I have a equation $$Ax=b,$$ where matrix $A$ is invertible, $b$ is a constant vector, and $x$ is the unknown vector. To obtain $x$, it is obvious $x=A^{-1}b$. Alternatively, if $A$ is Hurwitz, one can build a differential equation $$\dot{x}=Ax-b$$ whose stable equilibrium point is $x=A^{-1}b$. My question is that if $A$ is an arbitrary invertible matrix, are we still able to build a similar differential equation as the above one to obtain $x$? Thanks!

  • $\begingroup$ If A is a simple scalar matrix (1x1) of 1, this equation will not have a stable equilibrium, so I would say no $\endgroup$ – bobuhito Apr 21 '13 at 21:00
  • $\begingroup$ If A=1, we can still do it by building $\dot{x}$=b-Ax, right? $\endgroup$ – Wei Apr 21 '13 at 21:12
  • $\begingroup$ Wei: It will be unstable. $\endgroup$ – Misha Apr 21 '13 at 21:32

I think this is more appropriate for mathSE, however;

Recall that a linear differential equation can have fixed points as nodes, saddles and centers.

Your question requires the fixed point to be a node of some variety, so the real parts of the eigenvalues of $A$ must be the same sign.

If they all have negative real parts, the node is stable and you will get the solution you are after.

If they all have positive real parts, simply reverse the direction of time and solve as you figured in your comment above.

If they are of alternating sign or have zero real part, this approach will no longer work.


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