Partial Differential Equations/The Malgrange-Ehrenpreis theorem

For ${\displaystyle x_{1},\ldots ,x_{n}}$ pairwise different (i. e. ${\displaystyle x_{k}\neq x_{m}}$ for ${\displaystyle k\neq m}$) matrix is invertible, as the following theorem proves:

Proof:

We prove that ${\displaystyle \mathbf {B} \mathbf {A} =\mathbf {I} _{n}}$, where ${\displaystyle \mathbf {I} _{n}}$ is the ${\displaystyle n\times n}$ identity matrix.

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

${\displaystyle x_{m}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{m})=\sum _{j=1}^{n}x_{m}^{j}{\begin{cases}\sum _{1\leq l_{1}<\cdots <l_{n-j}\leq n \atop l_{1},\ldots ,l_{n-j}\neq k}(-1)^{j-1}x_{l_{1}}\cdots x_{l_{n-j}}&j<n\\1&j=n\end{cases}}}$.

Therefore, if ${\displaystyle \mathbf {c} _{k,m}}$ is the ${\displaystyle k,m}$-th entry of the matrix ${\displaystyle \mathbf {B} \mathbf {A} }$, then by the definition of matrix multiplication

${\displaystyle \mathbf {c} _{k,m}=\sum _{j=1}^{n}{\frac {x_{m}^{j}{\begin{cases}\sum _{1\leq l_{1}<\cdots <l_{n-j}\leq n \atop l_{1},\ldots ,l_{n-j}\neq k}(-1)^{j-1}x_{l_{1}}\cdots x_{l_{n-j}}&j<n\\1&j=n\end{cases}}}{x_{k}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{k})}}={\frac {x_{m}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{m})}{x_{k}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{k})}}={\begin{cases}1&k=m\\0&k\neq m\end{cases}}}$.${\displaystyle \Box }$

Proof:

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

${\displaystyle {\begin{pmatrix}x_{1}&\cdots &x_{n}\\x_{1}^{2}&\cdots &x_{n}^{2}\\\vdots &\ddots &\vdots \\x_{1}^{n}&\cdots &x_{n}^{n}\end{pmatrix}}}$,

we end up with the equation

${\displaystyle {\begin{pmatrix}y_{1}\\\vdots \\y_{n}\end{pmatrix}}=\mathbf {B} {\begin{pmatrix}0\\\vdots \\0\\1\end{pmatrix}}}$.

Calculating the last expression directly leads to the desired formula.${\displaystyle \Box }$