Partial Differential Equations/The Malgrange-Ehrenpreis theorem

Vandermonde's matrix

For $x_{1}, \dots, x_{n}$ 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 $𝐁 𝐀 = 𝐈_{n}$ , where $𝐈_{n}$ is the $n \times n$ identity matrix.

Let $1 \leq k, m \leq n$ . We first note that, by direct multiplication,

x_{m} \prod_{\binom{1 \leq l \leq n}{l \neq k}} (x_{l} - x_{m}) = \sum_{j = 1}^{n} x_{m}^{j} {\begin{matrix} \sum_{\binom{1 \leq l_{1} < \dots < l_{n - j} \leq n}{l_{1}, \dots, l_{n - j} \neq k}} (- 1)^{j - 1} x_{l_{1}} \dots x_{l_{n - j}} & j < n \\ 1 & j = n \end{matrix}

.

Therefore, if $𝐜_{k, m}$ is the $k, m$ -th entry of the matrix $𝐁 𝐀$ , then by the definition of matrix multiplication

𝐜_{k, m} = \sum_{j = 1}^{n} \frac{x_{m}^{j} {\begin{matrix} \sum_{\binom{1 \leq l_{1} < \dots < l_{n - j} \leq n}{l_{1}, \dots, l_{n - j} \neq k}} (- 1)^{j - 1} x_{l_{1}} \dots x_{l_{n - j}} & j < n \\ 1 & j = n \end{matrix}}{x_{k} \prod_{\binom{1 \leq l \leq n}{l \neq k}} (x_{l} - x_{k})} = \frac{x_{m} \prod_{\binom{1 \leq l \leq n}{l \neq k}} (x_{l} - x_{m})}{x_{k} \prod_{\binom{1 \leq l \leq n}{l \neq k}} (x_{l} - x_{k})} = {\begin{matrix} 1 & k = m \\ 0 & k \neq m \end{matrix}

.

◻

Proof:

We multiply both sides of the equation by $𝐁$ on the left, where $𝐁$ is as in theorem 10.2, and since $𝐁$ is the inverse of

(\begin{matrix} x_{1} & \dots & x_{n} \\ x_{1}^{2} & \dots & x_{n}^{2} \\ ⋮ & ⋱ & ⋮ \\ x_{1}^{n} & \dots & x_{n}^{n} \end{matrix})

,

we end up with the equation

(\begin{matrix} y_{1} \\ ⋮ \\ y_{n} \end{matrix}) = 𝐁 (\begin{matrix} 0 \\ ⋮ \\ 0 \\ 1 \end{matrix})

.

Calculating the last expression directly leads to the desired formula. $◻$