Let $\mathbb{F}$ be a field, $n\geq 2\in\mathbb{N}, f_1,\ldots,f_n$ polynomials such that the degree of $f_i\leq n-2$. Show that, for all $x_1,\ldots,x_n\in\mathbb{F}$, $$\det\begin{pmatrix} f_1(x_1) & f_1(x_2) & \ldots & f_1(x_n) \\ f_2(x_1) & f_2(x_2) & \ldots & f_2(x_n)) \\ \vdots & \vdots& \ddots & \vdots \\ f_n(x_1) & f_n(x_2) & \ldots & f_n(x_n)\end{pmatrix}=0.$$

I have tried a proof via induction and the basis step is fairly simple, since we get that $\det A_2=f_1(x_1)f_2(x_2)-f_2(x_1)f_1(x_2)=0$, as for $n=2$, $f_1,f_2$ have degree less than or equal to zero.

My problem: The induction step seems very hard, since for $n+1$, every polynomial has degree less than or equal to $n-1$, so we can't just simply use the induction hypothesis.

  • $\begingroup$ Isn't this false? Take $(x+a)(y+b) -(x+b)(y+a)= bx +ay - yb -ax$ which is not always zero. In this case $f_1 = x+a$ and $f_2= x+b$ $\endgroup$ Feb 3, 2020 at 15:46
  • $\begingroup$ @HelloDarkness Sorry, degree $n-1$ was a typo, I meant degree less than or equal to $n-2$. $\endgroup$
    – user731634
    Feb 3, 2020 at 15:53
  • 2
    $\begingroup$ The dimension of the space of polynomials of degree less than $n-1$ is $n-1$. Hence the rows are linearly dependent, so the determinant is zero. $\endgroup$ Feb 3, 2020 at 15:56

1 Answer 1


Hint Consider the vector space $$V=\{ P \in \mathbb F[X] : \deg (P) \leq n-2 \}$$ Then $V$ is a vector space over $\mathbb F$ of dimension $n-1$.

Since you are given $n$ polynomials in $V$, they are linearly dependent over $\mathbb F$.


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