# Prove that the determinant of polynomials is zero

Prove that this determinant is zero (this matrix is $n\times n$): $$\begin{vmatrix} f_1(a_1) & f_1(a_2) & \cdots & f_1(a_n) \\ f_2(a_1) & f_2(a_2) & \cdots & f_2(a_n) \\ \vdots & \vdots & \ddots & \vdots_\strut \\ f_n(a_1) & f_n(a_2) & \cdots & f_n(a_n) \end{vmatrix}$$ Every $f_i$ is a polynomial of degree $n-2$ maximum.

I had an idea that, because there are $n$ polynomials and the degree of each one is at most $n-2$, the lines of the matrix must be linearly dependent and therefore the determinant is $0$? I don't know how to write a formal and clear solution to this, I think my idea is pretty rough and need to be explained clearly.. can anyone help me please?

• Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. – Zev Chonoles Aug 13 '13 at 22:17

Consider the matrix $A$ where the entry $a_{ij}$ is the coefficient of $x^{j-1}$ in the polynomial $f_i$.
What is the maximal rank of this $n \times (n-1)$ matrix $A$?