tensor product and matrix multiplication distributive properties

Question

I am trying to find partial trace of some matrix of the form

$M = (A \otimes B)\times (A^{T*} \otimes B^{T*})$

in which $\otimes$ is tensor product, $\times$ is matrix multiplication, $T*$ is conjugate transpose of the matrix. $A$ and $B$ are vectors (but can be matrices too).

I was wondering if I could use the following relation?

$(A \otimes B)\times (A^{T*} \otimes B^{T*}) = (A \times A^{T*})\otimes (B \times B^{T*})$

It checks out for $A=\left(\begin{matrix} a\\b \end{matrix}\right)$ and $B=\left(\begin{matrix} c\\d \end{matrix}\right)$.

if this relation is general, what about this one (for an arbitrary vectors (or matrices) C and D)

$(A \otimes B)\times (C \otimes D) = (A \times C)\otimes (B \times D)$

although I don't know how to trace this matrix. I would appreciate any help. thank you in advance

Answer 1

Consider matrices $A,B,C,D$ of sizes such that the products $AC$ and $AD$ can be formed. We can use block matrix multiplication to show that $(A\otimes B)\,(C\otimes D)=(AC)\otimes(BD)$.

We will use the notation $A\otimes B = (a_{ij} B)_{ij}$ to denote block matrices, where indices are always supposed to range approriately. Then \begin{align*} (A\otimes B)\,(C\otimes D) &= (a_{ij} B)_{ij}\, (c_{ij} D)_{ij} \\ &= \left(\sum_k (a_{ik} B)(c_{kj} D)\right)_{ij} \\ &= \left( \left(\sum_k a_{ik} c_{kj}\right) BD\right)_{ij.} \end{align*} Note that $\sum_k a_{ik} c_{kj}$ is the $i,j$-th entry of $AC$ so the result is indeed equal to $(AC)\otimes (BD)$.

Since traces of Kronecker products are given as $\operatorname{Tr}(A\otimes B)=\operatorname{Tr}(A) \operatorname{Tr}(B)$, this yields $$ \operatorname{Tr}\left((A\otimes B)\,(C\otimes D)\right) = \operatorname{Tr}(AC) \operatorname{Tr}(BD). $$ In your case that gives $$ \operatorname{Tr}\left((A\otimes B)\,(\overline{A}^T\otimes \overline{B}^T)\right) = \operatorname{Tr}(A\overline{A}^T) \operatorname{Tr}(B\overline{B}^T) = \|A\|_F^2\, \|B\|_F^2, $$ where $\|\cdot\|_F$ denotes the Frobenius norm.

Answer 2

The equality in the last part of your question is true. One can prove it easier if we look at a matrix as a linear map and look at a matrix product as a composition of linear maps. Furthermore we consider the equality $$T\otimes S(v\otimes w)=T(v)\otimes S(w)$$ which is an obvious definition of tensor product of two linear maps.

So your equality becomes

$$(A \otimes B)\circ (C \otimes D) = (A \circ C)\otimes (B \circ D)$$

We can easilly evalute both side of the equation on a typical simple tensor $x\otimes y$.

So the validity of the equality under discussion is a result of the concept of tensor product of vector spaces and tensor product of linear maps.

Remark: I think perhaps a proof of this equality in its matrix setting is not short or straithforward but we should keep in mind that the following fact is necessary for our argument. Some books give it as a theorem and some other books present it as an exercize:

Fact: If matrix $A$ is matrix representatio of a linear map $T$ and $B$ is a matrix representation of a linear map $S$ then the tensor product matrix $A\otimes B$ is the matrix representation of $T\otimes S$.