Editing Polynomial interpolation (section)

===Second proof===
Write out the interpolation polynomial in the form
{{NumBlk||<math display="block">p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0. </math>|{{EquationRef|1}}}}
Substituting this into the interpolation equations <math>p(x_j) = y_j</math>, we get a [[system of linear equations]] in the coefficients <math>a_j</math>, which reads in matrix-vector form as the following [[Matrix multiplication|multiplication]]:
<math display="block">\begin{bmatrix}
x_0^n  & x_0^{n-1} & x_0^{n-2} & \ldots & x_0 & 1 \\
x_1^n  & x_1^{n-1} & x_1^{n-2} & \ldots & x_1 & 1 \\
\vdots & \vdots    & \vdots    &        & \vdots & \vdots \\
x_n^n  & x_n^{n-1} & x_n^{n-2} & \ldots & x_n & 1
\end{bmatrix}
\begin{bmatrix} a_n \\ a_{n-1} \\ \vdots \\ a_0 \end{bmatrix}  =
\begin{bmatrix} y_0 \\ y_1 \\ \vdots \\ y_n \end{bmatrix}.</math>

An interpolant <math>p(x)</math> corresponds to a solution <math>A = (a_n,\ldots,a_0)</math> of the above matrix equation <math>X \cdot A = Y</math>.  The matrix ''X'' on the left is a [[Vandermonde matrix]], whose determinant is known to be <math>\textstyle \det(X) = \prod_{1 \le i < j \le n} (x_j - x_i), </math> which is non-zero since the nodes <math>x_j</math> are all distinct. This ensures that the matrix is [[Invertible matrix|invertible]] and the equation has the unique solution <math>A = X^{-1}\cdot Y</math>; that is, <math>p(x)</math> exists and is unique.