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While reading about interpolation I came across the following equation in Norlund. It involves determinants and I don't understand it in full yet. I do know how Lagrange and Newton follow by using the Laplace expansion.

$$ P_n=-\det \begin{bmatrix} 1 & x_0 & \dots & x_0^n & y_0\\ 1 & x_1 & \dots & x_1^n & y_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & x_n & \dots & x_n^n & y_n\\ 1& x & \dots & x^n & 0\\ \end{bmatrix} : \det \begin{bmatrix} 1 & x_0 & \dots & x_0^n\\ 1 & x_1 & \dots & x_1^n\\ \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & \dots & x_n^n\\ \end{bmatrix} $$

Where $P_n$ is the interpolation polynomial. Let's define the following for the interpolation polynomial $P_n(x_i)=y_i$ for $i\in[0,1,...,n]$. But this implies, after converting this equation into the determinant of a single matrix $A$, the following:

$$ \det(A)=\det \begin{bmatrix} 1 & x_0 & \dots & x_0^n & y_0\\ 1 & x_1 & \dots & x_1^n & y_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & x_n & \dots & x_n^n & y_n\\ 1& x & \dots & x^n & P_n\\ \end{bmatrix} =0\quad\quad\quad\quad\quad\quad\quad\quad $$

(Question) What is the geometric interpretation of this statement? My best guess is something along the lines of infinite solution because $x\in\mathbb{R}$.

Thanks in advance.

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In the hypotesis, you have that

$$\det \begin{bmatrix} 1 & x_0 & \dots & x_0^n\\ 1 & x_1 & \dots & x_1^n\\ \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & \dots & x_n^n\\ \end{bmatrix} \neq 0$$

It means that

$$\ \text{rank} \begin{bmatrix} 1 & x_0 & \dots & x_0^n\\ 1 & x_1 & \dots & x_1^n\\ \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & \dots & x_n^n\\ \end{bmatrix}\ = n+1 = \text{rank} \begin{bmatrix} 1 & x_0 & \dots & x_0^n\\ 1 & x_1 & \dots & x_1^n\\ \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & \dots & x_n^n\\ 1 & x & \ldots & x_n\\ \end{bmatrix}\ $$ are both maximum. Now

$$ \det \begin{bmatrix} 1 & x_0 & \dots & x_0^n & y_0\\ 1 & x_1 & \dots & x_1^n & y_1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & \dots & x_n^n & y_n\\ 1& x & \dots & x^n & P_n\\ \end{bmatrix} =0 $$

means that

$$ n+1 \leq \text{rank} \begin{bmatrix} 1 & x_0 & \dots & x_0^n & y_0\\ 1 & x_1 & \dots & x_1^n & y_1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & \dots & x_n^n & y_n\\ 1& x & \dots & x^n & P_n\\ \end{bmatrix} < n+2 \Rightarrow \text{rank} \begin{bmatrix} 1 & x_0 & \dots & x_0^n & y_0\\ 1 & x_1 & \dots & x_1^n & y_1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & \dots & x_n^n & y_n\\ 1& x & \dots & x^n & P_n\\ \end{bmatrix} = n+1$$

so the last column is linearly dependent from the other columns: $\exists \lambda_0, \ldots, \lambda_n$ (the coefficients of the interpolation polynomial!) such that $\sum_{i = 0}^{n} \lambda_i x_j^i = y_j$ for all $j = 0, \ldots, n$ and $\sum_{i = 0}^{n} \lambda_i x^i = P_n$.

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  • $\begingroup$ Thanks! I like your reasoning, and I understand that the rank of the matrix $A$ is $n+1$ But could you elaborate your conclusion? $\endgroup$ – Max Mar 17 '19 at 11:25
  • $\begingroup$ In this case, the rank is the maximum number of linerly independent columns; adding the "$y$-column" the rank doesn't increase, so the added column must be linearly dependent from the others. $\endgroup$ – dcolazin Mar 17 '19 at 11:27
  • $\begingroup$ Thanks for your elaboration, I think I'll be able to understand it much better! PS. After refreshing linear dependance etc. $\endgroup$ – Max Mar 17 '19 at 11:30
  • $\begingroup$ @Max how do you define the rank? $\endgroup$ – dcolazin Mar 17 '19 at 12:05
  • $\begingroup$ The number of linearly independent columns... So thanks! $\endgroup$ – Max Mar 17 '19 at 14:13

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