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For a square matrix $A$. Define $exp(A)=I+\sum_{n}A^{n}/(n!)$ . I need to prove two things

  1. exp(A) converges and is invertible.
  2. Its inverse is given by exp(-A).

Second part is straightforward. Can anyone help on first part?

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Can you use Schur decomposition? – Tunococ Nov 27 '12 at 20:22
ok, you mean to prove convergernce with the resulting triangular matrix? – dineshdileep Nov 28 '12 at 2:39
up vote 5 down vote accepted

We can show convergence by normal convergence.

$\sum_{k=0}^{\infty} ||\frac{A^k}{k!}|| \le \sum_{k=0}^{\infty} \frac{||A||^k}{k!} = e^{||A||}$. Where $||\cdot||$ is a sub-multiplicative norm (e.g. operator norm)

For invertibility, there's probably a nicer way to show this, but I would just show that if $A$ and $B$ commute, then $e^{A+B} = e^{A}e^{B}$.

$e^{A+B} = \sum_{k=0}^{\infty} \frac{(A+B)^k}{k!} = \sum_{k=0}^{\infty} \sum_{i=0}^k \frac{A^i B^{k-i}}{k!} \binom{k}{i} = \sum_{k=0}^{\infty} \sum_{i=0}^k \frac{A^i B^{k-i}}{i! (k-i)!} = \sum_{k=0}^{\infty} \sum_{i+h=k} \frac{A^i B^{h}}{i! h!} = \sum_{i=0}^{\infty} \sum_{h=0}^{\infty} \frac{A^i B^{h}}{i! h!} = e^A e^B$

$A$ and $-A$ commute so $e^A e^{-A}= e^{A - A} = e^{0} = I$

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+1 Beat me to it. For convergence work we must assume that the entries of $A$ are either real or complex numbers, but that is not unreasonable. It fails for e.g. $A=I$ in the $p$-adics :-) – Jyrki Lahtonen Nov 27 '12 at 20:57

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