EDIT: My initial problem was to find the kind of matrices, for which the exponential of is diagonal (and later addition to conditions - unitary), i.e. find all $A$, with non-zero entries, such that $e^A$ is diagonal (and unitary).

I, mostly experimentally, found that the matrices of the following form \begin{equation} A = \begin{bmatrix} z & k \pi i \\ k \pi i & z \end{bmatrix}, \quad \text{where} \quad z\in \mathbb{C} \quad \text{and} \quad k \in \mathbb{Z} \end{equation}

give \begin{equation} e^A = \begin{bmatrix} (-1)^ke^z & 0 \\ 0 & (-1)^ke^z \end{bmatrix}. \end{equation}

Something similar I found for $A\in \mathbb{C}^{3 \times 3}$, but these seem to me partial cases, and if they are not, what do you think would be a good approach to prove that? Any directions towards good literature for this or published papers would be very helpful.

After that, must be fairly easy to prove (I presume, taken that the statement is correct) that:

If $A$ is skew-Hermitian and $e^A$ is diagonal, then $A$ can be expressed as

\begin{equation} A = \begin{bmatrix} bi & k \pi i \\ k \pi i & bi \end{bmatrix}, \quad \text{where} \quad b\in \mathbb{R} \quad \text{and} \quad k \in \mathbb{Z}. \end{equation}

P.S. I am not sure if this is already a proven result or if someone before has worked on this; so would really appreciate if someone who knows could share with that kind of information.

Thanks in advance!


This is not true. Take, for instance, $A=\left[\begin{smallmatrix}2\pi i&0\\0&0\end{smallmatrix}\right]$. It is skew-Hermitian and $e^A=\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$. However, $A$ is not of that form that you mentioned.

  • $\begingroup$ Sorry, I forgot to mention that I also impose a requirement for the second and third entries of the matrix to be non-zero. $\endgroup$ – Elen Khachatryan Jun 24 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.