Logarithm of a matrix $A \in M_2 (\mathbb{C})$

Question

If ($\{a,b,c,d\}\subset \mathbb{R}$)$$ e^A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$$ how do I get the matrix $A\in M_2(\mathbb{C})$? And if is $b=c=0$?

Answer 1

If $A$ is real and if you are so lucky that the eigenvalues of $A$ are not in $(-\infty,0]$, then there is a real matrix $B$ s.t. $e^B=A$ and $B$ is a polynomial in $A$; morover we can easily calculate it.

Proof. We consider the principal logarithm

$\log:re^{i\theta}\in \mathbb{C}\setminus (-\infty,0]\rightarrow \log(r)+i\theta$ where $\theta\in(-\pi,\pi)$.

Case 1. $A$ has two $>0$ distinct eigenvalues $\lambda_j$; let $P(x)=ax+b$ be the real polynomial that sends $\lambda_j$ to $\log(\lambda_j)$.

Then take $B=aA+bI$.

Case 2. $A$ has two non-real conjugate eigenvalues $r^{i\theta},re^{-i\theta}$ where $r>0,\theta\not= 0$. In the same way as above, we find the real coefficients.

$a=\theta)/(r\sin(\theta)),b=\log(r)-(\theta\cos(\theta))/\sin(\theta)$.

Case 3. $A$ has a double $>0$ eigenvalue $\lambda$. Then $A=\lambda(I_2+N)$ where $N^2=0$.

Take $B=\log(\lambda)I+N=\log(\lambda)I+(1/\lambda) A-I$.