弗罗比尼乌斯内积

在数学中，弗罗比尼乌斯内积（英語：Frobenius inner product）是一种基于两个矩阵的二元运算，结果是一个数值。它常常被记为 ⟨ A , B ⟩ F {\displaystyle \langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }} 。这个运算是一個將矩陣視為向量的逐元素内积。参与运算的两个矩阵必须有相同的维度、行数和列数，但不局限于方阵。 给定两个n×m维複矩阵 A和B： A = ( A 11 A 12 ⋯ A 1 m A 21 A 22 ⋯ A 2 m ⋮ ⋮ ⋱ ⋮ A n 1 A n 2 ⋯ A n m ) , B = ( B 11 B 12 ⋯ B 1 m B 21 B 22 ⋯ B 2 m ⋮ ⋮ ⋱ ⋮ B n 1 B n 2 ⋯ B n m ) {\displaystyle \mathbf {A} ={\begin{pmatrix}A_{11}&A_{12}&\cdots &A_{1m}\\A_{21}&A_{22}&\cdots &A_{2m}\\\vdots &\vdots &\ddots &\vdots \\A_{n1}&A_{n2}&\cdots &A_{nm}\\\end{pmatrix}}\,,\quad \mathbf {B} ={\begin{pmatrix}B_{11}&B_{12}&\cdots &B_{1m}\\B_{21}&B_{22}&\cdots &B_{2m}\\\vdots &\vdots &\ddots &\vdots \\B_{n1}&B_{n2}&\cdots &B_{nm}\\\end{pmatrix}}} 弗罗比尼乌斯内积定义为如下的矩阵元素求和 ⟨ A , B ⟩ F = ∑ i , j A i j ¯ B i j = t r ( A T ¯ B ) {\displaystyle \langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }=\sum _{i,j}{\overline {A_{ij}}}B_{ij}=\mathrm {tr} \left({\overline {\mathbf {A} ^{T}}}\mathbf {B} \right)} 其中上划线表示复数和複矩阵的共轭操作。若將定義詳細寫出，則有 ⟨ A , B ⟩ F = A ¯ 11 B 11 + A ¯ 12 B 12 + ⋯ + A ¯ 1 m B 1 m + A ¯ 21 B 21 + A ¯ 22 B 22 + ⋯ + A ¯ 2 m B 2 m ⋮ + A ¯ n 1 B n 1 + A ¯ n 2 B n 2 + ⋯ + A ¯ n m B n m {\displaystyle {\begin{aligned}\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }=&{\overline {A}}_{11}B_{11}+{\overline {A}}_{12}B_{12}+\cdots +{\overline {A}}_{1m}B_{1m}\\&+{\overline {A}}_{21}B_{21}+{\overline {A}}_{22}B_{22}+\cdots +{\overline {A}}_{2m}B_{2m}\\&\vdots \\&+{\overline {A}}_{n1}B_{n1}+{\overline {A}}_{n2}B_{n2}+\cdots +{\overline {A}}_{nm}B_{nm}\\\end{aligned}}} 此計算與點積十分相似，所以是一個內積的範例。 弗罗比尼乌斯内积是半双线性形式。给定複矩阵A, B, C, D, 以及复数a和b，我们有 ⟨ a A , b B ⟩ F = a ¯ b ⟨ A , B ⟩ F {\displaystyle \langle a\mathbf {A} ,b\mathbf {B} \rangle _{\mathrm {F} }={\overline {a}}b\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }} ⟨ A + C , B + D ⟩ F = ⟨ A , B ⟩ F + ⟨ A , D ⟩ F + ⟨ C , B ⟩ F + ⟨ C , D ⟩ F {\displaystyle \langle \mathbf {A} +\mathbf {C} ,\mathbf {B} +\mathbf {D} \rangle _{\mathrm {F} }=\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }+\langle \mathbf {A} ,\mathbf {D} \rangle _{\mathrm {F} }+\langle \mathbf {C} ,\mathbf {B} \rangle _{\mathrm {F} }+\langle \mathbf {C} ,\mathbf {D} \rangle _{\mathrm {F} }} 并且，交换複矩阵的次序所得到的是原来结果的共轭矩阵： ⟨ B , A ⟩ F = ⟨ A , B ⟩ F ¯ {\displaystyle \langle \mathbf {B} ,\mathbf {A} \rangle _{\mathrm {F} }={\overline {\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }}}} 对于相同的矩阵，有 ⟨ A , A ⟩ F ≥ 0 . {\displaystyle \langle \mathbf {A} ,\mathbf {A} \rangle _{\mathrm {F} }\geq 0\,.} 给定实矩阵： A = ( 2 0 6 1 − 1 2 ) , B = ( 8 − 3 2 4 1 − 5 ) {\displaystyle \mathbf {A} ={\begin{pmatrix}2&0&6\\1&-1&2\end{pmatrix}}\,,\quad \mathbf {B} ={\begin{pmatrix}8&-3&2\\4&1&-5\end{pmatrix}}} 则： ⟨ A , B ⟩ F = 2 ⋅ 8 + 0 ⋅ ( − 3 ) + 6 ⋅ 2 + 1 ⋅ 4 + ( − 1 ) ⋅ 1 + 2 ⋅ ( − 5 ) = 16 + 12 + 4 − 1 − 10 = 21 {\displaystyle {\begin{aligned}\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }&=2\cdot 8+0\cdot (-3)+6\cdot 2+1\cdot 4+(-1)\cdot 1+2\cdot (-5)\\&=16+12+4-1-10\\&=21\end{aligned}}} 给定复矩阵 A = ( 1 + i − 2 i 3 − 5 ) , B = ( − 2 3 i 4 − 3 i 6 ) {\displaystyle \mathbf {A} ={\begin{pmatrix}1+i&-2i\\3&-5\end{pmatrix}}\,,\quad \mathbf {B} ={\begin{pmatrix}-2&3i\\4-3i&6\end{pmatrix}}} 那么它们的共轭 (非转置) 矩阵为 A ¯ = ( 1 − i + 2 i 3 − 5 ) , B ¯ = ( − 2 − 3 i 4 + 3 i 6 ) {\displaystyle {\overline {\mathbf {A} }}={\begin{pmatrix}1-i&+2i\\3&-5\end{pmatrix}}\,,\quad {\overline {\mathbf {B} }}={\begin{pmatrix}-2&-3i\\4+3i&6\end{pmatrix}}} 因此， ⟨ A , B ⟩ F = ( 1 − i ) ⋅ ( − 2 ) + ( + 2 i ) ⋅ 3 i + 3 ⋅ ( 4 − 3 i ) + ( − 5 ) ⋅ 6 = ( − 2 + 2 i ) + − 6 + 12 − 9 i + − 30 = − 26 − 7 i {\displaystyle {\begin{aligned}\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }&=(1-i)\cdot (-2)+(+2i)\cdot 3i+3\cdot (4-3i)+(-5)\cdot 6\\&=(-2+2i)+-6+12-9i+-30\\&=-26-7i\end{aligned}}} 但注意 ⟨ B , A ⟩ F = ( − 2 ) ⋅ ( 1 + i ) + ( − 3 i ) ⋅ ( − 2 i ) + ( 4 + 3 i ) ⋅ 3 + 6 ⋅ ( − 5 ) = − 26 + 7 i {\displaystyle {\begin{aligned}\langle \mathbf {B} ,\mathbf {A} \rangle _{\mathrm {F} }&=(-2)\cdot (1+i)+(-3i)\cdot (-2i)+(4+3i)\cdot 3+6\cdot (-5)\\&=-26+7i\end{aligned}}} A、B与其本身的弗罗比尼乌斯内积分别为 ⟨ A , A ⟩ F = 2 + 4 + 9 + 25 = 40 {\displaystyle \langle \mathbf {A} ,\mathbf {A} \rangle _{\mathrm {F} }=2+4+9+25=40} ⟨ B , B ⟩ F = 4 + 9 + 25 + 36 = 74 {\displaystyle \langle \mathbf {B} ,\mathbf {B} \rangle _{\mathrm {F} }=4+9+25+36=74} 从弗罗比尼乌斯内积我们可以诱导出弗罗比尼乌斯范数 ‖ A ‖ F = ⟨ A , A ⟩ F . {\displaystyle \|\mathbf {A} \|_{\mathrm {F} }={\sqrt {\langle \mathbf {A} ,\mathbf {A} \rangle _{\mathrm {F} }}}\,.} 阿達瑪乘積 (矩陣) 希尔伯特-施密特算子 克罗内克积 矩陣分析（英语：Matrix analysis） 矩陣乘法 矩陣範數