Compute the Jordan form of a particular matrix

Question

Compute $e^S$, where $$ S = \pmatrix{ \frac 12 \ln(\alpha^2 + \beta^2) & -\arctan(\frac{\beta}{\alpha})\\ \arctan(\frac{\beta}{\alpha}) & \frac 12 \ln(\alpha^2 + \beta^2) } . $$

In order to do so I need to calculate the Jordan form of $S$. Can anybody help me with that?

Answer 1

Note that $S = D+J$ where $D = \frac 12 \ln(\alpha^2 + \beta^2) I$ and $$ J = \arctan(\beta/\alpha) \pmatrix{0&-1\\1&0} $$ since $D$ and $J$ commute, we have $e^S = e^De^J$. You can calculate both directly (via the power series) to find $$ e^D = \exp[\frac 12 \ln(\alpha^2 + \beta^2)]I = \sqrt{\alpha^2 + \beta^2}I\\ e^J = \pmatrix{\cos \arctan(\beta /\alpha) & -\sin \arctan(\beta/\alpha)\\ \sin \arctan (\beta/\alpha) & \cos \arctan (\beta/\alpha)} = \frac{1}{\sqrt{\alpha^2 + \beta^2}} \pmatrix{\alpha & -\beta\\\beta & \alpha} $$ Putting it all together, we calculate $$ e^S = \pmatrix{\alpha & -\beta \\ \beta & \alpha} $$

Answer 2

Here's a solution that uses some basic facts about Lie groups to bypass computing the Jordan canonical form (and actually, more or less avoid explicit computation altogether):

From a first complex analysis course, we recognize the entries $\tfrac{1}{2}\log(\alpha^2 + \beta^2)$ and $\arctan \frac{\beta}{\alpha}$ respectively as the real and imaginary parts of $\log (\alpha + i \beta)$ (at least for $a > 0$, which we assume for the moment, and for a suitable branch cut).

Now, the map $\Phi: \Bbb C^{\times} \to GL(2, \Bbb R)$ defined by $$x + i y \mapsto \pmatrix{x & -y \\ y & x}$$ is an Lie group homomorphism, and its derivative, $\phi : \Bbb C \to {\frak gl}(2, \Bbb R)$ is an Lie algebra homomorphism. Since $\Phi$ is the restriction of an ($\Bbb R$-)linear map, $\phi$ is given by the same formal formula, and in particular, $S = \phi(\log(\alpha + i \beta))$.

As for any Lie group homomorphism, the corresponding exponential maps $\Bbb C \to \Bbb C^{\times}$ (which is just the usual complex exponentiation map) and ${\frak gl}(2, \Bbb R) \to GL(2, \Bbb R)$ (which is just matrix exponentiation) are related by $$\exp \circ \,\phi = \Phi \circ \exp.$$ Putting this all together, we get $$\exp S = \exp \phi(\log(\alpha + i \beta)) = \Phi(\exp \log (\alpha + i \beta)) = \Phi(\alpha + i \beta) = \pmatrix{\alpha & -\beta \\ \beta & \alpha} .$$

Now, if $\alpha < 0$, $\arctan \frac{\beta}{\alpha}$ differs from $\arg(\alpha + i \beta) = \Im \log(\alpha + i \beta)$ by $\pi$ (again for a suitable choice of branch cut), the upshot of which is that for $\alpha < 0$ we have $$\exp S = -\pmatrix{\alpha & -\beta \\ \beta & \alpha} .$$