Partial Differential Equations/The Malgrange-Ehrenpreis theorem

Vandermonde's matrix

For $x_1,\dots ,x_n$ pairwise different (i. e. $x_k\ne x_m$ for $k\ne m$) matrix is invertible, as the following theorem proves:

Proof:

We prove that $𝐁𝐀={𝐈}_n$, where ${𝐈}_n$ is the $n\times n$ identity matrix.

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

$x_m\prod _{\genfrac{}{}{0ex}{}{1\le l\le n}{l\ne k}}(x_l-x_m)=\sum _{j=1}^n{x}_m^j\{\begin{array}{cc}\sum _{\genfrac{}{}{0ex}{}{1\le l_1<\cdots <l_{n-j}\le n}{l_1,\dots ,l_{n-j}\ne k}}(-1{)}^{j-1}{x}_{l_1}\cdots x_{l_{n-j}}& j<n\\ 1& j=n\end{array}$.

Therefore, if ${𝐜}_{k,m}$ is the $k,m$-th entry of the matrix $𝐁𝐀$, then by the definition of matrix multiplication

${𝐜}_{k,m}=\sum _{j=1}^n\frac{x_m^j\{\begin{array}{cc}\sum _{\genfrac{}{}{0ex}{}{1\le l_1<\cdots <l_{n-j}\le n}{l_1,\dots ,l_{n-j}\ne k}}(-1{)}^{j-1}{x}_{l_1}\cdots x_{l_{n-j}}& j<n\\ 1& j=n\end{array}}{x_k\prod _{\genfrac{}{}{0ex}{}{1\le l\le n}{l\ne k}}(x_l-x_k)}=\frac{x_m\prod _{\genfrac{}{}{0ex}{}{1\le l\le n}{l\ne k}}(x_l-x_m)}{x_k\prod _{\genfrac{}{}{0ex}{}{1\le l\le n}{l\ne k}}(x_l-x_k)}=\{\begin{array}{cc}1& k=m\\ 0& k\ne m\end{array}$.$◻$

The Malgrange-Ehrenpreis theorem

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

$\left(\begin{array}{ccc}{x}_1& \cdots & x_n\\ x_1^2& \cdots & x_n^2\\ ⋮& \ddots & ⋮\\ x_1^n& \cdots & x_n^n\end{array}\right)$,

we end up with the equation

$\left(\begin{array}{c}{y}_1\\ ⋮\\ y_n\end{array}\right)=𝐁\left(\begin{array}{c}0\\ ⋮\\ 0\\ 1\end{array}\right)$.

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