Exercise 2.4.1

Proof. Let $a_1,a_2,\cdots ,a_n$ be elements of a field $F$, and

We show by induction on $n,n\ge 2$ that

$\det <!--\; FUNCTION\; APPLICATION\; -->(A_2)=\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill \end{array}\right|=a_2-a_1=\underset{1\le i<j\le 2}{\prod <!--\; FUNCTION\; APPLICATION\; -->}(a_j-a_i).$

Suppose that this formula is true for the integer $n-1,n\ge 3$. We will show that it is true for the integer $n$.

If there exists a pair $(i,j),i\ne j$ such that $a_i=a_j$, then two columns in $A_n$ are identical, so $\det <!--\; FUNCTION\; APPLICATION\; -->(A_n)=0=\underset{1\le i<j\le n}{\prod <!--\; FUNCTION\; APPLICATION\; -->}(a_j-a_i).$

We can so suppose that the $a_i,1\le i\le n$ are distinct.

Let the polynomial $P\in F[X]$ given by

Then $\det <!--\; FUNCTION\; APPLICATION\; -->(A_n)=P(a_n)$, and $P(a_1)=P(a_2)=\cdots =P(a_{n-1})=0$. As $a_1,a_2,\cdots ,a_{n-1}$ are distinct roots of $P$, with $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(P)=n-1$, $P$ is factored as

where $k$ is the coefficient of $X^{n-1}$ in $P$, so $k$ is the cofactor of $X^{n-1}$ in $\det <!--\; FUNCTION\; APPLICATION\; -->(P)$: so

by the induction hypothesis.

Therefore

which completes the induction.

The matrix

is obtained from $A_n$ by $\frac{n(n-1)}{2}$ transpositions of rows : $n-1$ to put the last row in first position, then $n-2$ to put which is now the last row in second position, and so on.

Thus $\det <!--\; FUNCTION\; APPLICATION\; -->(B_n)={(-1)}^{(n(n-1))∕2}\det <!--\; FUNCTION\; APPLICATION\; -->(A_n)$.

As the number of factors in $\underset{1\le i<j\le n}{\prod <!--\; FUNCTION\; APPLICATION\; -->}(a_j-a_i)$ is $\frac{n(n-1)}{2}$,

Consequently,

Applying this result in the field $F(x_1,\cdots ,x_n)$, we obtain that

If $A=\left(\begin{array}{cccc}\hfill x_1^{n-1}\hfill & \hfill x_2^{n-1}\hfill & \hfill \cdots \hfill & \hfill x_n^{n-1}\hfill \\ \hfill x_1^{n-2}\hfill & \hfill x_2^{n-2}\hfill & \hfill \cdots \hfill & \hfill x_n^{n-2}\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \text{⋱}\hfill & \hfill ⋮\hfill \\ \hfill x_1\hfill & \hfill x_2\hfill & \hfill \cdots \hfill & \hfill x_n\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill \cdots \hfill & \hfill 1\hfill \\ \hfill \hfill \end{array}\right)$, then $A^t=\left(\begin{array}{ccccc}\hfill x_1^{n-1}\hfill & \hfill x_1^{n-2}\hfill & \hfill \cdots \hfill & \hfill x_1\hfill & \hfill 1\hfill \\ \hfill x_2^{n-1}\hfill & \hfill x_2^{n-2}\hfill & \hfill \cdots \hfill & \hfill x_2\hfill & \hfill 1\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill x_{n-1}^{n-1}\hfill & \hfill x_{n-1}^{n-2}\hfill & \hfill \cdots \hfill & \hfill x_{n-1}\hfill & \hfill 1\hfill \\ \hfill x_n^{n-1}\hfill & \hfill x_n^{n-2}\hfill & \hfill \cdots \hfill & \hfill x_n\hfill & \hfill 1\hfill \\ \hfill \hfill \end{array}\right)$

thus

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