proof of Taylor’s formula for matrix functions

Theorem.

Let $p$ be a polynomial and suppose $\mathrm{A}$ and $\mathrm{B}$are squared matrices of the same size, then $p\mathit{}\mathrm{(}\mathrm{A}\mathrm{+}\mathrm{B}\mathrm{)}\mathrm{=}{\displaystyle \mathrm{\sum }_{k\mathrm{=}\mathrm{0}}^n}{\displaystyle \frac{\mathrm{1}}{k\mathrm{!}}}\mathit{}{p}^{\mathrm{(}k\mathrm{)}}\mathit{}\mathrm{(}\mathrm{A}\mathrm{)}\mathit{}{\mathrm{B}}^k$where $n\mathrm{=}\mathrm{deg}\mathit{}\mathrm{(}p\mathrm{)}$.

Proof.

Since $p$ is a polynomial, we can apply the Taylor expansion:

$$p(x)=\sum _{k=0}^n\frac{1}{k!}{p}^{\left(k\right)}\left(x_0\right){\left(x-x_0\right)}^k$$

where $n=\mathrm{deg}(p)$. Now let $x=𝐀+𝐁$ and$x_0=𝐀$.

The Taylor expansion can be checked as follows: let$p\left(x\right)={\sum }_{k=0}^n{a}_k{x}^k$ for coefficients $a_k$(note that this coefficients can be taken from the space of squarematrices defined over a field). We define the formal derivative ofthis polynomial as $p^{\left(1\right)}\left(x\right)=\frac{dp}{dx}={\sum }_{k=1}^n{a}_{k}k{x}^{k-1}$ and we define$p^{\left(k\right)}=\frac{d{p}^{\left(k-1\right)}}{dx}$.

Then$p^{\left(k\right)}\left(x\right)={\sum }_{i=k}^n{a}_i\frac{i!}{\left(i-k\right)!}{x}^{i-k}$and we have$\frac{1}{k!}{p}^{\left(k\right)}\left(x_0\right)={\sum }_{i=k}^n{a}_i\frac{i!}{\left(i-k\right)!k!}{\left(x_0\right)}^{i-k}$.Now consider

${\displaystyle \sum _{k=0}^n}{\displaystyle \frac{1}{k!}}{p}^{\left(k\right)}\left(x_0\right){\left(x-x_0\right)}^k={\displaystyle \sum _{k=0}^n}\left({\displaystyle \sum _{i=k}^n}{a}_i{\displaystyle \frac{i!}{\left(i-k\right)!k!}}{\left(x_0\right)}^{i-k}{\left(x-x_0\right)}^k\right)$

$=$${\displaystyle \sum _{i=0}^n}{a}_i{\left(x_0\right)}^i+{\displaystyle \sum _{i=1}^n}{a}_{i}i{\left(x_0\right)}^{i-1}\left(x-x_0\right)+\mathrm{\dots }+{\displaystyle \sum _{i=j}^n}{a}_i{\displaystyle \frac{i!}{\left(i-j\right)!j!}}{\left(x_0\right)}^{i-j}{\left(x-x_0\right)}^j+\mathrm{\dots }+a_n{\left(x-x_0\right)}^n$

$=$$a_0+a_1\left(x\right)+\mathrm{\dots }+a_i\left({\displaystyle \sum _{j=0}^i}{\displaystyle \frac{i!}{\left(i-j\right)!j!}}{\left(x_0\right)}^{i-j}{\left(x-x_0\right)}^j\right)+\mathrm{\dots }+a_n\left({\displaystyle \sum _{j=0}^n}{\displaystyle \frac{n!}{\left(n-j\right)!j!}}{\left(x_0\right)}^{n-j}{\left(x-x_0\right)}^j\right)$

$=$${\displaystyle \sum _{k=0}^n}{a}_k{x}^i=p\left(x\right)$

since${\sum }_{j=0}^i\frac{i!}{\left(i-j\right)!j!}{\left(x_0\right)}^{i-j}{\left(x-x_0\right)}^j={\left(x\right)}^i$.∎