Prove that $e^{-A} = (e^{A})^{-1}$ Let $A, B \in R^{n \times n}$. Prove that $e^{-A} = (e^{A})^{-1}$.
($R$ is the real numbers)
I've tried messing around with both sides, evaluated as sums. I just can't get the two to match up.
Any ideas?
 A: Another less elegant way:
$$e^A = I+ A + \frac{A^2}{2!}+ \frac{A^3}{3!}+\cdots$$
$$e^{-A} = I - A + \frac{A^2}{2!}-\frac{A^3}{3!}+\cdots$$
$$e^A \, e^{-A} = I + \sum_{k=1}^\infty b_k A^k$$
where $$b_k =\sum_{j=0}^k \frac{(-1)^j}{j! (k-j)!}= \frac{1}{k!}\sum_{j=0}^k (-1)^j {k \choose j}=\frac{1}{k!}(1-1)^k =0$$
A: You should know that if $X,Y\in \Bbb R^{n\times n}$ and $XY=YX$, then $e^{X+Y}=e^Xe^Y$.
With this in mind, simply compute $e^Ae^{-A}$.
A: Well, if you allow that $e^{a + b} = e^a e^b$ and $e^0 = 1$, which are general properties of exponential functions and don't even really require anything special about $e$, in the sense that they're true for any base, then you don't need power series and you can simply note that
$e^x e^{-x} = e^{(x + (-x))} =  e^0 = 1, \tag{1}$
which immediately yields, upon multiplication by $(e^x)^{-1}$,
$(e^x)^{-1} = e^{-x}. \tag{2}$
If you don't yet have $e^{a + b} = e^a e^b$, I would try using the series $e^x =\sum_i (x^i / i!)$ to show it is true; it's an easier path, I suspect.  
A proof of $e^{a + b} = e^a e^b$ is given in Differential Equations, Dynamical Systems, and an,Introduction to Chaos, by Hirsch, Smale, and Devaney, chapter 6.  They do it for commuting matrices, but the series estimates are basically the same.
Note added in edit, Thursday 5 December 2013 3:14 PM PST: My answer was entered before the question was altered to address matrices, and I'm not going to re-type it now; but if anyone wants to see something which resolves the matrix case as well, check out my answer to $M,N\in \Bbb R ^{n\times n}$, show that $e^{(M+N)} = e^{M}e^N$ given $MN=NM$
End of note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
