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If I have a symmetric matrix $\boldsymbol{B} = e^{\boldsymbol{\Theta}t}e^{\boldsymbol{\Theta}^{T}t}$ where $\boldsymbol{\Theta}$ is a matrix with transpose $\boldsymbol{\Theta}^{T}$, is it possible to say that $\boldsymbol{B} = e^{\boldsymbol{\alpha}t}$ where $\boldsymbol{\alpha}$ is also a symmetric matrix?

Hello
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  • Is $\Theta$ symmetric? Is it normal? – Leo Aug 21 '19 at 17:42
  • Hello, no $\boldsymbol{\Theta}$ is any square constant matrix, and has no particular properties apart that it is real – Hello Aug 21 '19 at 17:43
  • $e^Xe^Y=e^{X+Y}$ so $\alpha = \Theta +\Theta^T$ – saulspatz Aug 21 '19 at 17:51
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    @saulspatz: this only holds if $XY = YX$. – Robert Lewis Aug 21 '19 at 17:54
  • @saulspatz this is only true when $X$ and $Y$ commute, which is precisely when $\Theta$ is normal. – Leo Aug 21 '19 at 17:54
  • This works only if they commute, however, I would like to probe this if $\boldsymbol{\Theta}$ does not commute with its transpose. Perhaps I could say that since $\boldsymbol{B}$ is symmetric, then it can be decomposed as $\boldsymbol{B} = \boldsymbol{P}\boldsymbol{D}\boldsymbol{P}^{-1}$ where $\boldsymbol{D}$ is a diagonal matrix and then $\boldsymbol{D} = e^{\boldsymbol{U}}$ but I would need all elements of $\boldsymbol{D} \geq 0$. Could this always be the case? – Hello Aug 21 '19 at 17:54
  • @RobertLewis Yes, my mistake. Thanks. – saulspatz Aug 21 '19 at 17:55
  • @Leo Thanks. Careless of me. – saulspatz Aug 21 '19 at 17:57

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