For a diagonal matrix $M$, what is $e^M$? For a diagonal matrix
$$
M=\left(\begin{array}{ccc}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right)
$$
show that
$$
e^M=\left(\begin{array}{ccc}
e^a & 0 & 0 \\
0 & e^b & 0 \\
0 & 0 & e^c
\end{array}\right).
$$
 A: Hint: $e^M$  is defined as the Taylor series. What does $M^2$ look like? Plug the powers of $M$ into the Taylor series.
A: $\operatorname{diag}:\begin{array}{ll}\Bbb R^n \to M_n\left(\Bbb R\right)\\\left(x_1,\dots,x_n\right)\mapsto \begin{pmatrix}x_1& 0 & \dots&\dots & 0\\
0&x_2&\ddots&&\vdots\\
\vdots&\ddots&\ddots &\ddots&\vdots\\
\vdots&&\ddots&x_{n-1}&0\\
0&\dots&\dots&0&x_n\end{pmatrix}\end{array}$

I'll hide the proofs so that you can try to show the results by yourself, just move your mouse over the blank to show them.


 $\forall \left(x_1,\dots,x_n\right),\left(y_1,\dots,y_n\right)\in \Bbb R^n, \forall \lambda \in \Bbb R,\\\operatorname{diag}\left(x_1\lambda y_1,\dots,x_n+\lambda y_n\right)\\=\begin{pmatrix}x_1+\lambda y_1& 0 & \dots&\dots & 0\\0&x_2+\lambda y_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&x_{n-1}+\lambda y_{n-1}&0\\0&\dots&\dots&0&x_n+\lambda y_n\end{pmatrix}\\=\begin{pmatrix}x_1& 0 & \dots&\dots & 0\\0&x_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&x_{n-1}&0\\0&\dots&\dots&0&x_n\end{pmatrix}+\lambda\begin{pmatrix}y_1& 0 & \dots&\dots & 0\\0&y_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&y_{n-1}&0\\0&\dots&\dots&0&y_n\end{pmatrix}\\=\operatorname{diag}\left(x_1,\dots,x_n\right)+\lambda\operatorname{diag}\left(y_1,\dots,y_n\right)$  

$\boxed{\operatorname{diag}\in \mathcal L\left(\Bbb R^n,M_n\left(\Bbb R\right)\right)}\tag{1}$


$\forall \left(x_1,\dots,x_n\right),\left(y_1,\dots,y_n\right)\in \Bbb R^n,\\\operatorname{diag}\left(x_1,\dots,x_n\right)\operatorname{diag}\left(y_1,\dots,y_n\right)\\=\begin{pmatrix}x_1& 0 & \dots&\dots & 0\\0&x_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&x_{n-1}&0\\0&\dots&\dots&0&x_n\end{pmatrix}\begin{pmatrix}y_1& 0 & \dots&\dots & 0\\0&y_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&y_{n-1}&0\\0&\dots&\dots&0&y_n\end{pmatrix}\\=\begin{pmatrix}x_1y_1& 0 & \dots&\dots & 0\\0&x_2y_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&x_{n-1}y_{n-1}&0\\0&\dots&\dots&0&x_ny_n\end{pmatrix}\\=\operatorname{diag}\left(x_1y_1,\dots,x_ny_n\right)$

$\boxed{\forall \left(x_1,\dots,x_n\right),\left(y_1,\dots,y_n\right)\in \Bbb R^n,\operatorname{diag}\left(x_1y_1,\dots,x_ny_n\right)=\operatorname{diag}\left(x_1,\dots,x_n\right)\operatorname{diag}\left(y_1,\dots,y_n\right)}\tag{2}$


 By recurrence and $(2)$:

$\boxed{\forall \left(x_1,\dots,x_n\right)\in \Bbb R^n, \forall p \in \Bbb N,\operatorname{diag}\left(x_1,\dots,x_n\right)^p=\operatorname{diag}\left(x_1^p,\dots,x_n^p\right)}\tag{3}$

$\exp:\begin{array}{ll}M_n\left(\Bbb R\right)\to M_n\left(\Bbb R\right)\\M \mapsto \sum\limits_{k=0}^{+\infty}\cfrac{M^k}{k!}\end{array}$


 $\forall \left(x_1,\dots,x_n\right)\in \Bbb R^n,\\\exp\left(\operatorname{diag}\left(x_1,\dots,x_n\right)\right)\\=\sum\limits_{k=0}^{+\infty}\cfrac{\operatorname{diag}\left(x_1,\dots,x_n\right)^k}{k!}\\=\lim\limits_{m\to +\infty}\sum\limits_{k=0}^{m}\cfrac{\operatorname{diag}\left(x_1,\dots,x_n\right)^k}{k!}\\=\lim\limits_{m\to +\infty}\sum\limits_{k=0}^{m}\cfrac{\operatorname{diag}\left(x_1^k,\dots,x_n^k\right)}{k!}\\=\lim\limits_{m\to +\infty}\operatorname{diag}\left(\sum\limits_{k=0}^{m}\cfrac{x_1^k}{k!},\dots,\sum\limits_{k=0}^{m}\cfrac{x_n^k}{k!}\right)\\=\operatorname{diag}\left(\lim\limits_{m\to +\infty}\sum\limits_{k=0}^{m}\cfrac{x_1^k}{k!},\dots,\lim\limits_{m\to +\infty}\sum\limits_{k=0}^{m}\cfrac{x_n^k}{k!}\right)\\=\operatorname{diag}\left(\sum\limits_{k=0}^{+\infty}\cfrac{x_1^k}{k!},\dots,\sum\limits_{k=0}^{+\infty}\cfrac{x_n^k}{k!}\right)\\=\operatorname{diag}\left(\exp\left(x_1\right),\dots,\exp\left(x_n\right)\right)$

Justification for each equality (try to find them yourself after reading the proof)

 The first inequality is the definition of $\exp$, the second is the definition of $\sum\limits_{k=0}^{+\infty}$, the third is $(3)$, the fourth is $(1)$, the fifth is the continuity of $\operatorname{diag}$ which comes from $(1)$ and the fact that both $\Bbb R^n$ and $M_n\left(\Bbb R\right)$ are finite dimensional, the sixth is the definition of $\sum\limits_{k=0}^{+\infty}$ and the seventh is the definition of $\exp$ over the real numbers.

$\boxed{\forall \left(x_1,\dots,x_n\right)\in \Bbb R^n,\exp\left(\operatorname{diag}\left(x_1,\dots,x_n\right)\right)=\operatorname{diag}\left(\exp\left(x_1\right),\dots,\exp\left(x_n\right)\right)}$
