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Say that we have a vector field on $GL(n,R)$, given by $A \rightarrow (A,A^2)$ (I mean the matrix power here). If we try to find the flow of this vector field, I get that it should be: $X(A,t) = e^{tA^2}A$, where $e^{tA^2}$ is the matrix exponential here. However, since I feel somewhat unsure about the material, I am not sure that my approach is correct. Could someone either give some confirmation as to that this method is correct, and if so, does it follow that our vector field will be a complete one? I strongly suspect that such is the case.

If I'm wrong I'd love some hint that might put me in the right direction.

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    When you say "I get that it should be", do you mean that you know from an authoritative source that this is a correct solution, or that you feel intuitively that this ought to be the solution?2011-09-29
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    Just my own limited knowledge on solving matrix differential equations. Nothing more than that. So if you have any thoughts on this, I'd love to hear it.2011-09-30

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I have tried a bit of googling to find a sense of "the flow of a vector field" where $X(A,t)=e^{tA^2}A$ would be a correct solution, but in vain. You require $\frac{d}{dt}X(A,t) = X(A,t)^2$ and $X(A,0)=A$, right?

In that case, for $GL(1,\mathbb R)$ it reduces to an ordinary differential equation $\frac{dy}{dt}=y^2$ which is easily solved to find $y=\frac{1}{y_0^{-1}-t}$ which is significantly different from $y=e^{ty_0^2}y_0$

On the other hand, intuitively and by analogy to the real case, $X(A,t) = (A^{-1}-tI)^{-1}$ ought to work. If $A$ has a positive real eigenvalue this makes sense only for $t<1/\lambda$ where $\lambda$ is the largest such eigenvalue, but that parallels the vertical blowup of the real solution and so is to be expected. (And, applying Cramer's rule we see that the inverse really does escape to infinity before it stops making sense).

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    I can't seem to comment, but yes, You seem right. Where could one read more about solving differential equations like this for matrices?2011-09-30
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    @Shaf_math, You accidentally created a duplicate account; with fewer than 50 reputation points, [an account can only comment on its own questions and answers](http://meta.stackexchange.com/questions/19756/how-do-comments-work/19757#19757). I have now merged your duplicate account into your current account (the one used to ask the question), so you will now be able to comment here (and anywhere, once you have >50 points). If you have trouble logging in, or if you accidentally create duplicate accounts, simply flag one of your own questions for moderator attention, and we will help out.2011-10-01
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    @Shaf: I don't actually know how to do this systematically myself, so I cannot recommend literature, sorry. The general procedure I follow when I have to is to pretend everything are reals, be careful about the order of factors, and hope there's a natural way to generalize the solution back to matrices when I'm done. Proving the generalized solution correct must then be done in a separate step.2011-10-01