# What is the geometric meaning of this null-determinant?

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.

## 1 Answer

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 $$\sum_{i = 0}^{n} \lambda_i x^i = P_n$$.

• Thanks! I like your reasoning, and I understand that the rank of the matrix $A$ is $n+1$ But could you elaborate your conclusion? – Max Mar 17 at 11:25
• 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. – dcolazin Mar 17 at 11:27
• Thanks for your elaboration, I think I'll be able to understand it much better! PS. After refreshing linear dependance etc. – Max Mar 17 at 11:30
• @Max how do you define the rank? – dcolazin Mar 17 at 12:05
• The number of linearly independent columns... So thanks! – Max Mar 17 at 14:13