Is there a real matrix A such that (Exponential of matrices) Is there a real matrix A such that
$$\exp(A) = \begin{bmatrix} -\alpha & 0 \\ 0 & -\beta \end{bmatrix}, \text{ where }\alpha,\beta>0?$$
(Hint): In two dimensions the exponential matrix of
$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
is given by:
$$\exp(A) = e^{\delta}\cos(\Delta)\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + e^{\delta}\frac{\sin(\Delta)}{\Delta}\begin{bmatrix} \phi & b \\ c & -\phi \end{bmatrix},$$
where $\delta=\dfrac{a+d}{2}$, $\phi=\dfrac{a-d}{2}$, $\Delta=\sqrt{\phi^2+bc}$.
Does this exercise only reduce the multiplication of the matrices given in the tip?
 A: $$
A =
\left(
\begin{array}{cc}
0 & \pi \\
- \pi & 0
\end{array}
\right)
$$
The point is this: by straightforward summing, we can find, for real number $t,$ given
$$
B =
\left(
\begin{array}{cc}
0 & t \\
- t & 0
\end{array}
\right) \; \; ,
$$
we get
$$
e^B =
\left(
\begin{array}{cc}
\cos t & \sin t \\
- \sin t & \cos t
\end{array}
\right) \; \; .
$$
That is, even powers of $B$ are diagonal, and odd powers of $B$ are off-diagonal, in fact skew symmetric. We are looking for
$$ e^B = I + B + \frac{1}{2} B^2 +  \frac{1}{6} B^3 +  \frac{1}{24} B^4+  \frac{1}{120} B^5+  \frac{1}{720} B^6 + \cdots  $$
We get the cosines on the diagonal from
$$  I +  \frac{1}{2} B^2 +    \frac{1}{24} B^4+  \frac{1}{720} B^6 + \cdots  $$
and the sines skew-symmetric from
$$  B +   \frac{1}{6} B^3 +   \frac{1}{120} B^5 + \cdots  $$
A: Let $A = PJP^{-1}$ be the complex Jordan normal form of $A$. Then, we have $\exp(A) = P\exp(J) P^{-1}$. 
The Jordan block structure is (up to permutation) unique. Hence, $\exp(J)$ is also diagonal and (up to permutation) has the diagonal $-\alpha, -\beta$. In particular $J$ is also diagonal with non real valued diagonal component. 
As $A$ is real, it follows both eigenvalues are not real, conjugated to each other, and distinct. 
Let $\lambda = x + yi$ denote an eigenvalue. Than, we have
$$ \exp(\lambda) = \underbrace{\exp(x)}_{>0} \exp(yi), $$
which is real and negative if and only if $\exp(yi) = -1$, that is $y=\pi \bmod 2\pi$.
Using the same logic on $\bar \lambda$, we obtain that $\alpha$ needs to be $\beta$.
So there is a solution $A$ if and only if $\alpha = \beta$. In that case $A$ is given by
$$ A = \log(\alpha) I + \log(-I), $$
where $\log(-I)$ is multi-valued.
Aside: 


*

*A canonic choice for $\log(-I)$ is given in @WillJagy's answer.

*Once you established $\alpha=\beta$, you can just interpret $A$ as a complex number $z = -\alpha + 0i$ and the question reduces to the complex logarithm.
A: The actual recipe given in Teschl is this, page 63. See that there are explicit notes about what to do when the square root will be of a negative quantity, $$  \cosh  i \Delta  = \cos \Delta \; ,  $$
$$ \frac{\sinh i \Delta}{i \Delta} = \frac{\sin \Delta}{\Delta} \; . $$
Using the symbols below, namely matrix $A,$ then $\delta = \frac{a+d}{2} \; , \; $ $\gamma = \frac{a-d}{2} \; . \; $ With real $t \neq 0,$
$$  \gamma^2 + bc < 0 \; ,  \; \; \mbox{LET} \; \;  t = \sqrt { \;- \left( \gamma^2 + bc\right) \; \; }  $$ 
THEN
$$
e^A =
e^\delta \cos t \; I \; + \; \frac{e^\delta \sin t}{t} \;
\left(
\begin{array}{cc}
\gamma & b \\
c & -\gamma
\end{array}
\right)
$$
Uses 
$$
\frac{1}{2ict}
\left(
\begin{array}{cc}
\gamma - it & \gamma + it \\
c & c
\end{array}
\right)
\left(
\begin{array}{cc}
\delta - it & 0 \\
0 & \delta + it
\end{array}
\right)
\left(
\begin{array}{cc}
-c & \gamma + it \\
c & - \gamma + it
\end{array}
\right) =
\left(
\begin{array}{cc}
a & b \\
c & d
\end{array}
\right)
$$

