2x2 subdeterminants of a matrix - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:37:35Z http://mathoverflow.net/feeds/question/119066 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119066/2x2-subdeterminants-of-a-matrix 2x2 subdeterminants of a matrix Carlo Mantegazza 2013-01-16T13:05:10Z 2013-01-21T18:22:35Z <p>If N>2, it is well known that if two invertible NxN matrices A and B have the same determinants of any 2x2 corresponding submatrices, then A=B or A=-B.<br> Given then all these 2x2 determinants of an invertible matrix A, is there an "explicit" way to recover/write down A?<br> If N=3 it is easy, as you can get the determinant of A (up to the sign) and all its cofactors, so you can obtain the inverse matrix of A or -A, but when N>3?</p> http://mathoverflow.net/questions/119066/2x2-subdeterminants-of-a-matrix/119071#119071 Answer by Nick Gill for 2x2 subdeterminants of a matrix Nick Gill 2013-01-16T13:30:52Z 2013-01-16T14:23:39Z <p>By "corresponding submatrices" I presume you mean those $2\times2$ minors obtained by deleting $n-2$ colums and $n-2$ rows, where these columns and rows have the same $n-2$ indices. Once you've calculated the determinants of these submatrices you recover the action of $A$ on the exterior square $\Lambda^2 V$.</p> <p>Now the following paper: "An algorithm for recognising the exterior square of a matrix" by Catherine Greenhill explains how to then obtain the original matrix $A$. Here's the relevant quote:</p> <blockquote> <p>One computational problem which presents itself immediately is this: how can we determine whether a given matrix $Y$ is equal to the exterior square of another matrix $X$? In particular, if such an $X$ exists then we would like to construct one. A polynomial-time algorithm which solves this problem is described in Section 5.</p> </blockquote> <p>The paper can be downloaded <a href="http://web.maths.unsw.edu.au/~csg/papers/ext-matrix.pdf" rel="nofollow">here</a>.</p> <p>One needs to be slightly careful here, because the exterior square does not quite determine the matrix $X$ uniquely. Here is another quote from the paper:</p> <blockquote> <p>We prove in Section 4 that two matrices $X$, $X'$ with rank at least three have the same exterior square if and only <code>$X'\in \{X, -X\}$</code>.</p> </blockquote> <p>So if the rank is at least three (which it is, since you are assuming invertibility), then we are pretty much done. I'm guessing that the situation where the rank is $\leq 2$ would be easy enough to resolve but in any case that's outside the scope of the question...</p>