# How can I prove that the determinant of a matrix formed of polynomials of degree n-2 or smaller is zero?

The following matrix is formed by polynomiais of degree $$n-2$$ or smaller and $$a_1,\cdots,a_n$$ belong to $$\mathbb{R}$$. How can I prove that it’s determinant is zero? I thought about using the fact that it is a matrix of even degree $$n$$.

Would really appreciate any help.

$$\begin{vmatrix}p_1(a_1) & p_1(a_2) & \cdots& p_1(a_n) \\ p_2(a_1) & p_2(a_2) & \cdots &p_2(a_n) \\ \vdots & \vdots & \ddots & \ \vdots \\ p_n(a_1) & p_n(a_2) & \cdots &p_n(a_n) \\ \end{vmatrix}$$

matrix

• I don't understand what a "degree" of a matrix is supposed to be, nor do I see any quantity realted to the matrix you are mentioning which may reasonably be presumed even.
– user239203
May 30, 2020 at 7:44
• Since they are not independent. May 30, 2020 at 7:49

## 2 Answers

Hints: The space of polynomials of degree $$n-2$$ or less is a vector space of dimension $$n-1$$. Therefore, any set of $$n$$ polynomials from this space is linearly dependent. Hence there exist scalars $$c_{1}, \ldots, c_{n}$$, not all zero, such that $$c_{1}p_{1} + \cdots + c_{n}p_{n} = 0$$ (the zero polynomial).

So $$c_{1}p_{1}(x) + \cdots + c_{n}p_{n}(x) = 0$$ for all $$x\in \Bbb{R}$$. Thus $$c_{1} p_{1}(a_i) + \cdots + c_{n}p_{n}(a_{i}) = 0$$

for all $$i = 1,\ldots,n$$.

Can you take it from here? (Try writing the above $$n$$ equations in matrix form.)

An alternative approach:

Suppose $$p_k(x) = c_k(0) + c_k(1)x+\cdots + c_k(n-2) x^{n-2}$$.

Then the above matrix can be written as $$\begin{bmatrix} c_1(0) & c_1(1)& \cdots & c_1(n-2) & 0 \\ \vdots & \vdots & & \vdots & \vdots \\ c_n(0) & c_n(1)& \cdots & c_n(n-2) & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 & \cdots & 1 \\ a_1 & a_2 &\cdots & a_n \\ \vdots & \vdots &\cdots & \vdots \\ a_1^{n-1} & a_2^{n-1} &\cdots & a_n^{n-1} \end{bmatrix}$$ from which we can see that the determinant must be zero.