From Serge Lang's Linear Algebra:
Let $x_1$, $x_2$, $x_3$ be numbers. Show that:
$$\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 &
x_3^2 \end{vmatrix}=(x_2-x_1)(x_3-x_1)(x_3-x_2)$$
The matrix presented above seems to be the specific case of Vandermonde determinant:
$$
\begin{vmatrix}
1 & x_1 & … & x_1^{n-1}\\
1 &x_2 & … & x_2^{n-1}\\
… & … & … & …\\
1 & x_n & … & x_n^{n-1}
\end{vmatrix}=\prod_{i, j}(x_i – x_j), \forall (1 \leq i \leq n) \land (1 \leq j \leq n)
$$
I'm trying to prove the specific case to then generalize it for arbitrary Vandermonde matrices.
My incomplete "proof"
Since determinant is a multilinear alternating function, it can be seen that adding a scalar multiple of one column (resp. row) to other column (resp. row) does not change the value (I omitted the proof to avoid too much text).
Thus considering that $x_1$ is a scalar, we can multiply each column but the last one of our specific Vandermonde matrix by $x_1$ and then starting from right to left subtract $n-1$th column from $n$:
$$\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 &
x_3^2 \end{vmatrix}=\begin{vmatrix}
x_1 & 0 & 0 \\
x_1 & x_2 – x_1 & x^{2}_2 – x^{2}_1\\
x_1 & x_3 – x_1 & x^{2}_3 – x^{2}_1
\end{vmatrix}$$
Then using the expansion rule along the first row (since all the elements in it but $x_1$ are zero):
$$… =x_1\begin{vmatrix}
x_2 – x_1 & x^{2}_2 – x^{2}_1\\
x_3 – x_1 & x^{2}_3 – x^{2}_1
\end{vmatrix}=(x_1x_2-x^2_1)(x^2_{3}-x^2_1)-(x^{2}_2x_1 – x^{3}_1)(x_3x_1 – x^2_1)$$
The first expansion seems interesting because it contains $x_2 – x_1$ and $x_3 – x_1$ (which are first two factors of specific Vandermonde matrix), but further expansion does not give satisfying results.
Question:
Is this a good simple start of inductively "proving" relation between Vandermonde matrix and its factors? If so what does it lack to show the complete result? Did I make mistake during evaluation?
Thank you!
Best Answer
The general proof is not difficult.
From the definition of a determinant (sum of products), the expansion must be a polynomial in $x_1,x_2,\cdots x_n$, of degree $0+1+2+\cdots n-1=\dfrac{(n-1)n}2$, and the coefficient of every term is $\pm1$.
On another hand, the determinant cancels whenever $x_j=x_k$, so that the polynomial must be a multiple of
$$(x_1-x_2)(x_1-x_3)(x_1-x_4)\cdots(x_1-x_n)\\ (x_2-x_3)(x_2-x_4)\cdots(x_2-x_n)\\ (x_3-x_4)\cdots(x_3-x_n)\\ \cdots\\ (x_n-x_{n-1})$$ ($\dfrac{(n-1)n}2$ factors).
Hence the determinant has no other choice than being $\pm$ this product.
For the $3\times3$ case,
$$\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix}= \begin{vmatrix} 1 & x_1 & x_1^2\\ 0 &x_2-x_1 & x_2^2-x_1^2\\ 0 & x_3-x_1 & x_3^2-x_1^2 \end{vmatrix}=\begin{vmatrix} x_2-x_1 & x_2^2-x_1^2\\ x_3-x_1 & x_3^2-x_1^2 \end{vmatrix}=(x_2-x_1)(x_3-x_1)\begin{vmatrix} 1&x_2+x_1 \\1& x_3+x_1 \end{vmatrix}=(x_2-x_1)(x_3-x_1)(x_3-x_2).$$