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Help me please to prove that matrix exponential which defined as:

$e^{A}=\sum\limits_{k=0}^{\infty }\frac{A^{k}}{k!}$ converges for all matrix $A$

Thanks beforehand.

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Since the space $E:=\mathcal M_n(\mathbb C)$ of all $n \times n$ complex matrices is a finite-dimensional vector space, all norms define the same topology. So we can take a sub-multiplicative norm, that is, a norm $\lVert\cdot\rVert$ such that $\lVert AB\rVert\leq \lVert A\rVert \cdot\lVert B\rVert$. (For example, we can take $\lVert\cdot\rVert$ to be the operator norm on $E$.) As a finite dimensional vector space, $E$ is complete, so it's enough to show normal convergence. We have that, for each integer $n \geq 0$, $$0\leq \lVert\frac{A^n}{n!}\rVert\leq \frac{\lVert A\rVert^n}{n!},$$ and we know that, for each real number $x$, the series $\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ converges (it defines the exponential function). Therefore, for any $A \in E$, the series $\sum_{n=0}^{+\infty}\frac{A^n}{n!}$ converges. (We also got the additional result that $\lVert e^A\rVert\leq e^{\lVert A\rVert}$ for any $A \in E$.)

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Hint: Show that $\sum \|A_n\|<\infty$ implies that $\sum A_n$ is convergent.

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  • $\begingroup$ It will show that the sequence of the partial sum is Cauchy, since it is a Cauchy sequence in a Banach Space $\sum A_{n}$ will converge. $\endgroup$ – Sushant Dec 2 '20 at 14:33
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HINT: Use one of the sub-multiplicative matrix norms to show that the series of real numbers in each position of $\sum_{k\ge 0}\frac{A^k}{k!}$ converges.

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