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With a diagonal matrix $D$ and exponential power series expansions:

\begin{align*} D &= \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \cdots & \lambda_n \\ \end{pmatrix} \\ e^{Dt} &= \sum\limits_{n=0}^\infty \frac{1}{n!} (Dt)^n \\ e^{\lambda_i t} &= \sum\limits_{n=0}^\infty \frac{1}{n!} (\lambda_i t)^n \\ \end{align*}

Show that:

\begin{align*} e^{Dt} &= \begin{pmatrix} e^{\lambda_1 t} & 0 & \cdots & 0 \\ 0 & e^{\lambda_2 t} & \cdots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \cdots & e^{\lambda_n t} \\ \end{pmatrix} \\ \end{align*}

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  • $\begingroup$ Hint: what happens when you take the $n$th power of a diagonal matrix? $\endgroup$ – Theo Bendit Mar 25 '18 at 1:20
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This can be shown by direct calculation. You have used n twice and that has caused a little confusion. I changed the order of $D$ to $k$. $$D^n = \begin{pmatrix} \lambda_1^n & 0 & \cdots & 0 \\ 0 & \lambda_2^n & \cdots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \cdots & \lambda_k^n \\ \end{pmatrix} \\$$ So we have $$e^{Dt} = \sum\limits_{n=0}^\infty \frac{1}{n!} (Dt)^n = \sum\limits_{n=0}^\infty \frac{1}{n!} \begin{pmatrix} (\lambda_1t)^n & 0 & \cdots & 0 \\ 0 & (\lambda_2t)^n & \cdots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \cdots & (\lambda_kt)^n \\ \end{pmatrix}$$ Since k is finite, we can move the infinite sum inside our matrix, which yields $$e^{Dt} = \begin{pmatrix} \sum\limits_{n=0}^\infty \frac{1}{n!}(\lambda_1t)^n & 0 & \cdots & 0 \\ 0 & \sum\limits_{n=0}^\infty \frac{1}{n!}(\lambda_2t)^n & \cdots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \cdots & \sum\limits_{n=0}^\infty \frac{1}{n!}(\lambda_kt)^n \\ \end{pmatrix}=\begin{pmatrix} e^{\lambda_1 t} & 0 & \cdots & 0 \\ 0 & e^{\lambda_2 t} & \cdots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \cdots & e^{\lambda_k t} \\ \end{pmatrix}$$

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