Is there a quick way to generate the characteristic polynomial of a Vandermonde matrix?

This came up on an exam recently as extra credit. The first part was to find the characteristic polynomial, $f_A = \text{det(}A - xI_n)$ where $I_n$ is the n by n identity matrix, of $A = \left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array} \right|.$ and that's easy enough with some basic calculations and brute force. But the second part asks for a general n x n Vandermonde matrix like $$B = \left| \begin{array}{ccc} 1 & & 1 \\ a_1 & ... & a_n \\ ... & & ... \\ a_1^{n-1} & & a_n^{n-1} \end{array} \right|.$$

I cant find anything in our book or the internet about the second part. Is there a "good" way to find the characteristic polynomial of B?

I know nothing of the values $a_i$ in the matrix if that makes any difference at all.

Thanks in Advance for any assistance.

EDIT: It turns out that this question was put on the exam last minute and the Professor thought it should be easy because the determinant is easy; Its not. So I don't even know if this has an answer.

• It is an interesting question simply because it seems deceptively easy at first glance. I could conceive of an iterative method which would basically be brute force calculation of the characteristic polynomial, but I could not say that it would be anything "good". Though it would make some interesting use of the polynomial aspect of the Vandermonde matrix. Commented Apr 17, 2013 at 16:27

1 Answer

It seems that the problem does not have a nice closed form solution, and is indeed looks deceptively easy. Even with $4$ variables things get messy. The characteristic polynomial is:

$$-x^3+x^4-x^2 a_1-x^2 a_1^2-x^2 a_1^3+x^2 a_2-x^3 a_2+x a_1^2 a_2+x a_1^3 a_2-x a_1 a_2^2-x a_1 a_2^3-x a_1^2 a_3+x a_2^2 a_3-x^2 a_2^2 a_3+a_1^3 a_2^2 a_3-a_1^2 a_2^3 a_3+x^2 a_3^2-x^3 a_3^2+x a_1 a_3^2+x a_1^3 a_3^2-x a_2 a_3^2+x^2 a_2 a_3^2-a_1^3 a_2 a_3^2+a_1 a_2^3 a_3^2-x a_1^2 a_3^3+a_1^2 a_2 a_3^3-a_1 a_2^2 a_3^3-x a_1^3 a_4-a_1^3 a_2^2 a_4+x a_2^3 a_4-x^2 a_2^3 a_4+a_1^2 a_2^3 a_4+a_1^3 a_3^2 a_4-a_2^3 a_3^2 a_4+x a_2^3 a_3^2 a_4-a_1^2 a_3^3 a_4+a_2^2 a_3^3 a_4-x a_2^2 a_3^3 a_4-x a_1^3 a_4^2+a_1^3 a_2 a_4^2-a_1 a_2^3 a_4^2-a_1^3 a_3 a_4^2+a_2^3 a_3 a_4^2-x a_2^3 a_3 a_4^2+x a_3^3 a_4^2-x^2 a_3^3 a_4^2+a_1 a_3^3 a_4^2-a_2 a_3^3 a_4^2+x a_2 a_3^3 a_4^2+x^2 a_4^3-x^3 a_4^3+x a_1 a_4^3+x a_1^2 a_4^3-x a_2 a_4^3+x^2 a_2 a_4^3-a_1^2 a_2 a_4^3+a_1 a_2^2 a_4^3+a_1^2 a_3 a_4^3-a_2^2 a_3 a_4^3+x a_2^2 a_3 a_4^3-x a_3^2 a_4^3+x^2 a_3^2 a_4^3-a_1 a_3^2 a_4^3+a_2 a_3^2 a_4^3-x a_2 a_3^2 a_4^3$$

It does not factor as a general rule. Mathematica's FullSimplify does not give anything visibly simpler. The coefficient at $x^0$ is the famous determinant, but the coefficients at $x$ and $x^2$ are : $$a_2 \left(a_2-a_3\right) a_3+a_2^2 \left(-a_3^3+a_2 \left(1+a_3^2\right)\right) a_4+a_3 \left(-a_2^3+\left(1+a_2\right) a_3^2\right) a_4^2+\left(-a_2+a_2^2 a_3-\left(1+a_2\right) a_3^2\right) a_4^3+a_1 \left(-a_2^2 \left(1+a_2\right)+a_3^2+a_4^3\right)+a_1^2 \left(a_2-a_3 \left(1+a_3^2\right)+a_4^3\right)+a_1^3 \left(a_2+a_3^2-a_4 \left(1+a_4\right)\right)$$ $$-a_1 \left(1+a_1+a_1^2\right)+a_2-a_2^2 a_3+a_3^2+a_2 a_3^2-a_2^3 a_4-a_3^3 a_4^2+\left(1+a_2+a_3^2\right) a_4^3$$ and they don't look as if they were anything simpler. (Neither me nor Wolfram know how to simplify these further). There is no symmetry to exploit in this problem (note that changing columns changes the characteristic polynomial non-trivially). If I am any judge, the polynomial for more variables is nothing nice simply.

• I went ahead and marked your answer as accepted. I haven't found anything to make me believe a good method exists. I have heard from another student that there might be a research paper on the subject but I've not been able to locate such a thing and it would almost surely be beyond what the class covered to that point. Commented Apr 24, 2013 at 5:59