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

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


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

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

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


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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.