<s>[INST] 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?
 [/INST] $$
A =
\left(
\begin{array}{cc}
0 & \pi \\
- \pi & 0
\end{array}
\right)
$$
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
$$  B +   \frac{1}{6} B^3 +   \frac{1}{120} B^5 + \cdots  $$
</s>