Has the ODE $$ \boldsymbol{y}' = \boldsymbol{\mathsf{M}}\cdot\boldsymbol{y} $$ (with $\boldsymbol{y}\in R^3$ and $\boldsymbol{\mathsf{M}}$ a constant real-valued 3x3 matrix) a closed solution? If so what is it?

I know the answer for the cases that $\boldsymbol{\mathsf{M}}$ is symmetric or anti-symmetric. But what if $\boldsymbol{\mathsf{M}}=\boldsymbol{\mathsf{S}}+\boldsymbol{\mathsf{A}}$ with $\boldsymbol{\mathsf{S}}^t=\boldsymbol{\mathsf{S}}$ and $\boldsymbol{\mathsf{A}}^t=-\boldsymbol{\mathsf{A}}$?

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    $\begingroup$ The answer is given as $$y(t) = y(0)e^{Mt}$$ where $e^{Mt}$ represents the matrix exponential $\endgroup$ Sep 28, 2014 at 15:43
  • $\begingroup$ thanks. So I learned about the matrix exponential. Great! (I would accept this comment morphed into an answer ...) $\endgroup$
    – Walter
    Sep 29, 2014 at 18:43

1 Answer 1


Per my comment: the answer is given as $$ y(t) = y(0)e^{Mt} $$ where $e^{Mt}$ represents the matrix exponential, defined by $$ e^{Mt} = \sum_{k=0}^\infty \frac{1}{k!}M^kt^k $$ where $M^0 := I$. An equally valid definition is that $e^{Mt}$ is the unique matrix function satisfying $$ \frac{d}{dt} e^{Mt} = M e^{Mt} = e^{Mt} M\\ e^{M\cdot 0} = e^{0} = I $$


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