Partial Differential Equations/The Malgrange-Ehrenpreis theorem

Vandermonde's matrix
For $$x_1, \ldots, x_n$$ pairwise different (i. e. $$x_k \neq x_m$$ for $$k \neq m$$) matrix is invertible, as the following theorem proves:

Proof:

We prove that $$\mathbf B \mathbf A = \mathbf I_n$$, where $$\mathbf I_n$$ is the $$n \times n$$ identity matrix.

Let $$1 \le k, m \le n$$. We first note that, by direct multiplication,
 * $$x_m \prod_{1 \le l \le n \atop l \neq k} (x_l - x_m) = \sum_{j=1}^n x_m^j \begin{cases}

\sum_{1 \le l_1 < \cdots < l_{n-j} \le n \atop l_1, \ldots, l_{n-j} \neq k} (-1)^{j-1} x_{l_1} \cdots x_{l_{n-j}} & j < n \\ 1 & j = n \end{cases}$$.

Therefore, if $$\mathbf c_{k, m}$$ is the $$k, m$$-th entry of the matrix $$\mathbf B \mathbf A$$, then by the definition of matrix multiplication
 * $$\mathbf c_{k, m} = \sum_{j=1}^n \frac{x_m^j \begin{cases}

\sum_{1 \le l_1 < \cdots < l_{n-j} \le n \atop l_1, \ldots, l_{n-j} \neq k} (-1)^{j-1} x_{l_1} \cdots x_{l_{n-j}} & j < n \\ 1 & j = n \end{cases}}{x_k \prod_{1 \le l \le n \atop l \neq k} (x_l - x_k)} = \frac{x_m \prod_{1 \le l \le n \atop l \neq k} (x_l - x_m)}{x_k \prod_{1 \le l \le n \atop l \neq k} (x_l - x_k)} = \begin{cases} 1 & k = m \\ 0 & k \neq m \end{cases}$$.

The Malgrange-Ehrenpreis theorem
Proof:

We multiply both sides of the equation by $$\mathbf B$$ on the left, where $$\mathbf B$$ is as in theorem 10.2, and since $$\mathbf B$$ is the inverse of
 * $$\begin{pmatrix}

x_1 & \cdots & x_n \\ x_1^2 & \cdots & x_n^2 \\ \vdots & \ddots & \vdots \\ x_1^n & \cdots & x_n^n \end{pmatrix}$$, we end up with the equation
 * $$\begin{pmatrix}

y_1 \\ \vdots \\ y_n \end{pmatrix}

=

\mathbf B

\begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix}$$. Calculating the last expression directly leads to the desired formula.