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Let $V(x_1,..,x_n)$ be the Vandermonde matrix induced by $x_1,..,x_n$, and let $\tilde{V} := V(\frac{x_1}{h},...,\frac{x_n}{h})$. My intuition says that the condition number should be invariant under such scaling of the nodes at least for some special cases of node configurations. My question is then:

  1. Are there constraints on the nodes $x_1,..,x_n$ (i.e symmetric, positive, equally spaced or others) for which ${\cal K}(\tilde{V})= {\cal K}(V)$, where ${\cal K}(V)$ is the condition number for a matrix $V$?
  2. Are there known lower and upper bounds which relate the singular values of $V$ and $\tilde{V}$ to one another?
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    $\begingroup$ For any configuration of nodes, as $h\to \infty$, $\tilde{V}$ tends to a singular matrix while remaining bounded in norm, so $\mathcal K(\tilde{V})\to \infty$. $\endgroup$ – Mike Jury Nov 12 '15 at 15:44
  • $\begingroup$ This comment pretty much sums it up. If you can repost it as an answer, I can accept and close. Thanks. $\endgroup$ – gil Nov 18 '15 at 12:45
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For any configuration of nodes, as $h\to \infty$, $\tilde{V}$ tends to a singular matrix while remaining bounded in norm, so $\mathcal K(\tilde{V})\to \infty$.

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Straight from the definition of Vandermonde matrices we have

$$\tilde{V}D = V,$$

where $D = diag(1,h,h^2\ldots,h^{n-1})$. But the 2-norm condition number of a matrix is not invariant under scaling of the columns. Thus, $\mathcal{K}(\tilde{V})\neq\mathcal{K}(V)$.

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  • $\begingroup$ Clearly the singular values are not invariant under scaling but this does not mean the ratio between the highest and the lowest values is not kept. see clarification in the question. $\endgroup$ – gil Sep 6 '15 at 7:52

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