# Sum of the squares of the minors of a matrix with orthonormal column vectors = 1?

Let $A$ be an $m \times n$ ($n \leq m$) matrix with real entries and orthonormal column vectors.

Claim: For $1 < k \leq n$, the sum of the squares of the $k\times k$ minors of $A$ is always $1$.

I have written some Matlab code and tried numerous cases and it seems to me that this claim is true. On the other hand, I could not find a reference for it nor be able to prove it. So, if you have a reference or an idea for a proof, I'd appreciate it immensely.

This result implies that the norm of the wedge product of $1< k \leq n$ orthonormal vectors in $\mathbb R^n$ is equal to $1$, case $k = n$ is trivial though.

Thanks.

EDIT: As indicated by the first response below this claim is obviously not true as stated. The claim should have stated instead that it holds for the sum of the squares of the $n \times n$ minors of any $m \times n$ matrix with orthonormal column vectors. The size of the minors should always match the number of vectors being wedged. Actually this gives a pretty good idea about a proof.

• This doesn't sound right. Consider $A=I_3$. The sum of squares of all 2x2 minors is 3. – user1551 Jan 4 '14 at 1:25
• True, and Claim corrected. I think The corrected claim holds since the wedge of any k orthonormal vectors in $\mathbb R^m$ has norm that equals the volume of the parallelepiped supported by the k vectors which is obviously 1. Writing down the wedge product in terms of the standard basis $e_i_1 \wedge ... \wedge e_i_k$ yields that the sum of squares of the $k \times k$ minors of the $m \times k$ matrix formed by the orthonormal vectors as columns equals 1. Right? – user2502771 Jan 5 '14 at 16:55