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In mathematics, the Frobenius inner product is a binary operation that takes two matrices and returns a number. It is often denoted $\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} )$ . The operation is a component-wise inner product of two matrices as though they are vectors. The two matrices must have the same dimension - same number of rows and columns, but are not restricted to be square matrices.

Definition

Given two complex number-valued n×m matrices A and B, written explicitly as

$\mathbf {A} ={\begin{pmatrix}A_{11}&A_{12}&\cdots &A_{1m}\\A_{21}&A_{22}&\cdots &A_{2m}\\\vdots &\vdots &\ddots &\vdots \\A_{n1}&A_{n2}&\cdots &A_{nm}\\\end{pmatrix))\,,\quad \mathbf {B} ={\begin{pmatrix}B_{11}&B_{12}&\cdots &B_{1m}\\B_{21}&B_{22}&\cdots &B_{2m}\\\vdots &\vdots &\ddots &\vdots \\B_{n1}&B_{n2}&\cdots &B_{nm}\\\end{pmatrix))$ the Frobenius inner product is defined as,

$\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }=\sum _{i,j}{\overline {A_{ij))}B_{ij}\,=\mathrm {Tr} \left({\overline {\mathbf {A} ^{T))}\mathbf {B} \right)\equiv \mathrm {Tr} \left(\mathbf {A} ^{\!\dagger }\mathbf {B} \right)$ where the overline denotes the complex conjugate, and $\dagger$ denotes Hermitian conjugate. Explicitly this sum is

{\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)) The calculation is very similar to the dot product, which in turn is an example of an inner product.

Relation to other products

If A and B are each real-valued matrices, the Frobenius inner product is the sum of the entries of the Hadamard product. If the matrices are vectorised (i.e., converted into column vectors, denoted by "$\mathrm {vec} (\cdot )$ "), then

$\mathrm {vec} (\mathbf {A} )={\begin{pmatrix}A_{11}\\A_{12}\\\vdots \\A_{21}\\A_{22}\\\vdots \\A_{nm}\end{pmatrix)),\quad \mathrm {vec} (\mathbf {B} )={\begin{pmatrix}B_{11}\\B_{12}\\\vdots \\B_{21}\\B_{22}\\\vdots \\B_{nm}\end{pmatrix))\,,$ $\quad {\overline {\mathrm {vec} (\mathbf {A} )))^{T}\mathrm {vec} (\mathbf {B} )={\begin{pmatrix}{\overline {A))_{11}&{\overline {A))_{12}&\cdots &{\overline {A))_{21}&{\overline {A))_{22}&\cdots &{\overline {A))_{nm}\end{pmatrix)){\begin{pmatrix}B_{11}\\B_{12}\\\vdots \\B_{21}\\B_{22}\\\vdots \\B_{nm}\end{pmatrix))$ Therefore

$\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }={\overline {\mathrm {vec} (\mathbf {A} )))^{T}\mathrm {vec} (\mathbf {B} )\,.$ Properties

It is a sesquilinear form, for four complex-valued matrices A, B, C, D, and two complex numbers a and b:

$\langle a\mathbf {A} ,b\mathbf {B} \rangle _{\mathrm {F} }={\overline {a))b\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} )$ $\langle \mathbf {A} +\mathbf {C} ,\mathbf {B} +\mathbf {D} \rangle _{\mathrm {F} }=\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} }+\langle \mathbf {A} ,\mathbf {D} \rangle _{\mathrm {F} }+\langle \mathbf {C} ,\mathbf {B} \rangle _{\mathrm {F} }+\langle \mathbf {C} ,\mathbf {D} \rangle _{\mathrm {F} )$ Also, exchanging the matrices amounts to complex conjugation:

$\langle \mathbf {B} ,\mathbf {A} \rangle _{\mathrm {F} }={\overline {\langle \mathbf {A} ,\mathbf {B} \rangle _{\mathrm {F} )))$ For the same matrix,

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

The inner product induces the Frobenius norm

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

Real-valued matrices

For two real-valued matrices, if

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

{\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)\\&=21\end{aligned)) Complex-valued matrices

For two complex-valued matrices, if

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

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

{\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)) The Frobenius inner products of A with itself, and B with itself, are respectively

$\langle \mathbf {A} ,\mathbf {A} \rangle _{\mathrm {F} }=2+4+9+25=40$ $\qquad \langle \mathbf {B} ,\mathbf {B} \rangle _{\mathrm {F} }=4+9+25+36=74$ 