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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!

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    $\begingroup$ The formulae do not reflect the operations you describe. $\endgroup$ – Bernard Aug 17 at 19:24
  • $\begingroup$ @Bernard I apologize, I can't understand your statement properly, do you mean that I incorrectly applied expansion formula along the first row? $\endgroup$ – ShellRox Aug 17 at 19:26
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    $\begingroup$ No. I mean the operations you describe do not lead to what is in the right matrix (first equation). $\endgroup$ – Bernard Aug 17 at 19:29
  • $\begingroup$ I think I made a mistake in description, I should've said multiply all columns but the final one with $x_1$, and then other operations described would lead to the right matrix, correct? $\endgroup$ – ShellRox Aug 17 at 19:34
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    $\begingroup$ I don't think so. What you should do is subtract $x_1\times$ the first column from the second column, and subtract $x_1^2\times $ the first column from the third. $\endgroup$ – Bernard Aug 17 at 19:41
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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,

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

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    $\begingroup$ Note: This requires some knowledge about the polynomial ring in $n$ variables -- either that it is a UFD, or at least that common multiples of distinct $x_i-x_j$'s (with $i<j$) are multiples of their product. $\endgroup$ – darij grinberg Aug 17 at 20:25
  • $\begingroup$ Thank you for the response. I don't posses sufficient knowledge of group structures that @darijgrinberg mentioned (thank you very much for the prerequisites), but how is alternating property of determinant related to factors of polynomial? Perhaps due to consequence that Darij mentioned or something else? (I apologize if I misunderstood this). $\endgroup$ – ShellRox Aug 17 at 21:36
  • $\begingroup$ I'm sorry, I didn't pay attention to correlation between alternating property and polynomial vanishing. Now it's understandable, thank you! $\endgroup$ – ShellRox Aug 18 at 7:37
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"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}

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    $\begingroup$ IMHO, there is little profit to do this kind of rather complicated calculation by hand in the $3 \times 3$ case, in the idea to extend it further in the general $n \times n$ case, which is doubtful... $\endgroup$ – Jean Marie Aug 17 at 20:25
  • $\begingroup$ @JeanMarie I agree. I did the calculation more to show how to get to the result in this case than to generalize it by induction. $\endgroup$ – Monadologie Aug 17 at 20:29
  • $\begingroup$ Thank you for the response. This answer clearly showed exactly where I made mistake with expansion formula, but I'm not completely able to answer the last part (finding the simplest possible factors). What does $L$ exactly imply? Is it a linear transformation that matrix is associated to? $\endgroup$ – ShellRox Aug 17 at 21:21
  • $\begingroup$ @ShellRox $L_i \leftarrow L_i - \lambda L_k$ is an elementary row operation (here $L_i$ is the $i$-th row). Doing elementary operations simplifies the matrix and changes the determinant by a multiplicative factor (see https://proofwiki.org/wiki/Effect_of_Elementary_Row_Operations_on_Determinant). $\endgroup$ – Monadologie Aug 18 at 6:50
  • $\begingroup$ Oh, now it is understandable, wasn't aware of the notation. Thank you very much for the simple specific solution, I assume this is can't be optimally generalized to $n \times n$ determinants, correct? $\endgroup$ – ShellRox Aug 18 at 7:57

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