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Both Vandermonde and Cauchy matrices with $n$ rows and $k$ ($n \geq k$) columns have the property that any $k$ rows are linearly independent (assuming the coefficient are independent). It seems to me that if you concatenates the rows of the identity matrix $I_k$ then the $k$-linear independence is preserved in the resulting ($k+n$ x $k$) matrix.

I want to use this claim in a paper and I want to know if there is already a formal work that I can cite to back up my (supposed true) claim. Otherwise I will do it in the annex section and post it here too.

thanks

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At least in the Vandermonde case, you are (implicitly) assuming that the nodes are all distinct, yes? Otherwise you have (obvious) linear dependence... –  J. M. May 2 '12 at 18:47
    
yes I make this implicit assumption. –  UmNyobe May 2 '12 at 18:49
    
If by "add up" you mean that you want to concatenate the matrices into a single matrix, changing the wording might make the statement a little clearer. –  rschwieb May 2 '12 at 19:20
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1 Answer 1

It is only true for Cauchy matrices, not true for general cases with Vandermonde ones.

The condition of k-linear independence in the resulting (k+n x k) matrix is: every square sub-matrix of (n x k) matrix (that is either Vandermonde or Cauchy in your question) MUST be invertible. It is easily to prove this by considering m columns of the identity and (k-m) of the (n x k) matrix, these k columns should be independent.

Hence, in general, all matrices whose every square sub-matrix is invertible fulfill your question.

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