I am not sure if I can remove the Kronecker operator in the following formula $$\mathrm{trace((\Sigma^{-1}\otimes S^{-1})A}),$$ where $\Sigma,S$ are positive-semidefinite and symmetric, and $A$ is symmetric. Any help would be appreciated.


@greg claimed in the answer below that we can "exactly represent $A$ as a finite sum of Kronecker products." I am wondering how to obtain this exact representation. To be more specific, I have $$A=\left(\begin{array}{cc} 0 & 0\\ 0 & R \end{array}\right), $$ where $R$ is a symmetric and positive semi-definite matrix, and its dimension is $k \times k$, $k=1,\dots,\mathrm{ncol}(A)$. The ultimate goal is to calculate the derivative $$\frac{\partial \mathrm{trace((\Sigma^{-1}\otimes S^{-1})A})}{\partial \Sigma}.$$ This is related to question: Derivative involving the trace of a Kronecker product. I did read Van Loan and Pitsianis (1993), but did not find a solution.

  • $\begingroup$ I'm just beginning to read about Tensor analysis on Manifolds. Do you have any tips? $\endgroup$ – jm324354 Feb 5 '15 at 22:14
  • $\begingroup$ @bd1251252, sorry I only know Kronecker operator. $\endgroup$ – Randel Feb 6 '15 at 17:10
  • $\begingroup$ Found the solution in Section 6 here. $\endgroup$ – Randel Jan 7 '17 at 22:49

You can write the function in terms of the Frobenius product as $$ f = A^T:(\Sigma^{-1}\otimes S^{-1}) $$ If you had a Kronecker Product factorization for $A=B\otimes C$, where $B,C$ are the same dimensions as $\Sigma,S$ respectively. Then you could re-write the function to eliminate the Kronecker product as follows $$ \eqalign { f &= (B^T\otimes C^T):(\Sigma^{-1}\otimes S^{-1}) \cr &= (B^T:\Sigma^{-1})\,(C^T:S^{-1}) \cr &= {\rm tr}(B\Sigma^{-1})\,\,{\rm tr}(CS^{-1}) \cr }$$ If you don't have an exact Kronecker factorization, a search for "Kronecker Product approximation" will turn up lots of software.

Further, you can exactly represent $A$ as a finite sum of Kronecker products $$\eqalign{ A &= \sum^r_{k=1} B_k\otimes C_k \cr f &= \sum^r_{k=1} {\rm tr}(B_k\Sigma^{-1})\,\,{\rm tr}(C_kS^{-1}) \cr }$$ But I'm not sure it's worth all of this effort simply to avoid a single Kronecker product.

  • $\begingroup$ Thanks a lot for this thorough answer! I found the question can be greatly simplified by letting $D$ be a scalar and $Z$ be a vector. $\endgroup$ – Randel Feb 6 '15 at 17:09

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