Let \(A\) be a 2x2 real matrix, with eigenvalues:
\[ \lambda _{\pm} = \frac{\text {Tr}A}{2}\pm \sqrt{\Delta}, \qquad \Delta \equiv \left( \frac{\text {Tr}A}{2} \right )^2-\det A \tag{1}\]
If \(\Delta \neq 0\), \(A\) has two distinct eigenvalues, and we can decompose \(A\) as follows:
\[ A = \lambda _+ P_++\lambda _-P_-, \tag{2}\]
with:
\[ P_\pm =\pm\dfrac{1}{\lambda _+ - \lambda _-}(A-\lambda _\mp), \tag{3}\]
that satisfy:
\[ (P_\pm)^2 = 1, \quad P_+ P_- = P_- P_+ = 0. \tag{4}\]
Using Equations 3 and 4, one can immediately compute the exponential:
\[ e^A = e^{\lambda _+} P_+ + e^{\lambda _-} P_- \qquad (\Delta \neq 0) \tag{5}\]
We can obtain the case \(\Delta = 0\) from the \(\Delta \to 0\) limit of Equation 5. Observing that:
\[ P_+ + P_- = 1, \qquad P_+-P_- = \frac{1}{\sqrt{\Delta}}(A-\frac{\text {Tr}A}{2}), \tag{6}\]
from Equation 5 we find:
\[ e^{-\frac{\text{Tr}A}{2}}e^A = A + 1-\frac{\text {Tr}A}{2}+O(\sqrt \Delta) \quad(\Delta \to 0), \tag{7}\]
that yields:
\[ e^A = e^{\frac{\text{Tr}A}{2}}\left(A + 1-\frac{\text {Tr}A}{2}\right) \quad(\Delta = 0). \tag{8}\]
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@online{gherardi2024,
author = {Gherardi, Valerio},
title = {Exponential of a 2x2 Real Matrix},
date = {2024-11-04},
url = {https://vgherard.github.io/posts/2024-11-04-exponential-of-a-2x2-matrix/exponential-of-a-2x2-real-matrix.html},
langid = {en}
}