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Let $A$ and $B$ two square positive definite matrices of order $k$. Is there any relation between $\operatorname{trace}(AB)$ and $\operatorname{trace}(ABAB)$ for generic $k$? So far I have found that $$\operatorname{trace}(ABAB)=\operatorname{trace}(AB)^{2}-2\left|A\right|\left|B\right|,$$ when $k=2$.

Adam Higgins
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Bounded
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  • Thanks very much for both replies. I am trying to find a way to compute $trace(ABAB)$ avoiding to compute $ABAB$ or even $AB$ as the matrices involved are kind of big, i.e., $n>1000$. I already have the eigenvalues of $A$, so maybe I can go through implicit SVD as pointed here?https://math.stackexchange.com/questions/67231/singular-value-decomposition-of-product-of-matrices – Bounded May 20 '19 at 07:52

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Let $X=AB$, then for a $2\times 2$ matrix we have the relationship that you deduced $$2\,\det(X) = {\rm Tr}(X)^2 - {\rm Tr}(X^2) $$ Systematically, such relationships can be derived by applying the Newton's Identities to the characteristic polynomial.

Relationships for $n\times n$ matrices involve complicated products of $${\rm Tr}(X^n),\,{\rm Tr}(X^{n-1}),\,{\rm Tr}(X^{n-2}),\,\ldots,\,{\rm Tr}(X^{1}),\,\det(X)$$ with large coefficients which grow like $n!$

For example, here are the relationships for $3\times 3$ and $4\times 4$ matrices. $$\eqalign{ 6\,\det(X) &= {\rm Tr}(X)^3 - 3\,{\rm Tr}(X)\,{\rm Tr}(X^2) + 2\,{\rm Tr}(X^3) \cr 24\,\det(X) &= {\rm Tr}(X)^4- 6\,{\rm Tr}(X)^2\,{\rm Tr}(X^2) + 3\,{\rm Tr}(X^2)^2 + 8\,{\rm Tr}(X){\rm Tr}(X^3) - 6\,{\rm Tr}(X^4) \cr }$$

greg
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