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Let:

$$M_n:=\left(\begin{array}{ccccc} 1 & 2 & 3 & \dots & n \\ 1^3 & 2^3 & 3^3 & \dots & n^3 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1^{2n-1} & 2^{2n-1} & 3^{2n-1} & \dots & n^{2n-1} \end{array}\right),$$

for $n\in\mathbb N$. How could I compute $\det M_n$? I tried computing the first few and seeing if I saw an apparent pattern to try inducting, but the third one was already too much of a mess and no pattern was evident. This was an exercise in an exam, so there must be a smart way of doing it. Is there? Which is it?

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$$M_n:=\left(\begin{array}{ccccc} 1 & 2 & 3 & \dots & n \\ 1^3 & 2^3 & 3^3 & \dots & n^3 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1^{2n-1} & 2^{2n-1} & 3^{2n-1} & \dots & n^{2n-1} \end{array}\right) = \left(\begin{array}{ccccc} 1 \times 1^0 & 2 \times 2^0 & 3 \times 3^0 & \dots & n \times n^0 \\ 1 \times 1^2 & 2 \times 2^2 & 3 \times 3^2 & \dots & n \times n^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 \times 1^{2n-2} & 2 \times 2^{2n-2} & 3 \times 3^{2n-2} & \dots & n \times n^{2n-2} \end{array}\right)$$

Factor out $n!$ (in the form of $j$ from the $j$th column):

$$\text{det} = n! \left \vert \left(\begin{array}{ccccc} 1^0 & 2^0 & 3^0 & \dots & n^0 \\ 1^2 & 2^2 & 3^2 & \dots & n^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1^{2n-2} & 2^{2n-2} & 3^{2n-2} & \dots & n^{2n-2} \end{array}\right) \right \vert$$

This is a Vandermonde matrix with $a_i = i^2$; the determinant of that matrix is $$\prod_{1 \leq i < j \leq n} (i^2 - j^2)$$

Therefore $$\text{det}(M_n) = n! \prod_{1 \leq i < j \leq n} (i+j)(i-j)$$

If you want to "simplify" it a bit more, you can do the following to get a feel for it, but it doesn't yield a nice closed form that I can see.

First, let $n$ be odd. In the $i+j$ terms, $3$ and $4$ turn up once; $5$ and $6$ turn up twice; $7$ and $8$ three times; and so on up to $n$ and $n+1$, which occurs $\frac{n-1}{2}$ times. Then $n+2$ turns up $\frac{n-1}{2}$ times, $n+3$ and $n+4$ turn up $\frac{n-3}{2}$ times, and so on down to $2n-2$ and $2n-1$ which happen once. Therefore $$\prod_{1 \leq i < j \leq n} (i+j) = 3 \times 4 \times 5^2 \times 6^2 \times \dots \times n^{(n-1)/2} \times (n+1)^{(n-1)/2} \times (n+2)^{(n-1)/2} \times (n+3)^{(n-1)/2-1} \times \dots \times (2n-2) \times (2n-1)$$

You can do similar things if $n$ is even, and then analyse the $i-j$ terms in a similar way. (The $i-j$ terms lead to a product of factorials.)

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    $\begingroup$ "Factor out the $n!$ from each column" should read "Factor out $j$ from the column $j$". $\endgroup$ – A.Γ. Oct 1 '15 at 12:03
  • $\begingroup$ And is that determinant really negative? No because if you sum over $i<j$ $i^2-j^2$ is always negative… $\endgroup$ – MickG Oct 1 '15 at 12:15
  • $\begingroup$ This kind of confirms my impression that this exercise was totally inappropriate and WAY too hard for a linear algebra exam from an Economics degree course. $\endgroup$ – MickG Oct 1 '15 at 12:39
  • $\begingroup$ My friend who must take the exam has never heard of Vandermonde matrices before. Is there any way he could possibly have done this? $\endgroup$ – MickG Oct 1 '15 at 12:40
  • $\begingroup$ @MickG The determinant is not negative because there are an even number of terms. It is possible to prove the determinant of the Vandermonde matrix by induction, but I don't know of a way to guess it. It's basic linear algebra, though. (math.stackexchange.com/questions/1336408/… ). $\endgroup$ – Patrick Stevens Oct 1 '15 at 12:56

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