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:


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.

  • 2
    $\begingroup$ No: look at the trace. $\endgroup$
    – user296602
    Commented Apr 30, 2019 at 21:10
  • $\begingroup$ 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. $\endgroup$
    – zwim
    Commented Apr 30, 2019 at 21:31

1 Answer 1


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 $\exp(\bar \lambda) = -4$.


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