# Does there exist a real matrix $A$ for the given matrix $e^A$?

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.

• – user296602
Commented Apr 30, 2019 at 21:10
• 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 $$\exp\pmatrix{0&-\pi\\ \pi & 0} = \pmatrix{-1&0\\0&-1}$$ Commented Apr 30, 2019 at 21:12
• A wiki source for T's trace identity Commented Apr 30, 2019 at 21:17
• @T.Bongers the trace satisfies $e^{\operatorname{tr}(A)} = \det(e^A) = 4$; how does that help? Commented Apr 30, 2019 at 21:20
• Some theory on the subject: m-hikari.com/ija/ija-password-2008/ija-password1-4-2008/… In particular Theorem 3 seconds Omnom's answer.
– zwim
Commented Apr 30, 2019 at 21:31

## 1 Answer

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