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We are given vectors $\mathbf a_1,\dots,\mathbf a_N\in\mathbb C^n$. Let these vectors be columns of $n$-by-N matrix $\mathbf A$. We additionally know that any $n-1$ vectors are linearly independent. This makes possible to do the following: we consider all $n-1$ tuples $\mathbf a_{i_1},\dots,\mathbf a_{i_{n-1}}$ (there are $N$ choose $n-1$ of them) and for very $n-1$-tuple we take the vector (it is unique up to scaling factor) which is orthogonal to $\mathbf a_{i_1},\dots,\mathbf a_{i_{n-1}}$.

Let the new vectors form the matrix $\mathbf B$. Then $\mathbf B$ is $n$-by-$N$ choose $n-1$ matrix. (We also assume that the columns of $\mathbf B$ are ordered somehow)

The question is: is there exist some special name and the notation for the matrix $\mathbf B$? (some one told me that this operation relates somehow to the Hodge star dual)

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1 Answer 1

$B = \Lambda^{n-1}A$.

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To Sasha: The even(or odd) rows of $\Lambda^{n-1}\mathbf A$ must be multiplied by $-1$ also. But what is the name for $\mathbf B$? ($\Lambda^{n-1}\mathbf A$ is called $n-1$-th compound of $\mathbf A$ and is denoted by $\mathcal C_{n-1}(\mathbf A)$) –  don Bass Sep 21 '12 at 13:41
    
It is called $(n-1)$-st wedge power of $A$. –  Sasha Sep 21 '12 at 14:10
    
$(n−1)$-th wedge power of $\mathbf A$ = $(n−1)$-th exterior power of $\mathbf A$ = $(n−1)$-th compound of $\mathbf A$, but to get $B$ we must multiply from the left by antidiagonal matrix with 1,-1,1,-1,... on the antidiagonal. –  don Bass Sep 21 '12 at 14:14
    
The signs depend on your identification of $V = C^n$ with $\Lambda^{n-1}V$. –  Sasha Sep 21 '12 at 14:29

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