Exponential of a 2x2 real matrix

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:

\[ e^A = e^{\frac{\text{Tr}A}{2}}\left(A + 1-\frac{\text {Tr}A}{2}\right) \quad(\Delta = 0). \tag{8}\]