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Given a $n\times n$ real (complex) matrix $A$. Let me define:

$$\exp A=\sum_{n=0}^\infty \frac{A^n}{n!}$$

and

$$\ln A=\sum_{n=1}^\infty (-1)^{n+1}\frac{(A-I)^n}{n}$$

Let assume that the $2$ above series converge for $A=M$. How can I prove that:

$$\exp(\ln M)=M$$

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It's simplest for matrices with all eigenvalues distinct. If $v$ is an eigenvector of $M$ for eigenvalue $\lambda$, then $(\ln M) v = \sum_n (-1)^{n+1} \dfrac{(\lambda - 1)^n}{n} v$ converges, and what it converges to must be $\ln(\lambda) v$ (there's a subtle point here about convergence of power series on the boundary of the disk of convergence: see Abel's theorem). Thus $v$ is also an eigenvector of $\ln M$ for eigenvalue $\ln \lambda$. Then $\exp(\ln M) v = \exp(\ln \lambda) v = \lambda v = M v$. The eigenvectors forming a basis, we conclude that $\exp(\ln M) = M$.

In the more general case, you can try using Jordan canonical form, but I think it gets complicated. Alternatively, use the fact that matrices with distinct eigenvalues form a dense set.

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    $\begingroup$ Can I show that the infinite polynomial $\exp\ln(x)$ is equal to $x$ in the case of convergence? Then I just put $M$ into the polynomial... $\endgroup$ – quangtu123 Sep 12 '14 at 15:57
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Treating $\exp(x)$ and $\ln(1+x)$ as formal power series, since $\ln(1+x)$ has no constant term, one may compose these to obtain formal power series $\exp(\ln(1+x))$. Series for $\exp(x)$ and $\ln(1+x)$ have nonzero radius of convergence, therefore $\exp(\ln(1+x))$ has nonzero radius of convergence.

Moreover, we know that $\exp(\ln(1+x)) = 1+x$ if we treat these as functions from $\mathbb{R}$ to $\mathbb{R}$. Therefore the power series for $\exp(\ln(1+x))$ is just $1+x$.

What follows is that $\exp(\ln(1+A)) = 1 + A$ for all matrices $A$ in some neighbourhood of the zero matrix (i.e. in a ball of radius $r$ which is less than the radius of convergence of $\ln(1+x)$ and the radius of convergence of $\exp(\ln(1+x))$).

This argument holds for matrices as well as any other Banach algebras.

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