How do I see that the infinite sum corresponding to the matrix exponential converges?

Question

How do I see that the infinite sum corresponding to the matrix exponential converges ?

I know that the power series of $$e^z = \sum_{k = 0}^{\infty} \frac {z^k} {k!}$$ has infinite range of convergence.

But $$e^A = \sum_{k = 0}^{\infty} \frac {A^k} {k!} = I_n + \frac {A A} 2 + \frac {A A A} 6 \ldots$$ is not exactly on the above form or is it ?

I've seen this done using Jordan normal form. Write $A=PJP^{-1}$ then it can be shown fairly easily that $e^A=Pe^JP^{-1}$, then $e^A$ converges if $e^J$ converges. There is a standard way of calculating $J^n$ using Jordan Blocks, so this can be reduced to showing that each Jordan Block $e^{J_i}$ converges. This also gives an explicit formula for $e^A$. — Robert Chamberlain, Feb 27 '15 at 11:47
This seems to have a good solution: http://math.stackexchange.com/questions/33851/how-to-calculate-the-matrix-exponential-explicitly-for-a-matrix-which-isnt-diag — Robert Chamberlain, Feb 27 '15 at 11:48

score 4 · Answer 1 · answered Feb 27 '15 at 11:44

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The exponential of a matrix is defined exactly as: $$ e^A\triangleq\sum_{k\geq 0}\frac{A^k}{k!} $$ and it is converging since for any induced matrix norm $\| \cdot\|$ we have: $$ \| e^A \| \leq \sum_{k\geq 0}\frac{\|A\|^k}{k!} = e^{\|A\|}. $$

answered Feb 27 '15 at 11:44

Jack D'Aurizio

361,689

What if you don't know this norm, is there a way show to show algebraically that the sum converges entry-wise? – Andreas Lykke Iversen Feb 27 '15 at 12:13
@AndreasLykkeIversen: this norm is just any induced norm. As an alternative, you can just compute the Jordan normal form of $e^A$ given the Jordan normal form of $A$. The eigenvalues of $e^A$ are just the exponential of the eigenvalues of $A$, hence entry-wise convergence is pretty obvious. – Jack D'Aurizio Feb 27 '15 at 12:19
How can one write equality as you do, if one doesn't know if $e^A$ actually exist ? – Andreas Lykke Iversen Feb 27 '15 at 12:43
@AndreasLykkeIversen: consider $$\sum_{k=0}^{N}\frac{A^k}{k!}$$ and show that inequality holds for every $N$. – Jack D'Aurizio Feb 27 '15 at 12:59
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You only have proven that the partial sums stay bounded in the matrix norm. – Christian Blatter Feb 27 '15 at 15:21
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@ChristianBlatter: but that gives that $\sum_{k=0}^{N}\frac{A^k}{k!}$ is a Cauchy sequence. Total convergence implies uniform convergence. – Jack D'Aurizio Feb 27 '15 at 15:24

score 2 · Answer 2 · answered Feb 27 '15 at 14:58

To be more precise, a way to show it is to show that $(S_n)_{n\geq 0}=\left(\sum_{k\leq n}\frac{A^k}{k!}\right)_{n\geq 0}$ is a Cauchy sequence.

Let $\|.\|$ be any submultiplicative matrix norm (for example an induced norm like the spectral norm). If $p<q$, the triangle inequality and the submultiplicativity lead to

$$ \| S_q - S_p \| = \|\sum_{k=p+1}^q\frac{A^k}{k!}\| \leq \sum_{k=p+1}^q\frac{\|A\|^k}{k!} \longrightarrow_{(p,q\rightarrow \infty)} 0 $$

since $\left(\sum_{k\leq n}\frac{\|A\|^k}{k!}\right)_{n\geq 0}$ is Cauchy. Therefore, since the set of matrices is complete (with any of its equivalent norms), $(S_n)_{n\geq 0}$ will converge to a limit that we can name $e^A$.

How do I see that the infinite sum corresponding to the matrix exponential converges?

2 Answers2