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I can follow the proof given in (2nd proof, or the induction proof), until the sentence: "From the Expansion Theorem for Determinants‎, we can see that the coefficient of $x_k$ is:".

I don't understand why it is only needed the determinant of the cofactor of $x_k$, since the calculation of a determinant involves taking one row and the multiplication of the cofactors in the chosen row.

Martin Sleziak
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user2820579
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2 Answers2

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It's talking about how you work out determinants you can "expand" them into a sum of 2x2 determinants

It's what I did here ( Vandermonde determinant by induction ) 2 years ago.

Community
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Alec Teal
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  • I understand your derivation and also checked it, thanks for the feedback. I don't understand the line that I wrote in ".." in the second proof given in the webàge I posted. I corrected the question. – user2820579 Jul 29 '15 at 22:36
  • @user2820579 it's literally in that question – Alec Teal Jul 29 '15 at 22:38
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It has been shown that the determinant is something like:

$$V_{k+1}(x) = f(x) = C(x-a_2)(x-a_3)\ldots(x-a_k)(x-a_{k+1})$$

Note that if you would expand this it would be of the form:

$$f(x) = Cx^k + B_{k-1}x^{k-1}+\ldots$$

Now consider the expansion theorem for the first row:

$$V_{k+1}(x) = \begin{vmatrix} a_2^{k-1} & \cdots & a_2^2 & a_2 & 1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ a_{k+1}^{k-1} & \cdots & a_{k+1}^2 & a_{k+1} & 1 \end{vmatrix}x^k +B_{k-1}x^{k-1}+\ldots$$

Where $B_{i}$ is some constant.

dietervdf
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