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For a diagonal matrix $$ M=\left(\begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right) $$ show that $$ e^M=\left(\begin{array}{ccc} e^a & 0 & 0 \\ 0 & e^b & 0 \\ 0 & 0 & e^c \end{array}\right). $$

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Welcome to MSE! It really helps readability to format questions using MathJax (see FAQ). Your question is very difficult to read in its current form. What have you tried and what are your thoughts? Is this homework? If so, it should be tagged as such. Regards –  Amzoti May 16 '13 at 15:14
    
Expanding the exponentials on the left and right hand side each might have gotten you pretty close. –  Brady Trainor May 16 '13 at 16:07
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2 Answers 2

Hint: $e^M$ is defined as the Taylor series. What does $M^2$ look like? Plug the powers of $M$ into the Taylor series.

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$\operatorname{diag}:\begin{array}{ll}\Bbb R^n \to M_n\left(\Bbb R\right)\\\left(x_1,\dots,x_n\right)\mapsto \begin{pmatrix}x_1& 0 & \dots&\dots & 0\\ 0&x_2&\ddots&&\vdots\\ \vdots&\ddots&\ddots &\ddots&\vdots\\ \vdots&&\ddots&x_{n-1}&0\\ 0&\dots&\dots&0&x_n\end{pmatrix}\end{array}$


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.

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

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