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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 $e^A$: $$65x^{2}-166xy+106y^{2}=e^2$$

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  • $\begingroup$ There may be a more sophisticated way to do this, but have you tried a few key points on the unit circle and see if there's any obvious behavior? $\endgroup$
    – David P
    Commented Jun 8, 2022 at 0:08
  • $\begingroup$ No, I'm not sure I I would know how to do that. $\endgroup$
    – thefool
    Commented Jun 8, 2022 at 0:10
  • $\begingroup$ For example, find the images of $(0,1), (1,0)$ etc. It also seems a mistake has been made somewhere, it seems like for your matrxi $A$, $\exp(A)=eA$, according to software: link. $\endgroup$
    – David P
    Commented Jun 8, 2022 at 0:22
  • $\begingroup$ Ohh, I see. So having that exp(A) I should find the images of some points and look for a pattern? $\endgroup$
    – thefool
    Commented Jun 8, 2022 at 0:25
  • $\begingroup$ Notice $\exp(A)=eA$ means that the action of $\exp(A)$ on any vector $v$ is basically the same as the action of $A$ on $v$, so what would $A$ do to a vector on the unit circle? Then scale it by factor of $e$. $\endgroup$
    – David P
    Commented Jun 8, 2022 at 0:28

3 Answers 3

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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 $t=1$.

enter image description here enter image description here


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

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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.

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  • $\begingroup$ I think I know what to do next. I will edit my original post soon. $\endgroup$
    – thefool
    Commented Jun 8, 2022 at 1:30
  • $\begingroup$ I got it, I will post full answer. $\endgroup$
    – thefool
    Commented Jun 8, 2022 at 3:02
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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 $TT^*$.

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