Timeline for Jacobi's equality between complementary minors of inverse matrices
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 18, 2017 at 5:42 | comment | added | darij grinberg | I suspect that your $Q$ is not "defined similarly" but rather is the inverse of the matrix defined similarly. But yes, this is essentially how it's done. | |
| Feb 17, 2017 at 15:05 | comment | added | Albert Gu | I believe the justification is not too bad. Assume the result is true for I=K, J=K, K=[k]. Then for any other I,J of cardinality k, we have $\det(A[I,J])=\det((PAQ)[K,K])=\det(PAQ)\det(Q^{-1}A^{-1}P^{-1}[K^c,K^c]) = \det(PAQ)\det(A^{-1}[J^c,I^c])$, where $P$ is the permutation matrix taking $I$ to $1,\dots,k$ and $I^c$ to $k+1,\dots,n$ (I believe you call this permutation $w(I)$), and $Q$ is defined similarly. So we incur a factor of $\text{sign}(P)\text{sign}(Q)$ which is $(-1)^{\sum I + \sum J}$ by a counting argument, as you proved. | |
| Feb 17, 2017 at 1:19 | comment | added | darij grinberg | Really nice... although a nontrivial part of my answer is justifying the "WLOG" in the first sentence. Still, this simplifies a lot. | |
| Feb 16, 2017 at 22:43 | review | Late answers | |||
| Feb 16, 2017 at 22:48 | |||||
| Feb 16, 2017 at 22:28 | review | First posts | |||
| Feb 16, 2017 at 22:42 | |||||
| Feb 16, 2017 at 22:26 | history | answered | Albert Gu | CC BY-SA 3.0 |