弗罗比尼乌斯内积

给定两个n×m维复矩阵 A和B：

${\displaystyle \mathbf {A} ={\begin{pmatrix}A_{11}&A_{12}&\cdots &A_{1m}\\A_{21}&A_{22}&\cdots &A_{2m}\\\vdots &\vdots &\ddots &\vdots \\A_{n1}&A_{n2}&\cdots &A_{nm}\\\end{pmatrix}}\,,\quad \mathbf {B} ={\begin{pmatrix}B_{11}&B_{12}&\cdots &B_{1m}\\B_{21}&B_{22}&\cdots &B_{2m}\\\vdots &\vdots &\ddots &\vdots \\B_{n1}&B_{n2}&\cdots &B_{nm}\\\end{pmatrix}}}$ 

弗罗比尼乌斯内积定义为如下的矩阵元素求和

${\displaystyle \langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }=\sum _{i,j}{\overline {A_{ij}}}B_{ij}=\mathrm {tr} \left({\overline {\mathbf {A} ^{T}}}\mathbf {B} \right)}$ 

其中上划线表示复数和复矩阵的共轭操作。若将定义详细写出，则有

${\displaystyle {\begin{aligned}\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }=&{\overline {A}}_{11}B_{11}+{\overline {A}}_{12}B_{12}+\cdots +{\overline {A}}_{1m}B_{1m}\\&+{\overline {A}}_{21}B_{21}+{\overline {A}}_{22}B_{22}+\cdots +{\overline {A}}_{2m}B_{2m}\\&\vdots \\&+{\overline {A}}_{n1}B_{n1}+{\overline {A}}_{n2}B_{n2}+\cdots +{\overline {A}}_{nm}B_{nm}\\\end{aligned}}}$ 

此计算与点积十分相似，所以是一个内积的范例。

性质

弗罗比尼乌斯内积是半双线性形式。给定复矩阵A, B, C, D, 以及复数a和b，我们有

${\displaystyle \langle a\mathbf {A} ,b\mathbf {B} \rangle _{\mathrm {F} }={\overline {a}}b\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }}$ 

${\displaystyle \langle \mathbf {A} +\mathbf {C} ,\mathbf {B} +\mathbf {D} \rangle _{\mathrm {F} }=\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }+\langle \mathbf {A} ,\mathbf {D} \rangle _{\mathrm {F} }+\langle \mathbf {C} ,\mathbf {B} \rangle _{\mathrm {F} }+\langle \mathbf {C} ,\mathbf {D} \rangle _{\mathrm {F} }}$ 

并且，交换复矩阵的次序所得到的是原来结果的共轭矩阵：

${\displaystyle \langle \mathbf {B} ,\mathbf {A} \rangle _{\mathrm {F} }={\overline {\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }}}}$ 

对于相同的矩阵，有

${\displaystyle \langle \mathbf {A} ,\mathbf {A} \rangle _{\mathrm {F} }\geq 0\,.}$ 

实矩阵

给定实矩阵：

${\displaystyle \mathbf {A} ={\begin{pmatrix}2&0&6\\1&-1&2\end{pmatrix}}\,,\quad \mathbf {B} ={\begin{pmatrix}8&-3&2\\4&1&-5\end{pmatrix}}}$ 

则：

${\displaystyle {\begin{aligned}\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }&=2\cdot 8+0\cdot (-3)+6\cdot 2+1\cdot 4+(-1)\cdot 1+2\cdot (-5)\\&=16+12+4-1-10\\&=21\end{aligned}}}$ 

复矩阵

给定复矩阵

${\displaystyle \mathbf {A} ={\begin{pmatrix}1+i&-2i\\3&-5\end{pmatrix}}\,,\quad \mathbf {B} ={\begin{pmatrix}-2&3i\\4-3i&6\end{pmatrix}}}$ 

那么它们的共轭 (非转置) 矩阵为

${\displaystyle {\overline {\mathbf {A} }}={\begin{pmatrix}1-i&+2i\\3&-5\end{pmatrix}}\,,\quad {\overline {\mathbf {B} }}={\begin{pmatrix}-2&-3i\\4+3i&6\end{pmatrix}}}$ 

因此，

${\displaystyle {\begin{aligned}\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }&=(1-i)\cdot (-2)+(+2i)\cdot 3i+3\cdot (4-3i)+(-5)\cdot 6\\&=(-2+2i)+-6+12-9i+-30\\&=-26-7i\end{aligned}}}$ 

但注意

${\displaystyle {\begin{aligned}\langle \mathbf {B} ,\mathbf {A} \rangle _{\mathrm {F} }&=(-2)\cdot (1+i)+(-3i)\cdot (-2i)+(4+3i)\cdot 3+6\cdot (-5)\\&=-26+7i\end{aligned}}}$ 

A、B与其本身的弗罗比尼乌斯内积分别为

${\displaystyle \langle \mathbf {A} ,\mathbf {A} \rangle _{\mathrm {F} }=2+4+9+25=40}$ 

${\displaystyle \langle \mathbf {B} ,\mathbf {B} \rangle _{\mathrm {F} }=4+9+25+36=74}$ 

从弗罗比尼乌斯内积我们可以诱导出弗罗比尼乌斯范数

${\displaystyle \|\mathbf {A} \|_{\mathrm {F} }={\sqrt {\langle \mathbf {A} ,\mathbf {A} \rangle _{\mathrm {F} }}}\,.}$ 

• 阿达玛乘积 (矩阵)
• 希尔伯特-施密特算子
• 克罗内克积
• 矩阵分析（英语：Matrix analysis）
• 矩阵乘法
• 矩阵范数