Determine the image of the unit circle $S^1$ by the action of the matrix $e^A$. We have:
$$e^{ \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} }$$
I need to determine the image of the unit circle $S^1$ by the action of the matrix $e^A$.
I think that I know how to calculate $e^A$:
I get the Jordan decomposition:
$$A = \begin{pmatrix}
  -5 & 9\\
  -4 & 7
 \end{pmatrix}  =
 \begin{pmatrix}
  -6 & 1\\
  -4 & 0
 \end{pmatrix}
 \cdot
 \begin{pmatrix}
  1 & 1\\
  0 & 1
 \end{pmatrix}
 \cdot
 \frac{1}{4}
 \begin{pmatrix}
  0 & -1\\
  1 & -6
 \end{pmatrix}
 $$
With eigenvalues: $\lambda$ = 1, algebraic multiplicity = 2, eigenvecotrs: $\left\{ \begin{pmatrix}
  1\\
  0
 \end{pmatrix}, \begin{pmatrix}
 0\\
 1
 \end{pmatrix} \right\}$
$$ \displaystyle e^A = \sum^{\infty}_{i = 0} \frac{A^i}{i!}$$
$$e^A =
 \begin{pmatrix}
  -6 & 1\\
  -4 & 0
 \end{pmatrix}
 \cdot
 \left(
 \begin{pmatrix}
  1 & 0\\
  0 & 1
 \end{pmatrix}
 +
 \displaystyle \sum^{\infty}_{i = 1} \frac{ \begin{pmatrix}
  1 & 1\\
  0 & 1
 \end{pmatrix}}{i!}
 \right)
 \cdot
 \frac{1}{4}
 \begin{pmatrix}
  0 & -1\\
  1 & -6
 \end{pmatrix}$$
$$e^A = \begin{pmatrix}
  -5 & 9\\
  -4 & 7
 \end{pmatrix}  =
 \begin{pmatrix}
  -6 & 1\\
  -4 & 0
 \end{pmatrix}
 \cdot
 \begin{pmatrix}
  \displaystyle \sum^{\infty}_{i = 1} \frac{1}{i!}& \displaystyle \sum^{\infty}_{i = 1} \frac{2^{i-1}}{i!}\\
  0 & \displaystyle \sum^{\infty}_{i = 1} \frac{1}{i!}
 \end{pmatrix}
 \cdot
 \frac{1}{4}
 \begin{pmatrix}
  0 & -1\\
  1 & -6
 \end{pmatrix}$$
Where: $$\displaystyle \sum^{\infty}_{i = 1} \frac{2^{i-1}}{i!} = \frac{1}{2} \sum^{\infty}_{i = 1} \frac{2^{i}}{i!} = \frac{1}{2}(e^2 - 1)
 $$
So:
$$e^A =
 \begin{pmatrix}
  -6 & 1\\
  -4 & 0
 \end{pmatrix}
 \cdot
 \begin{pmatrix}
  e & \displaystyle \frac{e^2}{2} - \displaystyle \frac{1}{2}\\
  0 & e
 \end{pmatrix}
 \cdot
 \frac{1}{4}
 \begin{pmatrix}
  0 & -1\\
  1 & -6
 \end{pmatrix} =
 \begin{pmatrix}
  \displaystyle \frac{-3e^2 + e + 3}{4} & \displaystyle \frac{9e^2 - 9}{2}\\
  \displaystyle \frac{-e^2 + 1}{2} & 3e^2 + e - 3
 \end{pmatrix}
 $$
Now, I don't know if I did it correctly up to this point and what I should do next - to operate on my unit circle.

Solution:
Because of @Oscar Lanzi we know that:
$$e^{\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}}=e\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}$$
Then because of that:
Equation of unit circle under linear transformation - can't understand role of inverse matrix
(answer by @Prototank)
We know that the image of unit circle in action of the matrix $A$ is given by: $$65x^{2}-166xy+106y^{2}=1$$
Now we need to scale by $e$ and we get the image of unit circle in action of the matrix $e^A$:
$$65x^{2}-166xy+106y^{2}=e^2$$
 A: This address the general question of shape of image of an unit circle under the action of $e^A$. We can think of a linear dynamic system $\dot x=Ax$, the solution of which will be $x=e^{tA}$. Then what you are seeking is the ending position of points starting from the unit circle, after following linear dynamics for $t=1$.



This animation has credit to Ella Batty. I TAed her class last semester and used this as a demo for 2d dynamic systems.
https://twitter.com/i/status/1319061743679799297
A: The matrix exponentiation is much simpler than it looks. When you find that $1$ is the only eigenvalue, render
$\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}=\begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}+\begin{pmatrix}-6 & 9\\-4 & 6\end{pmatrix}.$
The first matrix on the right just gives a factor of $e$ to the overall exponential. The second matrix is nilpotent and the series for its exponential is just
$\begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}+\begin{pmatrix}-6 & 9\\-4 & 6\end{pmatrix}=\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}.$
So
$e^{\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}}=e\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}.$
Continue from there.
A: The calculation won't be easy, but standard.
To understand the image of any invertible matrix $T$ acting on the unit circle, first do the SVD decomposition $$T=U\Sigma V = U\begin{pmatrix} \sigma_1 & \\ & \sigma_2\end{pmatrix}V$$ Note that $V$ send the circle back to itself, as it keeps norms. Now it should be clear that the image is just an ellipse with two principal axes being $\sigma_1, \sigma_2$ along the directions of the two column vectors of $U$.
Also to find $U$ and $\sigma_i$, we only have to diagonalize $TT^*$.
