# Patterns for eigenvalues of Vandermonde matrix

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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• Is this helpful? - mathoverflow.net/questions/155845/… – ancientmathematician Mar 8 '17 at 15:07
• @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.. – Widawensen Mar 8 '17 at 15:14
• 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). – ancientmathematician Mar 8 '17 at 15:20
• @ancientmathematician so the answer lies somewhere in 35-pages paper, hmm ... it can be rather a long way to understanding .. – Widawensen Mar 8 '17 at 15:22
• @ancientmathematician That's true :) Step by step makes a mile.. – Widawensen Mar 8 '17 at 15:32

## 1 Answer

If you assume $x_{i}<x_{j}$ for $i<j$, it will always be positive, https://en.wikipedia.org/wiki/Vandermonde_matrix

• Can you tell us what the "it" is? – ancientmathematician Mar 8 '17 at 14:58
• @shdp Why do you think so? Because the determinant is positive? – Widawensen Mar 8 '17 at 14:59