# Recover local systems from composite systems

Define $$A,B$$ as two linear operator of two local systems. Define $$C:= A \otimes B$$ as the composite systems. How to recover $$A$$ and $$B$$ given $$C$$? For example, we set \begin{align} A=\left[\begin{matrix} 1/2&1/4 \\ 1/4&1/2 \end{matrix} \right], \end{align} \begin{align} B=\left[\begin{matrix} 1/2&0 \\ 0&1/2 \end{matrix} \right]. \end{align} It is easily shown that \begin{align} A \otimes B=\left[\begin{matrix} 1/4&0&1/8&0 \\ 0&1/4&0&1/8 \\ 1/8&0&1/4&0 \\ 0&1/8&0&1/4 \end{matrix} \right]. \end{align} My question is how to recover $$A$$ and $$B$$ given the result of $$A \otimes B$$. The above content is an example. I would like to ask whether tensor product is invertible.

I think what you're looking for is the partial trace. Generally speaking, given two finite-dimensional Hilbert spaces $$\mathcal H_1$$ and $$\mathcal H_2$$, if $$\text{L}(\mathcal H_1)$$ and $$\text{L}(\mathcal H_2)$$ represent the spaces of linear operators on these two respective spaces, and $$A\in \text{L}(\mathcal H_1)$$ is a fixed linear operator, then we have a function $$A\otimes - ~ : \text{L}(\mathcal H_2) \to \text{L}(\mathcal H_1 \otimes \mathcal H_2)$$ which is the "take the tensor product with $$A$$" function, sending any operator $$B\in\text{L}(\mathcal H_2)$$ to the operator $$A\otimes B$$ on the space $$\mathcal H_1\otimes\mathcal H_2$$. The partial trace function $$\text{tr}_1 ~ : ~ \text{L}(\mathcal H_1 \otimes \mathcal H_2) \to \text{L}(\mathcal H_2)$$ sometimes referred to as "tracing out by the first subsystem", is a left-inverse to $$A\otimes -$$ for any fixed $$A$$ with trace equal to $$1$$. More generally, it satisfies $$\text{tr}_1(A\otimes B) = \text{tr}(A)B$$ so that when $$A$$ has trace equal to $$1$$, this just equals $$B$$, as desired.
However, keep in mind that $$A\otimes -$$, the "tensor with $$A$$" function, does not have a right-inverse generally. That is, there is no function $$f:\text{L}(\mathcal H_1 \otimes \mathcal H_2) \to \text{L}(\mathcal H_2)$$ such that $$A\otimes f(C) = C$$ for any given operator $$C$$ on $$\mathcal H_1\otimes \mathcal H_2$$, because some such operators are not equal to the tensor product of any pair of operators on the individual spaces. For instance $$C = \begin{bmatrix}1/3 & 0 & 0 & 0 \\ 0 & 1/3 & 0 & 0 \\ 0 & 0 & 1/3 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$$ is not equal to $$A\otimes B$$ for any operators $$A,B$$ on $$\mathcal H_1$$ and $$\mathcal H_2$$. But given a matrix $$C$$ that is known to be a tensor product, taking partial traces with respect to each of the two subsystems will allow you to recover the two halves of the tensor product.
Do you know one of either A or B? I suppose in the situation the calculation would be simple enough, but I'm not sure about otherwise. Without that info, you could look at the four quadrants of the tensor product matrix and say that you know B is some multiple of $$\begin{pmatrix}0.25&0\\0&0.25\end{pmatrix}$$ then get values for A based on that (i.e. the 00 entry of A is the number that will give you the first quadrant of B), but this will still give you some uncertainty up to a coefficient.