012.07.012.
8. {{Cite book 矩阵的 克洛内克积 对应于线性映射的抽象张量积。具体而言，如果“V”、“W”、“X”和“Y”分别具有基{v1, ..., vm}, {w1, ..., wn}, {x1, ..., xd},{y1, ..., ye}, ，如果矩阵“A”和“B”分别表示在适当的基上的线性变换S : VXT : WY;，则矩阵AB 表示二个映射的张量积，VW 上的ST : VWXY关于基{v1w1, v1w2, ..., v2w1, ..., vmwn}模板:Void of VW ；和具有同样性质AB(viwj) = (Avi) ⊗ (Bwj)类似定义的基（的映射） XY 其中，「i」及「j」为适当范围内的整数 This formula is also useful for representing 2D image processing operations in matrix-vector form. 这个公式对于以矩阵向量形式表示二维图像处理操作也是有用的。 | last=Dummit | first=David S. Another example is when a matrix can be factored as a Hadamard product, then matrix multiplication can be performed faster by using the above formula. This can be applied recursively, as done in the radix-2 FFT and the Fast Walsh–Hadamard transform. Splitting a known matrix into the Hadamard product of two smaller matrices is known as the "nearest Kronecker Product" problem, and can be solved exactly by using the SVD. To split a matrix into the Hadamard product of more than two matrices, in an optimal fashion, is a difficult problem and the subject of ongoing research; some authors cast it as a tensor decomposition problem. 另一个例子是当一个矩阵可以被分解为 哈达玛乘积Hadamard product时，使用上面的公式可以更快地执行矩阵乘法。这可以递归地应用，就像在基数-2 FFT 和快速沃尔什-阿达马变换。将一个已知矩阵分解为两个较小矩阵的 Hadamard 乘积被称为“最近克洛内克积”问题，并且可以用奇异值分解来精确求解。将一个矩阵以最优方式分解为两个以上矩阵的 哈达玛积是一个困难的问题，也是正在进行的研究课题; 一些作者将其称为张量分解问题。 | last2=Foote | first2=Richard M. In conjunction with the least squares method, the Kronecker product can be used as an accurate solution to the hand eye calibration problem. 结合最小二乘法，克洛内克积可以用作手眼校准问题的精确解决方案。 | title=Abstract Algebra | edition=2 | year=1999 Two related matrix operations are the Tracy–Singh and Khatri–Rao products, which operate on partitioned matrices. Let the matrix A be partitioned into the blocks Aij and matrix B into the }} blocks Bkl, with of course , , and q = q}}. 两个相关的矩阵操作是 Tracy-Singh 和 Khatri-Rao 产品，它们对分块矩阵进行操作。将矩阵 a 划分为 a < sub > ij 和 b }块 b < sub > kl ，当然，和 q = q }。 | publisher=John Wiley and Sons | location=New York | isbn=978-0-471-36857-1 The Tracy–Singh product is defined as 特雷西-辛格积被定义为 | pages=401–402 }}
9. Van Loan, C; Pitsianis, N (1992). Approximation with Kronecker Products. Ithaca, NY: Cornell University.
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12. King Keung Wu; Yam, Yeung; Meng, Helen; Mesbahi, Mehran (2016). "Kronecker product approximation with multiple factor matrices via the tensor product algorithm". 2016 IEEE International Conference on Systems, Man, and Cybernetics (SMC). pp. 004277–004282. doi:10.1109/SMC.2016.7844903. ISBN 978-1-5090-1897-0.
13. Dantas, Cássio F.; Cohen, Jérémy E.; Gribonval, Rémi (2018). "Learning Fast Dictionaries for Sparse Representations Using Low-Rank Tensor Decompositions". Latent Variable Analysis and Signal Separation. Lecture Notes in Computer Science. 10891. pp. 456–466. doi:10.1007/978-3-319-93764-9_42. ISBN 978-3-319-93763-2.
14. Algo Li, et al. "Simultaneous robot-world and hand-eye calibration using dual-quaternions and Kronecker product." International Journal of the Physical Sciences Vol. 5(10), pp. 1530-1536, 4 September 2010.
15. Algo Li, et al. "Simultaneous robot-world and hand-eye calibration using dual-quaternions and Kronecker product." International Journal of the Physical Sciences Vol. 5(10), pp. 1530-1536, 4 September 2010.