# Calculate determinant of Vandermonde using specified steps.

$V_n(a_1,a_2\dots, a_n)$ 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 $V_n(a_1,a_2,\dots a_n) = \prod^n_{1\leq i< j\leq n} (a_j - a_i)$

• You seem to have mixed up $\;a_i$'s and $\;z_i$'s ... – DonAntonio Apr 2 '14 at 3:46

(a) Show that $P(x)$ is a polynomial in $x$ of degree $\le n−1$.
Use the Leibniz formula for determinants, and show that each sum contains only one factor of $x^k$ from the top row.
(b) Show that $P(x)$ has $n−1$ distinct roots $a_2,~ \dots ,~ a_n$ and therefore has factorization $P(x)=k\prod^n_{i=2}(x−a_i)$ where the constant factor $k$ is the coefficient of $x^{n−1}. Plug$x = a_i$into your matrix, and compare the first row to the$i^{th}$row. Show that the determinant must be zero in that case. (c) Show that$k = (-1)^{n-1}V_{n-1}(a_2,\dots a_n)$This follows immediately from the Laplace expansion that you did in part (a). (d) Use parts (b) and (c) to deduce the recursion formula... Just plug$x=a_1$and part (c) into the formula in (b). (e) Use part (d) to deduce that... Use the fact that $$\bigg(\prod_{x \in X} f_x\bigg)\bigg(\prod_{y \in Y} g_y\bigg) = \prod_{(x,y) \in X \times Y} f_x \cdot g_y$$ 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 $$V(a_1,a_2,\ldots,a_n) =\left(\prod_{2\leq i<j\leq n}(a_j-a_i)\right)\prod_{j=2}^n(a_j-a_1) =\prod_{1\leq i<j\leq n}(a_j-a_i).$$