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.

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