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I seek for the following relationship (if there is one so):

$$C \otimes D = (A_1 \otimes B_1) + (A_2 \otimes B_2)$$

I would like to obtain $C = f(A_1,A_2)$ (in terms of $A$'s) and $D = g(B_1,B_2)$ (in terms of $B$'s). For simplicity, we can assume $A_i$ and $B_i$ are covariance matrices, so positive-definite, square, and symmetric.

Any help is greatly appreciated!

PS: For more simplification (if so), we can assume $\dim(A_1) = \dim(A_2)$ and same for $B_i$'s.

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Suppose $A=A_1\otimes B_1+ A_2\otimes B_2=C\otimes D$.

Let $F(A)=Id\otimes tr(\cdot)(A)=Ctr(D)=A_1tr(B_1)+A_2tr(B_2)$ and

$G(A)=tr(\cdot)\otimes Id(A)=Dtr(C)=B_1tr(A_1)+B_2tr(A_2)$.

Notice that $tr(A)\neq 0$, because $0\neq A$ is a positive semidefinite symmetric matrix. Observe that $tr(A)=tr(C)tr(D)$.

So $A=C\otimes D= \frac{1}{tr(C)tr(D)}G(A)\otimes F(A)=\frac{1}{tr(A)}G(A)\otimes F(A)$.

Thus, if $tr(A)\neq 0$ then $A=C\otimes D$ iff $A=\frac{1}{tr(A)}G(A)\otimes F(A)$.

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If $$ C \otimes D = (A_1 \otimes B_1) + (A_2 \otimes B_2), $$ then $A_1$ and $A_2$ are linearly dependent and so are $B_1$ and $B_2$. If $A_1$ and $B_1$ are nonzero, then $$ C \otimes D = \lambda(A_1 \otimes B_1) $$ for some $\lambda\in\mathbb R$ and so $C=\mu A_1$ and $D=\frac\lambda\mu B_1$ for any $\mu\neq0$. You cannot determine $C$ and $D$ uniquely from $C\otimes D$ since you always have the freedom to scale one up and the other one down.

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  • $\begingroup$ thanks! Actually I have $(A_1\otimes B_1) + (A_2 \otimes B_2)$ and want to come up with such a representation. Your answer starts with this, so if there is such a representation, then they are linearly dependent (right?). But if we don't know but want to come up with such a representation, I think we can not assume linear dependence. Is that right? so what are the conditions of existence? $\endgroup$ – user48547 Jul 4 '15 at 11:15
  • $\begingroup$ @oeda, If there is such a representation for $C\otimes D$, then we have linear dependence. And if the $A$s and $B$s are linearly dependent, we can always find $C$ and $D$. If you don't want the sum of the tensor products to be a tensor product, it suffices to assume that $A$s and $B$s are linearly independent. $\endgroup$ – Joonas Ilmavirta Jul 4 '15 at 11:18
  • $\begingroup$ Thank you. Matrices are square symmetric and pos. def., and of same size (A and B's). But I can't directly say anything about their dependence. $\endgroup$ – user48547 Jul 4 '15 at 11:29

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