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

• 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? Commented Jun 8, 2022 at 0:08
• No, I'm not sure I I would know how to do that. Commented Jun 8, 2022 at 0:10
• 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. Commented Jun 8, 2022 at 0:22
• Ohh, I see. So having that exp(A) I should find the images of some points and look for a pattern? Commented Jun 8, 2022 at 0:25
• 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 $e$. Commented Jun 8, 2022 at 0:28

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.

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.

• I think I know what to do next. I will edit my original post soon. Commented Jun 8, 2022 at 1:30
• I got it, I will post full answer. Commented Jun 8, 2022 at 3:02

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 $$TT^*$$.