Matrix exponential

  • #1
I'd like to do one of two things:

1: Find an example of a non-diagonal matrix whose matrix exponential (defined in terms of series) is diagonal.
2: Prove that no such examples exist.

I'm working with matrices over the complex field. My gut tells me that 2 is the way to go. I'd really appreciate any help with this.

Ian
 

Answers and Replies

  • #2
4
0
Let [tex] \sigma_x = \left( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right) [/tex], and consider things like [tex] e^{i \sigma_x \varphi} [/tex] where [tex] \varphi [/tex] is some angle. You should find that the matrix exponential of a non-diagonal matrix can, in fact, be diagonal.

(You can find a useful / interesting identity related to my suggestion at http://en.wikipedia.org/wiki/Pauli_matrices" [Broken]).
 
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  • #3
534
1
Or: consider [itex]A = \left( \begin{smallmatrix}0&-1\\1&0\end{smallmatrix} \right)[/itex]. Then [itex]e^{\theta A}[/itex] is a rotation by the angle [itex]\theta[/itex]. In particular, [itex]e^{n\pi A}[/itex] is diagonal for any integer [itex]n[/itex].
 
  • #4
Thanks to both of you, that's very helpful. The 2x2 examples are good, but using higher dimensional Pauli matrices is more satisfying; thanks Sando.

I found a complete answer to my question in Gantmacher's text "The theory of matrices" (chapter VIII, section 8) which gives a complete characterization of the inverse map of the matrix exponential, and in particular, said inverse map in the case of diagonal matrices.

Given a diagonal matrix D without any entries on the negative real axis, the set of "logarithms" (ie. the set {H} such that exp(H) = D) obviously includes all matrices where you just take the logarithm of each element of D, and possibly add a multiple of 2[tex]\pi[/tex]i. I had originally thought that this would be it. But as Gantmacher shows, you need to also include conjugation by anything that commutes with D. (I'm not attempting to be formal here.) Which is where the examples you provided come in.

Ian
 
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