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Suppose one has the inverse $A^{-1}$ of an $N\times N$ non-singular matrix $A$. Is there an ''efficient'' way to obtain $\det{A}$?

With ''efficient'' I mean anything that has a better scaling than the standard $\mathcal{O}(N^3)$. Naively, one can argue that $\det{A}$ is already implicitly incorporated in the inverse via the adjugate $A^{-1}=\displaystyle\frac{1}{\det{A}}\operatorname{adj}A$, so all the work has "already been done", and one only needs an efficient trick to distill the determinant.

E.g.: for $N=2$, we have $A=\left(\begin{array}{cc} a & b \\ c & d\end{array}\right)$ and $A^{-1}=\displaystyle\frac{1}{\det{A}}\left(\begin{array}{cc} d & -b \\ -c & a\end{array}\right)$. The determinant can efficiently be obtained from $\displaystyle\frac{A_{11}}{[A^{-1}]_{22}}=\det{A}$ in $\mathcal{O}(1)$.

Max Wong
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cortogantese
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  • It suffices to calculate one entry of $\operatorname{adj}(A)$, and yet I'm still not confident that this can be done in less than $O(N^3)$... – Ben Grossmann Jun 07 '17 at 14:02
  • @omnomnomnom Thanks. Well, I think indeed that each entry of $\textrm{adj}A$ has the same scaling as $\det{A}$, so this is not the "trick" I am interested in. – cortogantese Jun 07 '17 at 14:55
  • I am not an expert, but maybe you can use $A^{-1}$ to more efficiently compute some decomposition of $A$ as LU- or QR-decomposition or maybe you already have such a decomposition or you get it as a byproduct from wherever you got your inverse. The determinant $\det(A)$ is then just the product of the determinants of the factors and these might be very easy to compute. But I guess you already considered this. But out of curiosity: where do you got your inverse from? – M. Winter Jun 07 '17 at 15:28
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    If $A$ has the characteristic polynomial $p(x) = x^n + a_{n-1} x^{n-1} + \cdots + a_0$, then $A^{-1}$ has characteristic polynomial $q$ where $q(x) \det(A) = a_0 x^n + a_1 x^{n-1} + \cdots + 1$. So if the "middle" coefficients (as in $a_{n/2}$) of characteristic polynomials were "easy" to compute, this would give you the determinant. For example, doing this in the $3 \times 3$ case gives that $\det(A) = (tr(A)^2 - tr(A^2)) / tr(A^{-1})$. (Since $tr(A^2)$ is $O(n^2)$ this is "efficient"). However I do not know how to find these coefficients more efficiently than a determinant... – Joppy Jun 07 '17 at 16:17
  • @M. Winter: Thanks. My inverse was found constructively. I am working a lot with a set of matrices $X_{ij}$ ($i=1\dots N$) which fulfill so-called Yang-Baxter-Gaudin relations, e.g. $X_{ij}X_{jk}+X_{ki}X_{ij}+X_{ki}X_{jk}=0,\forall ijk$ and $X_{ij}+X_{ji}=0,\forall ij$. This example is solved by the Cauchy matrices, so, my question was pure academic. I had constructed the inverse of the Cauchy matrices from the YBG relations (and not so much from the typical Lagrange interpolation). The determinant can be constructed similarly, but the two constructions are notrelated. That struck me. – cortogantese Jun 08 '17 at 07:48
  • @Joppy. Thanks for the suggestion. I didn't know about the characteristic polynomial of the inverse. If one considers $p(1)$ and $q(1)$, then we have $\det(A)=p(1)/q(1)$ , and there is no need to calculate the coefficients $a_n$ explicitly. I am not sure yet if $p(1)$ is efficiently evaluated. – cortogantese Jun 08 '17 at 07:52
  • Why not calculate the determinant of $A^{-1}$ and then inverting it? Wikipedia lists fast matrix multiplication as algorithm with better complexity. – mvw Oct 17 '20 at 18:51

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This is strictly harder than a problem we already don't know how to solve: quickly taking the determinant of an orthogonal matrix.

If $Q$ is an orthogonal matrix, then $Q Q^{\mathsf T} = I$. In particular, if you're given $Q$, you are also automatically given its inverse. The determinant $\det(Q)$ is guaranteed to be $\pm 1$; however, determining which one it is appears to be just as hard as taking any other determinant.

If we could, in general, efficiently compute $\det(A)$ given both $A$ and $A^{-1}$, then in particular, for an orthogonal matrix $Q$, we could efficiently compute $\det(Q)$.

Misha Lavrov
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