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Is there any relationship that simplifies the Cayley transform

$g(X) = \left(X - \mathrm{i}I_N\right)^{-1}\left(X + \mathrm{i}I_N\right)$,

where $X \in \mathbb{C}^{N\times N}$ and $I_N$ denotes the identity matrix, applied to the Kronecker sum

$X = A \oplus B = A \otimes I_M + I_M \otimes B$,

where $N = M^2$, $\otimes$ denotes the Kronecker product and $A, B \in \mathbb{C}^{M\times M}$?

I guess that a decomposition in two matrices $C \otimes D$ is possible, but I cannot prove it. Additionally, the relationship may have certain constraints of course.

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  • $\begingroup$ @Cosmas Zachos I am fairly sure that X acts on $N=M^2$ due to the Kronecker products within the Kronecker sum. $\endgroup$ – carlosvalderrama Jun 25 '18 at 16:54
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$$g(X) = \left(X - \mathrm{i}I_N\right)^{-1}\left(X + \mathrm{i}I_N\right)\\= I_N -2i \left(X - \mathrm{i}I_N\right)^{-1},$$ which invites you to examine the geometric series involved, i.e., arbitrary powers of X, all completely symmetrized in A and B; the cross terms, of course, do not vanish.

For example, $$ X^2= A^2\otimes I_M+I_M\otimes B^2+ 2 ~A\otimes B , $$ etc.

I strongly doubt you could recast the Cayley transform to a simple tensor product. Consider selective eigenvalues of the respective factors in the coproduct.

As per request, I don't have anything elegant. Consider M=2 , A and B both diagonalizable. So, without loss of generality, in their respective subspaces, they are (a,b) meaning diag$(a,b)$ and (c,d), respectively: the diagonalizing similarity transformations in the subspaces tensor-factorize, so X is also diagonal, $$ X=(a+c, a+d,b+c,b+d). $$ I have utilized the "right-factor matrix into each element of large left-factor matrix" convention for the tensor products.

Consequently, $$ g(X)= \left (\frac{a+c +i}{ a+c -i},\frac{a+d +i}{ a+d -i},\frac{b+c +i}{ b+c -i}, \frac{b+d +i}{ b+d -i}\right ), $$ which does not look like a simple function of A and B. To be sure, as per above, only powers of the denominator matter, and maybe there are Newton identities to the rescue, but there is no compact answer I can see.

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  • $\begingroup$ The power series approach is an interesting approach, but indeed it seems that I cannot make any factorization. My initial guess is that a factorization is only possible for certain eigenvalues. Your last comment sounds promising, however, I am afraid I don't follow. Could you please elaborate on selective eigenvalues and the coproduct? $\endgroup$ – carlosvalderrama Jun 25 '18 at 20:42
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos Jun 26 '18 at 12:36

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