Inductively simplify specific Vandermonde determinant

Question

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:

$$ \begin{vmatrix} 1 & x_1 & ... & x_1^{n-1}\\ 1 &x_2 & ... & x_2^{n-1}\\ ... & ... & ... & ...\\ 1 & x_n & ... & x_n^{n-1} \end{vmatrix}=\prod_{i, j}(x_i - x_j), \forall (1 \leq i \leq n) \land (1 \leq j \leq n) $$

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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!

Answer 1

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.

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For the $3\times3$ case,

$$\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix}= \begin{vmatrix} 1 & x_1 & x_1^2\\ 0 &x_2-x_1 & x_2^2-x_1^2\\ 0 & x_3-x_1 & x_3^2-x_1^2 \end{vmatrix}=\begin{vmatrix} x_2-x_1 & x_2^2-x_1^2\\ x_3-x_1 & x_3^2-x_1^2 \end{vmatrix}=(x_2-x_1)(x_3-x_1)\begin{vmatrix} 1&x_2+x_1 \\1& x_3+x_1 \end{vmatrix}=(x_2-x_1)(x_3-x_1)(x_3-x_2).$$

Answer 2

"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) " is right. But $$ \begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix} \neq \begin{vmatrix} x_1 & 0 & 0 \\ x_1 & x_2 - x_1 & x^{2}_2 - x^{2}_1\\ x_1 & x_3 - x_1 & x^{2}_3 - x^{2}_1 \end{vmatrix} \neq (x_1x_2-x^2_1)(x^2_{3}-x^2_1)-(x^{2}_2x_1 - x^{3}_1)(x_3x_1 - x^2_1) $$ Remember that when you multiply a row or a column by $\lambda$, the determinant is multiplied by $\lambda$. And be careful when distributing $x_1$. We have \begin{align} \begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix} &= x_1 \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}\\ &= x_1^2 \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_1^2((x_2 - x_1)(x^{2}_3 - x^{2}_1) - (x^{2}_2 - x^{2}_1)(x_3 - x_1))\\ &\neq (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) \end{align} Keep in mind that we are trying to have the simplest possible factors. Here, you can do \begin{align} \begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix}&=_{L_3 \leftarrow L_3 - L_2 \text{ and } L_2 \leftarrow L_2 - L_1} \begin{vmatrix} 1 & x_1 & x_1^2\\ 0 &x_2 -x_1& (x_2 - x_1)(x_2+x_1)\\ 0 & x_3 - x_2 & (x_3 - x_2)(x_3+x_2) \end{vmatrix}\\ &=_{L_3 \leftarrow L_3 - L_2} (x_2 - x_1)(x_3-x_2) \begin{vmatrix} 1 & x_1 & x_1^2\\ 0 &1& x_2 + x_1\\ 0 & 0 & x_3 -x_1 \end{vmatrix}\\ &=(x_2 - x_1)(x_3-x_2)(x_3-x_1) \end{align}

Answer 3

Using induction on $n$ is maybe the most convenient way. Let's proof that if we assume that the result is true for $n - 1$ then the result is true for $n$. So, consider the Vandermonde matrix $$ V =\begin{pmatrix} 1 & x_{1} & x_{1}^{2} \cdots & x_{1}^{n-1} \\ 1 & x_{2} & x_{2}^{2} \cdots & x_{2}^{n-1} \\ & & \vdots \\ 1 & x_{n -1} & x_{n - 1}^{2} \cdots & x_{n-1}^{n-1} \\ 1 & x_{n} & x_{n}^{2} \cdots & x_{n}^{n-1} \end{pmatrix} $$ Now consider $\det(V)$ as a polynomial of degree $n-1$ in the variable $x_{n}$. We can compute $\det(V)$ using the $n$-th column by means of the expansion of $\det(V)$ in terms of cofactors, so $$ \det(V) = a_{n-1} x_{n}^{n-1} + a_{n-2} x_{n}^{n-2} + \cdots + a_{0} \implies a_{n-1} = \det \begin{pmatrix} 1 & x_{1} & x_{1}^{2} \cdots & x_{1}^{n-2} \\ 1 & x_{2} & x_{2}^{2} \cdots & x_{2}^{n-2} \\ & & \vdots \\ 1 & x_{n -1} & x_{n - 1}^{2} \cdots & x_{n-1}^{n-2} \end{pmatrix}\\ \text{(Note that $a_{n-1}$ is the leading coeficient of the polynomial $\det(V)$)} $$ In addition, $x_{1}, x_{2}, \dots, x_{n-1}$ are the roots of $\det(V)$, so we can write $$ \det(V) = a_{n-1}(x_{n}- x_{1}) (x_{n} - x_{2}) \cdots (x_{n} - x_{n-1}). $$ By induction hypothesis we have the $$ a_{n-1} = \prod_{1 \leq i < j \leq n-1} (x_{i} - x_{j}). $$ Hence, $$ \det(V) = \left[\prod_{1 \leq i < j \leq n-1}(x_{i} - x_{j})\right] (x_{n} - x_{1}) \dots (x_{n} - x_{n-1}) = \prod_{1 \leq i < j \leq n} (x_{i} - x_{j}). $$