Let be a field, , and be an injective map. Evaluate the determinant of the Vandermonde matrix

(In this exercise, matrix entries are indexed from 0.)

First, we define a map recursively as follows.

.

Now, for each , define a matrix . For example, if , then is just the Vandermonde matrix. For each , is a square matrix with dimension .

We will now prove some facts about that will be useful later.

Lemma 1: for all and . Proof: We proceed by induction on . For the base case , certainly . Suppose now the result holds for some such that and consider . We have . So the result holds for all and all .

Lemma 2: For all and all ,

Proof: We will prove this directly, handling the cases and separately. For the case , we have the following.

= | ||

= | ||

= | ||

= | . |

As desired. Suppose now that . Then we have the following.

= | ||

= | ||

= | ||

= | ||

= | ||

= | ||

= | ||

= | ||

= | ||

= | ||

= | , |

again as desired. So the lemma is proved.

Now we claim that, for all , the matrix is row and column equivalent to . Moreover, we claim that . To see this, we will perform row and column operations on , keeping track of their effect on the determinant.

By Lemma 1, the first column of consists of all 1s. So for each row , we subtract row 0 from row . This operation does not affect the determinant. Now the entry in row and column is . Column 0 has a 1 in row 0 and 0 elsewhere, so we may add an appropriate multiple of column 0 to the remaining columns so that row 0 has all 0s except in column 1. Finally, from each row in (that is, except the 0th), we divide . This has the effect of dividing the determinant by . By Lemma 2, the -minor matrix is now precisely .

So the determinant of is times the determinant of , which is precisely the determinant of . Note that , which certainly has determinant 1. By induction, then, .