$V_n(a_1,a_2\dots, a_n)$ is a $n\times n$ Vandermonde matrix =
$$\left|\begin{array}[cccc]
11&a_1&\cdots&a^{n-1}_1\\
1&a_2&\cdots&a^{n-1}_2\\
1&a_3&\cdots&a^{n-1}_3\\
\vdots&\vdots&\ddots&\vdots\\
1&a_n&\cdots&a_n^{n-1}
\end{array}\right|$$
Replace $a_1$ by $x$ so that the first row is $1,x, \dots ,x^{n-1}$
$$V_n(x,a_2\dots, a_n) = \left|\begin{array}[cccc]
11&x&\cdots&x^{n-1}\\
1&a_2&\cdots&a^{n-1}_2\\
1&a_3&\cdots&a^{n-1}_3\\
\vdots&\vdots&\ddots&\vdots\\
1&a_n&\cdots&a_n^{n-1}
\end{array}\right|$$
Let $P(x) = V_n(x,a_2\dots, a_n)$.
(a) Show that $P(x)$ is a polynomial in $x$ of degree $\leq n-1$.
$z = 1\det() -x\det() + x^2\det() -\cdots+ (-1)^{n-1}x^{n-1} \cdot \det()$, where each $\det()$ stands for the determinant of a smaller matrix after removing the appropriate columns, rows.
Is this right? But isn't $z$ a polynomial of degree $n$?
(b) Show that $P(x)$ has $n-1$ distinct roots $a_2, \dots a_n$ and therefore has factorization $P(x) = k \prod_{i=2}^n(x-a_i)$ where the constant factor $k$ is the coefficient of $x^{n-1}.$
I'll be honest, I have no clue how to do this. Not even sure where to start. Nor do I know where to start on the rest.
(c) Show that $k = (-1)^{n-1}V_{n-1}(a_2,\dots a_n)$.
(d) Use parts (b) and (c) to deduce the recursion formula $$V_n(a_1, \dots a_n) = \left(\prod_{i=2}^n(a_i-a_1)\right)V_{n-1}(a_2,\dots a_n)$$
(e) Use part (d) to deduce that $V_n(a_1,a_2,\dots a_n) = \prod^n_{1\leq i< j\leq n} (a_j – a_i)$
Best Answer
For part (a), this is just development (Laplace expansion) of the determinant by the first row. Actually the $\det()$ factors should have alternating signs. Since the only occurrences of $x$ are in that first row, all the $\det()$ expressions are constants, and one gets a polynomial of degree at most $n-1$ (from the final term) in $x$.
For part (b), that $P(a_i)=0$ for $i=2,3,\ldots,n$ is just the fact that $P(a_i)$ equals the determinant of the matrix obtained by substituting $a_i$ for $x$, so from the original matrix $a_1$ has been replaced by $a_i$, and as this matrix has its rows $1$ and $i$ identical, its determinant vanishes. all this uses is that the Laplace expansion used commutes with such substitution. Furthermore a polynomial of degree at most $n-1$ with $n-1$ specified roots $a_2,\ldots,a_n$ can only be a scalar multiple of $(x-a_2)\ldots(x-a_n)$.
For part (c), this is just remarking that the $\det()$ in question is $(-1)^{n-1}$ times the determinant of the lower-left $(n-1)\times(n-1)$ submatrix, which determinant precisely matches the definition of $V_{n-1}(a_2,\ldots,a_n)$.
For part (d) write $(-1)^{n-1}\prod_{i=2}^n(x-a_i)=\prod_{i=2}^n(a_i-x)$ to get $$ V(x,a_2,\ldots,a_n)= (-1)^{n-1}V_{n-1}(a_2,\ldots,a_n)\prod_{i=2}^n(x-a_i) =V_{n-1}(a_2,\ldots,a_n)\prod_{i=2}^n(a_i-x), $$ and then set $x=a_1$ to get $$ V(a_1,a_2,\ldots,a_n) =V_{n-1}(a_2,\ldots,a_n)\prod_{i=2}^n(a_i-a_1), $$
Part (e) applies induction on $n$ to $V_{n-1}(a_2,\ldots,a_n)$ (the starting case is $V_0()=1=\prod_{1\leq i<j\leq 0}1$, an empty product, or if you fear $n=0$ it is $V_1(a)=1=\prod_{1\leq i<j\leq 1}1$, still an empty product), to get $$ V(a_1,a_2,\ldots,a_n) =\left(\prod_{2\leq i<j\leq n}(a_j-a_i)\right)\prod_{j=2}^n(a_j-a_1) =\prod_{1\leq i<j\leq n}(a_j-a_i). $$