克洛内克积

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For the Kronecker product of representations of symmetric groups, see Kronecker coefficient.

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.

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.

在数学中， 克洛内克积Kronecker product，有时表示为⊗ ，是对任意大小的两个矩阵的运算，得到一个分块矩阵。它是 外积Outer product(用同一个符号表示)从向量到矩阵的推广，并给出了 张量积关于 基Basis的标准选择的矩阵。 克洛内克积将区别于通常的矩阵乘法，这是一个完全不同的操作。 克洛内克积有时也被称为 矩阵方向积。

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.

\end{bmatrix}, </math>

结束{ bmatrix } ，</math >

Definition定义

more explicitly:

更准确地说:

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

如果A是m × n 矩阵，B是p × q 矩阵，那么 克洛内克积A ⊗ B 是pm × qn块矩阵：

[math]\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}, }[/math]

  \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

[math]\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}. }[/math]

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

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

More compactly, we have

更简单地说，我们有

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

[math]\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} \\ }[/math]

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

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

Similarly

同样地

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

[math]\displaystyle{ \end{bmatrix}. }[/math]

(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 }.

</math>

More compactly, we have

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

使用恒等式 [math]\displaystyle{ i\%p = i - \lfloor i/p \rfloor p }[/math] ，其中 [math]\displaystyle{ i \% p }[/math] 表示 [math]\displaystyle{ i/p }[/math] 的余数，这可能以更对称的形式写为

[math]\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}. }[/math]

</math>

Similarly

同样，

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

如果 A 和 B 分别表示线性变换，则表示两个映射的张量积。

[math]\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}. }[/math]

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

使用标识[math]\displaystyle{ i\%p=i-\lfloor i/p\rfloor p }[/math]，其中[math]\displaystyle{ i\%p }[/math]表示[math]\displaystyle{ i/p }[/math]的余数，这可以用更对称的形式来写

 \begin{bmatrix}

开始{ bmatrix }

[math]\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 \\ }[/math]

 \end{bmatrix} \otimes

 \begin{bmatrix}

If A and B represent linear transformations V₁ → W₁ and V₂ → W₂, respectively, then A ⊗ B represents the tensor product of the two maps, V₁ ⊗ V₂ → W₁ ⊗ W₂.

如果“A”和“B”分别代表线性变换V₁ → W₁ 和 V₂ → W₂，则 A ⊗ B表示张量积的两个映射V₁ ⊗ V₂ → W₁ ⊗ W₂。

   0 & 5 \\

   0 & 5 \\

   6 & 7 \\

   6 & 7 \\

Examples示例

 \end{bmatrix} =

结束{ bmatrix } =

[math]\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 \\ }[/math]

 \end{bmatrix} =

Similarly:

类似地:

 \begin{bmatrix}

    0 &  5 &  0 & 10 \\

[math]\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 }[/math]

\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 \\

[math]\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 \\ }[/math]

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:

克洛内克积Kronecker product 是张量积的一个特例，所以它是双线性的和相关联的:

</math>

[math]\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} }[/math]

结束{ align } </math >

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

克洛内克积是张量积的特例，因此是[[双线性算符|双线性]和关联性的：

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

其中 A，B 和 C 是矩阵，0是零矩阵，k 是标量。

[math]\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}. }[/math]

 \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 .

如果 A 和 B 是方阵，甚至是排列相似的，这意味着我们可以。

\end{align}</math>

The matrices and are perfect shuffle matrices. The perfect shuffle matrix S_p,q can be constructed by taking slices of the I_r identity matrix, where [math]\displaystyle{ r=pq }[/math].

这些矩阵是 完全洗牌矩阵Perfect shuffle matrices。 完全洗牌矩阵S_p,q可以通过对 I_r恒等式矩阵进行切片来构造，其中[math]\displaystyle{ r=pq }[/math]。

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

其中，「A」、「B」及「C」为矩阵，「0」为一个零矩阵，「k」为一个标度。

[math]\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} }[/math]

结束{ bmatrix } </math >

| journal = Linear and Multilinear Algebra

| volume = 9

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

这里使用 MATLAB 冒号符号表示子矩阵，I_r 是单位矩阵。如果 [math]\displaystyle{ \mathbf{A} \in \mathbb{R}^{m_1 \times n_1} }[/math] 和 [math]\displaystyle{ \mathbf{B} \in \mathbb{R}^{m_2 \times n_2} }[/math]，那么

| pages = 271–288

| number = 4

[math]\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} }[/math]

| 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>

[math]\displaystyle{ \mathbf{B} \otimes \mathbf{A} = \mathbf{P} \, (\mathbf{A} \otimes \mathbf{B}) \, \mathbf{Q}. }[/math]

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

如果A、B、C和D是一定大小的矩阵，可以形成矩阵积AC和BD，则

If A and B are square matrices, then A ⊗ B and B ⊗ A are even permutation similar, meaning that we can take P = Q^T.

如果“A”和“B”是方阵，则 A ⊗ B 和 B ⊗ A 甚至是排列相似，这意味着我们可以取P = Q^T。

[math]\displaystyle{ (\mathbf{A} \otimes \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{AC}) \otimes (\mathbf{BD}). }[/math]

(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. The perfect shuffle matrix S_p,q can be constructed by taking slices of the I_r identity matrix, where [math]\displaystyle{ r=pq }[/math].

}}</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[math]\displaystyle{ r=pq }[/math]。

|4= Hadamard product (element-wise multiplication):

| 4 = Hadamard 乘积(元素级乘法) :

[math]\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}). }[/math]

但是如果你想要一个更好的数学家，那么你需要一个更好的数学家。数学

 \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}</math>

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

因此，当且仅当A和B都是可逆的，则其为可逆，在这种情况下，逆由以下给出

[math]\displaystyle{ (\mathbf{A} \otimes \mathbf{B})^{-1} = \mathbf{A}^{-1} \otimes \mathbf{B}^{-1}. }[/math]

< 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 I_r is the r × r identity matrix. If [math]\displaystyle{ \mathbf{A} \in \mathbb{R}^{m_1 \times n_1} }[/math] and [math]\displaystyle{ \mathbf{B} \in \mathbb{R}^{m_2 \times n_2} }[/math], then

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

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

穆尔－彭罗斯伪逆Moore–Penrose pseudoinverse的逆积性质也成立，即

[math]\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} }[/math]

[math]\displaystyle{ (\mathbf{A} \otimes \mathbf{B})^{+} = \mathbf{A}^{+} \otimes \mathbf{B}^{+}. }[/math]

< 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 Mat_F 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 I_n, and the tensor product is given by the Kronecker product.

在范畴论的语言中， 克洛内克积（和更一般的张量积）的 混合积Mixed-product性质表明，域F上矩阵的范畴Mat_F实际上是一个幺半范畴，对象自然数为n，映射为n×m矩阵，其项在F中，合成由矩阵乘法给出，单位箭头是单位矩阵I_n，张量积由 克洛内克积给出。

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

如果“‘A’、‘B’、‘C’、‘D’”等矩阵的大小足以形成矩阵积s’‘AC’、‘BD’”，则

Mat_F is a concrete skeleton category for the equivalent category FinVect_F 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.

Mat_{F╱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 基中 张量积的表示。}

[math]\displaystyle{ (\mathbf{A} \otimes \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{AC}) \otimes (\mathbf{BD}). }[/math]

|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:

转置和 共轭转置Conjugate transposition可分布于 克洛内克积上：

As immediate consequence,

因此，

[math]\displaystyle{ (\mathbf{A}\otimes \mathbf{B})^\textsf{T} = \mathbf{A}^\textsf{T} \otimes \mathbf{B}^\textsf{T} }[/math] and [math]\displaystyle{ (\mathbf{A}\otimes \mathbf{B})^* = \mathbf{A}^* \otimes \mathbf{B}^*. }[/math]

[math]\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}) }[/math].

|7= Determinant:

7 = 行列式:

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

尤其是，使用以下的“转置”属性，这意味着如果

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

设A为矩阵，B为矩阵。然后

[math]\displaystyle{ \mathbf{A} = \mathbf{Q} \otimes \mathbf{U} }[/math]

[math]\displaystyle{ \left| \mathbf{A} \otimes \mathbf{B} \right| = \left| \mathbf{A} \right| ^m \left| \mathbf{B} \right| ^n . }[/math]

右 | = 左 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 数学 >

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.

其中的指数是B的顺序，而 中的指数是A的顺序。

|8= Kronecker sum and exponentiation:

|8 = 克洛内克和与幂:

|4= Hadamard product (element-wise multiplication):模板:Paragraph

|4「（阿达玛积（矩阵）」（元素相乘）：」模板:段

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

如果A是，B是，而I_{k表示单位矩阵，则我们可以定义有时被称为的 克洛内克和Kronecker sum，⊕，由}

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

混合积性质Mixed-product property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

[math]\displaystyle{ \mathbf{A} \oplus \mathbf{B} = \mathbf{A} \otimes \mathbf{I}_m + \mathbf{I}_n \otimes \mathbf{B} . }[/math]

[math]\displaystyle{ (\mathbf{A} \otimes \mathbf{B}) \circ (\mathbf{C} \otimes \mathbf{D}) = (\mathbf{A} \circ \mathbf{C}) \otimes (\mathbf{B} \circ \mathbf{D}). }[/math]

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

这不同于两个矩阵的 直和。这个运算与 李代数Lie algebras上的张量积有关。

|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.

我们有以下矩阵指数的公式，这在一些数值计算中很有用。

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

由此可见， A ⊗ B为逆当且仅当'A'和'B'均为可逆，在这种情况下， 逆由以下给出

[math]\displaystyle{ (\mathbf{A} \otimes \mathbf{B})^{-1} = \mathbf{A}^{-1} \otimes \mathbf{B}^{-1}. }[/math]

[math]\displaystyle{ \exp({\mathbf{N} \oplus \mathbf{M}}) = \exp(\mathbf{N}) \otimes \exp(\mathbf{M}) }[/math]

如果你想知道更多的信息，请访问我们的网站: http://www.math.com/

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

可逆转积属性也适用于穆尔–彭罗斯伪逆， ，为

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

当考虑非相互作用系统的整体时， 克洛内克和Kronecker sums自然地出现在物理学中。设 Hⁱ为该系统的 哈密顿量。那么这个整体的 总哈密顿量是

[math]\displaystyle{ (\mathbf{A} \otimes \mathbf{B})^{+} = \mathbf{A}^{+} \otimes \mathbf{B}^{+}. }[/math]

[math]\displaystyle{ H_{\mathrm{Tot}}=\bigoplus_{i}H^{i} }[/math].

[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 Mat_F of matrices over a field F, is in fact a monoidal category, with objects natural numbers n, morphisms n → m are n-by-m matrices with entries in F, composition is given by matrix multiplication, identity arrows are simply n × n identity matrices I_n, and the tensor product is given by the Kronecker product.

在[[范畴理论]的语言中，Kronecker积（及更一般的张量积）的混合积性质表明，[[域（数学）[域]]上的矩阵的类别「Mat」、「sub」、「F」、「sub」、「sub」、「F」，实际上是一个 单项类型，其对象为自然数「n」，同态映射 n → m 是以「F」为输入项的「n」乘「m」矩阵，组合是以矩阵相乘的方式给出，单位箭头 是简单的n × n 标准矩阵 I_n，张量积由 克洛内克积Kronecker product给出。

Mat_F is a concrete skeleton category for the equivalent category FinVect_F of finite dimensional vector spaces over F, whose objects are such finite dimensional vector spaces V, arrows are F-linear maps L : V → W, 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 : V → W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:

|6= Transpose:模板:Paragraph

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

假设 A 和 B 分别是大小为 n 和 m 的方阵。设 A 的特征值为 λ < sub > 1 ，... ，λ < sub > n ，μ < sub > 1 ，... ，μ < sub > m 为 B 的特征值(按重数列出) 。则特征值是

Transposition and conjugate transposition are distributive over the Kronecker product:

转置和共轭转置分布于 克洛内克积Kronecker product：

[math]\displaystyle{ \lambda_i \mu_j, \qquad i=1,\ldots,n ,\, j=1,\ldots,m. }[/math]

1，ldots，n，，j = 1，ldots，m.数学

[math]\displaystyle{ (\mathbf{A}\otimes \mathbf{B})^\textsf{T} = \mathbf{A}^\textsf{T} \otimes \mathbf{B}^\textsf{T} }[/math] and [math]\displaystyle{ (\mathbf{A}\otimes \mathbf{B})^* = \mathbf{A}^* \otimes \mathbf{B}^*. }[/math]

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

因此， 克洛内克积的迹和行列式由以下给出

|7= Determinant:模板:Paragraph

[math]\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. }[/math]

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

设“A”为一个n × n 矩阵，“B”为一个 m × m矩阵。然后

|2= Singular values:

|2 = 单数值:

[math]\displaystyle{ \left| \mathbf{A} \otimes \mathbf{B} \right| = \left| \mathbf{A} \right| ^m \left| \mathbf{B} \right| ^n . }[/math]

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 r_A nonzero singular values, namely

如果A和B是矩形矩阵，则可以考虑其奇异值。假设A有 r_A个非零的奇数值，即

[math]\displaystyle{ \sigma_{\mathbf{A},i}, \qquad i = 1, \ldots, r_\mathbf{A}. }[/math]

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

类似地，用以表示 B 的非零奇数值

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

如果“A”是{nowrap{“n”x“n”}，则“B”是{nowrap |“m”x“m”和“I”表示{nowrap |“k”x“k”}[[身份矩阵]，则我们可以通过

[math]\displaystyle{ \sigma_{\mathbf{B},i}, \qquad i = 1, \ldots, r_\mathbf{B}. }[/math]

1，ldots，r _ mathbf { b }.数学

[math]\displaystyle{ \mathbf{A} \oplus \mathbf{B} = \mathbf{A} \otimes \mathbf{I}_m + \mathbf{I}_n \otimes \mathbf{B} . }[/math]

Then the Kronecker product has r_Ar_B nonzero singular values, namely

然后，克洛内克积有 r_Ar_B 非零奇异值，即

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

[math]\displaystyle{ \sigma_{\mathbf{A},i} \sigma_{\mathbf{B},j}, \qquad i=1,\ldots,r_\mathbf{A} ,\, j=1,\ldots,r_\mathbf{B}. }[/math]

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.

|4= Relation to products of graphs:

| 4 = 与图的乘积的关系:

[math]\displaystyle{ \exp({\mathbf{N} \oplus \mathbf{M}}) = \exp(\mathbf{N}) \otimes \exp(\mathbf{M}) }[/math]

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.

图的 邻接矩阵的 克洛内克积Kronecker product是张量积图的 邻接矩阵。图的邻接矩阵的 克洛内克和是邻接矩阵图的 笛卡儿积。

}}

}}

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

[math]\displaystyle{ H_{\mathrm{Tot}}=\bigoplus_{i}H^{i} }[/math].

}}

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.

克罗内克积可用来方便表示某些矩阵方程式。例如，考虑该方程式，其中A、B和C是给定的矩阵，而矩阵X是未知的。

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

我们可以使用“ 向量 技巧”来重写这个等式

Abstract properties 理论性质

[math]\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]

. math

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

假设「A」及「B」分别为大小为「n」及「m」的方阵。设λ₁, ..., λ_n 是 A 的特征值而 μ₁, ..., μ_m 是B（根据[[多重性（数学）|多重性]列出）的。则A ⊗ B 的[[特征值]为

[math]\displaystyle{ \lambda_i \mu_j, \qquad i=1,\ldots,n ,\, j=1,\ldots,m. }[/math]

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

在这里，vec (x)表示矩阵 x 的向量化，它是将 x 的列堆叠到一个列向量中形成的。

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

因此，一个Kronecker积的[[迹（矩阵）|迹]和行列式由以下给出

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 .

现在从克洛内克积的性质可以推出，方程有唯一的解，当且仅当 A 和 B 是非奇异的。

[math]\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. }[/math]

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

如果X和AXB分别按行顺序排列到列向量u和v中，则

|2= Singular values:模板:Paragraph

[math]\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}. }[/math]

.</math>

. math

The reason is that

原因是

Similarly, denote the nonzero singular values of B by

同样地，将非零的奇数值「「B」」表示为

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

\mathbf{v} =

2009年10月11日

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

Then the Kronecker product A ⊗ B has r_Ar_B nonzero singular values, namely

那么，克洛内克积 A ⊗ B 具有r_Ar_B 非零单数值，即

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

[math]\displaystyle{ \sigma_{\mathbf{A},i} \sigma_{\mathbf{B},j}, \qquad i=1,\ldots,r_\mathbf{A} ,\, j=1,\ldots,r_\mathbf{B}. }[/math]

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

左(mathbf { a }/times mathbf { b } ^ textsf { t }右)操作符名{ vec }(mathbf { x ^ textsf { t }) =

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

左(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

由于矩阵的阶数等于非零奇数值的个数，我们发现

.</math>

. math

[math]\displaystyle{ \operatorname{rank}(\mathbf{A} \otimes \mathbf{B}) = \operatorname{rank} \mathbf{A} \, \operatorname{rank} \mathbf{B}. }[/math]

|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.

关于这个公式应用的一个例子，见关于李亚普诺夫方程的文章。

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

这个公式也可以用来说明矩阵正态分布是多变量正态分布的一个特例。

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, ..., v_m}, {w₁, ..., w_n}, {x₁, ..., x_d}, and {y₁, ..., y_e}, respectively, and if the matrices A and B represent the linear transformations S : V → X and T : W → Y, respectively in the appropriate bases, then the matrix A ⊗ B represents the tensor product of the two maps, S ⊗ T : V ⊗ W → X ⊗ Y with respect to the basis {v₁ ⊗ w₁, v₁ ⊗ w₂, ..., v₂ ⊗ w₁, ..., v_m ⊗ w_n}模板:Void of V ⊗ W and the similarly defined basis of X ⊗ Y with the property that A ⊗ B(v_i ⊗ w_j) = (Av_i) ⊗ (Bw_j), where i and j are integers in the proper range.

[math]\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} }[/math]

如果你是一个数学家，那么你就是一个数学家，而不是数学家

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

当‘v’和‘w’是[[李代数] ] s时，并且 S : V → V 和 T : W → W是李代数同态s，“A”和“B”的 克罗内克和Kroncker sum表示 李代数同态因子Induced Lie algebra homomorphismsV ⊗ W → V ⊗ W。

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.

也就是说，积的第（ij）th子块是矩阵，其中第（k）th子块等于矩阵。本质上，特雷西–Singh积是二个矩阵中每对分区的配对Kronecker积。

这意味着乘积的(ij)-th 子块是矩阵，其中(k)-th 子块等于 }矩阵 }。本质上 Tracy-Singh 积就是两个矩阵中每一对分区的成对克洛内克积。

|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.:

例如，如果 A 和 B 都是分块矩阵。:

[math]\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[ . }[/math]

\begin{array} {c c | c}

开始{ array }{ 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}

结束{数组}

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

\right]

[右]

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

\begin{array} {c | c}

开始{ array }{ c | c }

The reason is that

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

11} & mathbf { b }{12}

[math]\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[ . }[/math]

\begin{array} {c | c c}

开始{ array }{ 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.

有关此公式应用的示例，请参阅有关Lyapunov方程式的文章。

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.

这个公式也很方便地说明[[矩阵正态分布]是多元正态分布的一个特例。

3 & 6 & 9

3 & 6 & 9

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

此公式亦适用于以矩阵形式表示二维图像处理操作。

\end{array}

结束{数组}

\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 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一哈达玛积时，使用上述公式可以更快地进行矩阵相乘。这可以被递推应用，如[[库勒–Tukey FFT算法|基数-2dit情况|基数-2fft]和[[快速沃尔什–阿达玛变换]所做。将一个已知矩阵分解为二个较小矩阵的Hadamard积称为“最近kroncker积”问题，该问题可以使用 SVD精确地解决 。将一个矩阵以最佳方式拆分为二个以上矩阵的阿达玛积是一个困难的问题，也是正在研究的课题；一些作者将其转化为张量分解问题。 ,

,

</math>

数学

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

与[[最小平方|最小平方法]相结合，Kronecker积可作为[[手眼校准问题]的精确解决方案。

we get:

我们得到:

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

[math]\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} }[/math]

   \end{array}\right]

结束{数组}右]

 ={} &\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 [math]\displaystyle{ \circ }[/math] B is the m_i p × n_j q matrix A_ij [math]\displaystyle{ \circ }[/math] B, of which the (k模板:Ell)-th subblock equals the m_i p_k × n_j q_模板:Ell matrix A_ij ⊗ B_k模板:Ell. Essentially the Tracy–Singh product is the pairwise Kronecker product for each pair of partitions in the two matrices.

这意味着{{nowrap{“mp’‘mp’’‘nq’”}}积的第 (ij)个子块A [math]\displaystyle{ \circ }[/math] B是m_i p × n_j q 矩阵的A_ij [math]\displaystyle{ \circ }[/math] B，其中第(k模板:Ell)-th ）子块等于m_i p_k × n_j q_模板:Ell 矩阵 A_ij &otimes；B_k模板:Ell。本质上， 特雷西-辛格积Tracy–Singh product是两个矩阵中每对分区的成对 克洛内克积Kronecker 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

        \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} \\

[math]\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} }[/math]

结束{ align } </math >

\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

混合产品属性 < br/>

1 & 4 & 7 \\

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

\hline

2 & 5 & 8 \\

[math]\displaystyle{ (\mathbf{A} \bull \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{A}\mathbf{C}) \bull (\mathbf{B} \mathbf{D}) }[/math],

3 & 6 & 9

\end{array}

Similarly:

类似地: < br/>

\right]

[math]\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}) }[/math],

(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 > ,

,

</math>

[math]\displaystyle{ c^\textsf{T} \bull d^\textsf{T} = c^\textsf{T} \otimes d^\textsf{T} }[/math], where [math]\displaystyle{ c }[/math] and [math]\displaystyle{ d }[/math] are vectors,

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

we get: 我们得到：

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

其中[math]\displaystyle{ c }[/math]和[math]\displaystyle{ d }[/math] 是向量，[math]\displaystyle{ \circ }[/math] 表示 哈达玛乘积Hadamard product

[math]\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), }[/math]

(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} \\

[math]\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 }[/math],

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

其中 [math]\displaystyle{ \star }[/math] 是向量卷积，而 [math]\displaystyle{ \mathcal F }[/math] 是傅里叶变换矩阵(这个结果是计数素描属性的演化) ,

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

   \end{array}\right]

[math]\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}) }[/math],

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

where [math]\displaystyle{ \ast }[/math] denotes the Column-wise Khatri–Rao product

这里 [math]\displaystyle{ \ast }[/math] 表示纵列的 卡特里—拉奥乘积Katri-Rao product< br/>

        \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:

类似地: < br/>

        \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} \\

[math]\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) }[/math],

        \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} \\

[math]\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) }[/math], where [math]\displaystyle{ c }[/math] and [math]\displaystyle{ d }[/math] are vectors

其中 [math]\displaystyle{ c }[/math]和 [math]\displaystyle{ d }[/math] 是向量

        \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}</math>

Khatri–Rao product 卡特里—拉奥积

Main article: Khatri–Rao product

• 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

范畴: 矩阵理论

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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