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In this question, they mentioned that $m\times n$ Vandermonde matrices have the property that any submatrix with $n$ rows has full rank.

Does that mean if the submatrix is $n\times r$, where $r\leq n$, then the submatrix has full column rank?

Lai Ye
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1 Answers1

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The determinant of any such submatrix formed by the rows $k_1,k_2,\dots k_n$ is $$ \prod\limits_{1\le j<i\le n}(x_{k_i}-x_{k_j}) $$ with $x_{k_i}\neq x_{k_j}$ for $i\neq j$, which is obviously non-zero.

GReyes
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  • A a confusion here, I thought they meant that any $n\times r$ submatrix has full column rank ($r\leq n$), does this still hold? I will add this to the question. – Lai Ye Nov 12 '23 at 00:15
  • Oh, I think the submatrix does have full column rank since all its columns are independent... – Lai Ye Nov 12 '23 at 00:24