# Show that $\det A_n$=0 for a polynomial matrix

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.

• 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 $f_2= x+b$ Feb 3, 2020 at 15:46
• @HelloDarkness Sorry, degree $n-1$ was a typo, I meant degree less than or equal to $n-2$.
– user731634
Feb 3, 2020 at 15:53
• 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. Feb 3, 2020 at 15:56

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 $$\mathbb F$$.