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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

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  • $\begingroup$ 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. $\endgroup$
    – user239203
    May 30, 2020 at 7:44
  • $\begingroup$ Since they are not independent. $\endgroup$
    – dust05
    May 30, 2020 at 7:49

2 Answers 2

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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.)

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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.

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