For a natural number $i>0$, let $p_i$ be the $i$th prime number, that is, $p_1=2, p_2=3, p_3=5,...$. Show that for all $n$, the following matrix has positive determinant

$$ \begin{pmatrix} 1^{p_1} & 2^{p_1} & \cdots & (n-1)^{p_1} & n^{p_1} \\ 1^{p_2} & 2^{p_2} & \cdots & (n-1)^{p_2} & n^{p_2} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 1^{p_n} & 2^{p_n} & \cdots & (n-1)^{p_n} & n^{p_n}\\ \end{pmatrix} $$

The hint provided is to use the fact that the polynomial $P(x)=a_nx^{p_n}+a_{n-1}x^{p_{n-1}} + \cdots + a_1x^{p_1}+a_0x^{p_1}$ has at most $n-1$ positive roots for all real constants. But I don't know how to use the fact to prove the question. Can anyone help me?

The source of the problem is here, question $2.5$.

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    $\begingroup$ What is $\;p_0\;$ ?? $\endgroup$ – Timbuc Jun 5 '15 at 9:14
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    $\begingroup$ How does $P$ have only $n-1$ roots? It is of degree $p_n\gg n$. - Concretely, with $n=2$, $P(x)=x^3-x^2$ has two roots $x=1$ and $x=0$. $\endgroup$ – Hagen von Eitzen Jun 5 '15 at 9:27
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    $\begingroup$ @Timbuc Because the poster provided a link to the source. $\endgroup$ – muaddib Jun 5 '15 at 13:12
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    $\begingroup$ I see...the question's been edited 6 times so far. Ok, thanks. $\endgroup$ – Timbuc Jun 5 '15 at 13:13
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    $\begingroup$ @HagenvonEitzen That counter eg does not disprove, there is only one positive root. $\endgroup$ – Macavity Jun 6 '15 at 10:37

Motivation: Attempting to calculate the determinant directly by cofactor expansion leads to a headache.
It is not obvious if we can decompose this matrix into a product or conjugation (change of basis).
Thus, we are left with the good old approach of elementary row operations into an upper triangular matrix.

There are several red herrings in this problem:
1) $p_n$ is a list of primes is irrelevant (we just need distinct positive integers)
2) The bases are the integers from 1 to $n$ is irrelevant (we just need distinct positive integers)
3) The hint should not have a $a_o x^{p_1} $ (but that's easy to guess)

Proof of hint: This is immediate from Descarte Rule of Signs. $\square$

We now prove the statement by induction on $n$. Let $M_n$ denote the $n \times n$ matrix.

Claim 1: We can perform a series of elementary row operations (ERO) which will make the matrix $M_n$ into an upper triangular matrix with positive diagonal coefficients.

Claim 2: We seek a polynomial of the form

$$ f(x) = 1 \times x ^{p_n} + a_{n-1} x^{p_{n-1}} + \ldots + a_1 x ^{p_1} $$

such that $ f(1), f(2) , \ldots f(n-1) = 0 $.

Proof of claim 2: The existence follows from $\det(M_{n-1} ) \neq 0 $. As such, there is a solution to $$ \begin{pmatrix} a_1 & a_2 & \ldots a_{n-1} \end{pmatrix} M_{n-1} = \begin{pmatrix} -1^{p_n} & -2^{p_n} & \ldots -(n-1)^{p_n} \end{pmatrix}. $$

It is obvious that $ a_i$ satisfy the conditions in the claim. $_\square$

Proof of claim 1: Applying the elementary row operation $e_n = e_n + a_{n-1} e_{n-1} + \ldots + a_1 e_1$, will transform the last matrix row into $ \left( f(1), f(2), \ldots, f(n-1), f(n) \right)$.
1. By construction, $f(1) = f(2) = \ldots f(n-1) = 0 $.
2. From the hint, this polynomial has $n-1$ positive real roots already, which means that there are no further positive real roots, and thus $f(n) > 0 $.

We now induct, and the rest of the matrix can be ERO to an upper triangular matrix.

Finally, proof of statement: Since it is ERO equivalent to a upper triangular matrix with positive diagonal elements, which has a positive determinant, hence we are done.


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