Does the matrix logarithm always converge for exponential matrices?

Question

We have defined functions on square matrices $X$:

$$e^X := I + X + \dfrac{X^2}{2!}+\dfrac{X^3}{3!}$$ and $$log(X):= (X-I)-\dfrac{(X-I)^2}{2} + \dfrac{(X-I)^3}{3}-...$$

The exponential converges for all matrices, but the logarithm converges for all $X$ with norm (maximum eigenvalue) less than or equal to 1. The two can be shown to be inverses of each other whenever the logarithm is defined. We thus have $$e^{log(X)} = X$$ whenever $log(X)$ is defined. But also $$log(e^X)=X$$ whenever $log(e^X)$ is defined.

My question is, is log defined for all matrices which can be written in the form $e^X$?

Answer 1

No, just consider the $1 \times 1$ case, i.e. real numbers. Let $X=10$ and the series doesn't converge.

Answer 2

@ Owen , it is a bad way to start the study of log.

1.With your definition, $\log(X)$ exists if, for instance, $||X-I||_2<1$. In particular, $X$ must be invertible.

2. $\log(e^X)$ and $X$ are not necessarily equal because, for instance $e^{0_2}=e^U$ where $U=\begin{pmatrix}0&2\pi\\2\pi&0\end{pmatrix}$.

3. Your question is non-sense because $\log(e^X)$ is defined if $||e^X-I||_2<1$, that is a strong condition about the eigenvalues of $X$; indeed, if $\lambda >0$ is an eigenvalue of $X$, then necessarily $\lambda<\log(2)$.