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Let $A$ be a Vandermonde type matrix

$A = \begin{bmatrix} 1 & 1 & \dots & 1 \\ x_1 & x_2 &\dots & x_n \\ \dots& \dots & \dots& \dots\\ x_1^{n-1} &x_2^{n-1} &\dots & x_n^{n-1} \end{bmatrix}$

When I was testing some such real matrices with positive entries for their eigenvalues I have noticed that eigenvalues are also real and positive.

  • How such property (if true in general case) can be explained ?

Moreover in examples which I tested (when it was assumed for $i<j$ that $x_i <x_j$) the greatest eigenvalue was always close to the greatest value in this matrix and the smallest one always less than $1$.

  • How to explain also these facts ?

Examples of $3 \times 3$ matrices generated from natural numbers

$A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 5 \\ 4 & 9 & 25 \end{bmatrix}$

Eigenvalues: $\{ 27.09 , \ 0.12 , \ 1.79 \} $

$A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 7 & 11 \\ 9 & 49 & 121 \end{bmatrix}$

Eigenvalues: $\{ 125.63, \ 0.37 , \ 3.03 \} $

Of course if two eigenvalues in the examples above are rather small then the third must be rather great from the trace of the matrix, but why these two must be small ?

  • If the answer for general case is hard to find let dimension of a matrix be specific i.e. $ 3 \times 3$ and entries only natural.
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    $\begingroup$ Is this helpful? - mathoverflow.net/questions/155845/… $\endgroup$ Mar 8, 2017 at 15:07
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    $\begingroup$ @ancientmathematician Yes, it provides some knowledge although not for everything I mentioned here. It seems from mathoverflow that the problem is rather difficult, I'm afraid that explanation there is rather hard for me. What does it mean "strictly totally positive" for the problem( this meaning of minors) I was thinking that the problem is easier.. $\endgroup$
    – Widawensen
    Mar 8, 2017 at 15:14
  • $\begingroup$ it means all the minors are positive. If one wants to understand the eigenvalues of the Vandermonde matrix, then I think you'd need to grapple with this stuff. (I may be wrong). $\endgroup$ Mar 8, 2017 at 15:20
  • $\begingroup$ @ancientmathematician so the answer lies somewhere in 35-pages paper, hmm ... it can be rather a long way to understanding .. $\endgroup$
    – Widawensen
    Mar 8, 2017 at 15:22
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    $\begingroup$ @ancientmathematician That's true :) Step by step makes a mile.. $\endgroup$
    – Widawensen
    Mar 8, 2017 at 15:32

1 Answer 1

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As such Vandermonde matrix $A$ has positive entries, then by Perron-Frobenius theorem it has positive eigenvalue $\lambda$ with positive eigenvector $v$ (with positive entries). This eigenvalue is the largest by modulus among other eigenvalues.

If you look on the last coordinate of the identity $Av = \lambda v$ you will see, that $$\lambda v_n = x_1^{n-1} v_1 + \dots + x_n^{n-1} v_n > x_n^{n-1} v_n$$ because $v_i$ are positive. Cancelling $v_n$ you get that $\lambda$ is greater than $x_n^{n-1}$.

As it was mentioned earlier in the link by @ancient mathematician, all minors of such matrix are positive and you can apply Perron-Frobenius theorem to exterior powers of original matrix and get, that all other eigenvalues are positive.

Because the largest eigenvalue is so big, other eigenvalues should be small by trace argument.

For bounding the largest eigenvalue from above you can use Gershgorin circles theorem and its strengthening, which gives in the case $x_1=1$, $x_2 = 7$, $x_3 = 11$ that the largest eigenvalue is not greater than 133.

Or you can look at the trace and find that the largest eigenvalue should be not greater than 129.

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