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. 

 A: 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.
