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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.
Given then all these 2x2 determinants of an invertible matrix A, is there an "explicit" way to recover/write down A?
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?

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Something is wrong, as $A$ and $-A$ have the same $2\times2$ minors. –  Denis Serre Jan 16 '13 at 15:44
    
You're right, I should have written A=B or A=-B. The same for the last sentence, you can get the determinant of A up to the sign. Sorry for the inaccuracy. –  Carlo Mantegazza Jan 16 '13 at 23:24
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up vote 20 down vote accepted

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 $\Lambda^2 V$.

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:

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.

The paper can be downloaded here.

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:

We prove in Section 4 that two matrices $X$, $X'$ with rank at least three have the same exterior square if and only $X'\in \{X, -X\}$.

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...

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Thanks, very interesting reference. Just a comment on my question: it comes from determining explicitely the unique (up to the sign) second fundamental form $A$ of a hypersurface of dimension $N>3$ in $R^{N+1}$. As $Riem=A*A$ (where $*$ is the Kulkarni-Nomizu product of two symmetric 2-forms) all the 2x2 determinants of the matrix associated to $A$ in an orthonormal basis are given by the sectional curvatures, hence they are all "intrinsic", so also $A$, up to the sign. –  Carlo Mantegazza Jan 16 '13 at 23:44
    
Sorry $N\geq3$ here above. –  Carlo Mantegazza Jan 17 '13 at 4:22
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