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Let $A$ be a square matrix. To show: Matrix exponential converges to some matrix $X$.

$$ \lim_{N \rightarrow \infty} \sum_{k=0}^{N}\frac{A^k}{k!} =X $$

In some proofs that I have seen it is stated that because (for a sub-multiplicative norm) $$ 0 \le \sum_{k=0}^{\infty} \left\Vert \frac{A^k }{k!} \right\Vert \le  \sum_{k=0}^{N} \frac{\Vert A \Vert ^k }{k!}<\infty , $$ then the series $\sum_{k=0}^{N}\frac{A^k}{k!}$ has to be convergent. That however isn't clear to me.

To me more intuitive way to show convergence would be to show that $$ \lim_{N \rightarrow \infty} \left\Vert \sum_{k=0}^{N} \frac{A^k}{k!} -X \right\Vert  =0$$ and use some intuitive matrix norm for which it is clear that all elements of $\frac{A^k}{k!} -X$ converge to zero.

Any hints?

Axel
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mathslover
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  • Does this help: https://math.stackexchange.com/questions/131013/matrix-exponential-convergence – Stuck Jun 04 '21 at 19:34
  • It's usually easier to show that a sequence is Cauchy (with respect to some complete metric) than to show it is convergent. – Hayden Jun 04 '21 at 19:34
  • The matrix $x$ is $e^A$ which is defined as the infinite sum, so you end up having to show that $\sum_{k=n}^\infty\frac{A^k}{k!}\to )$ as $n\to\infty$ – saulspatz Jun 04 '21 at 19:35
  • You have to convince yourself (convince yourself again and again!) that finite dimensional real spaces are complete... and that therefore Cauchy sequences converge. For this, you need to look in details at norms (that are all equivalent) and the completness of the reals. – mathcounterexamples.net Jun 04 '21 at 19:44

1 Answers1

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Consider the largest element of $A$ (in absolute value), let $b$, and consider the matrix $B$ filled with all $b$'s. We have $|A|\le B$ element-wise, then

$$|A^2|\le dbB$$

where $d$ is the order of the matrix, and by induction,

$$|A^n|\le d^{n-1}b^{n-1}B.$$

In other words,

$$|a_{ij}^{(n)}|\le d^{n-1}b^n$$ and $$\frac{|a_{ij}^{(n)}|}{k!}\le\frac{d^{n-1}b^n}{k!}.$$

Hence

$$\left|\sum_{k=0}^\infty\frac{a_{ij}^{(n)}}{k!}\right|\le\sum_{k=0}^\infty\frac{|a_{ij}^{(n)}|}{k!}\le\frac{e^{db}}d$$ converges for every $ij$.

  • Thanks! Can you clarify what $\left| A^2 \right| \le dbB$ means? Isn't $  \left| A^2 \right|$ a number and $B$ a matrix? That inequality doesn't make sense to me. – mathslover Jun 05 '21 at 06:56
  • @mathslover: I didn't mean $|A^2|$. Element-wise. –  Jun 05 '21 at 10:24