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Let $k,n$ be positive integers, $k\le n$. Let $v_1,\cdots,v_k$ be vectors in $\mathbb{R}^n$. Let $M$ be the $k\times n$ matrix with rows $v_1,\cdots,v_k$ in this order. The Gram determinant of $M$ is defined as the determinant of the $k\times k$ matrix $MM^*$.

For each subset $\sigma$ of $S=\{1,\cdots,n\}$ of $n-k$ elements, let $x_{\sigma}$ be the determinant of the submatrix of $M$ by deleting the $i$-th columns for $i\in \sigma$.

Can anyone show me where I can find the proof that the Gram determinant of $M$ is equal to the sum of $x_{\sigma}^2$ over all subsets $\sigma$ of $S$ of $n-k$ elements?

Note that for $k=2$ the assertion is the Lagrange identity.

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up vote 1 down vote accepted

I think this is Caucy-Binet formula whose proof can be found in:

http://www.lacim.uqam.ca/~lauve/courses/su2005-550/BS3.pdf

Or on page 9 of F.R.Gantmacher, Theory of Matrices.

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