2
$\begingroup$

Does there exist a matrix $A \in \Bbb R^{2 \times 2}$ such that the following holds?

$$e^A = \begin{pmatrix}-4 && 0 \\ 0 && -1\end{pmatrix}$$

I know that for a Jordan block

$$J=\begin{pmatrix} \lambda&1&0&0&\dots\\ 0&\lambda&1&0&\dots\\ 0&0&\lambda&1&\dots\\ 0&0&0&\lambda&\dots\\ &&\vdots&&\ddots \end{pmatrix}$$

$$e^{Jt}=\begin{pmatrix} 1&\frac{t}{1!}&\frac{t^2}{2!}&\frac{t^3}{3!}&\dots\\ 0&1&\frac{t}{1!}&\frac{t^2}{2!}&\dots\\ 0&0&1&\frac{t}{1!}&\dots\\ 0&0&0&1&\dots\\ \vdots&\vdots&\vdots&\vdots&\ddots \end{pmatrix}$$

And using the Jordan normal form of a matrix $M:$ $M=TJT^{-1}$ it holds:

$$e^{Mt}=Te^{Jt}T^{-1}$$

I think that such a matrix $A$ does not exist because the eigenvalues are negative and $\ln (-4)$, $\ln(-1)$ is undefined but I don't know how properly prove that such a matrix doesn't exist.

$\endgroup$
5
  • 2
    $\begingroup$ No: look at the trace. $\endgroup$ Commented Apr 30, 2019 at 21:10
  • 1
    $\begingroup$ The fact that $A$ has non-real eigenvalues (i.e. that $e^A$ has negative eigenvalues) is not enough on its own. For instance, we find that $$ \exp\pmatrix{0&-\pi\\ \pi & 0} = \pmatrix{-1&0\\0&-1} $$ $\endgroup$ Commented Apr 30, 2019 at 21:12
  • $\begingroup$ A wiki source for T's trace identity $\endgroup$ Commented Apr 30, 2019 at 21:17
  • $\begingroup$ @T.Bongers the trace satisfies $e^{\operatorname{tr}(A)} = \det(e^A) = 4$; how does that help? $\endgroup$ Commented Apr 30, 2019 at 21:20
  • $\begingroup$ Some theory on the subject: m-hikari.com/ija/ija-password-2008/ija-password1-4-2008/… In particular Theorem 3 seconds Omnom's answer. $\endgroup$ Commented Apr 30, 2019 at 21:31

1 Answer 1

Reset to default
4
$\begingroup$

If $\lambda$ is an eigenvalue of $A$, then $e^{\lambda}$ is an eigenvalue of $e^A$. So, since $e^A$ has eigenvalues $-1,-4$, $A$ has one eigenvalue satisfying $e^{\lambda_1} = -1$ and another satisfying $e^{\lambda_2} = -4$.

However, the strictly complex eigenvalues of any real matrix must come in conjugate pairs. That is, if $\lambda$ is a non-real eigenvalue of $A$, then the conjugate $\overline{\lambda}$ must also be an eigenvalue. This tells us that $A$ cannot be real, since there is no complex $\lambda$ satisfying $\exp(\lambda) = -1$ and $\exp(\bar \lambda) = -4$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.