In my Ordinary Differential Equations class, we've learned the following method for finding a general solution for $x'=Ax$ (where $A$ is a $2x2$ matrix), if it gives two complex eigenvalues, $\mu+i\upsilon$ and $\mu-i\upsilon$ respectively.

Picking either of these eigenvectors, we should be able to reach a general real solution. Let's say we pick the eigenvalue $\mu+i\upsilon$ and then find two eigenvectors, $a+ib$ and $a+ib$ (where $a$ and $b$ are $2x1$ vectors).

My textbook then says that we can then find a general real solution by picking either one of these eigenvalues, and plugging it into the following:

$x = c_1e^{\mu t}(a*cos(\upsilon t)-b*sin(\upsilon t))+c_2e^{\mu t}(a*sin(\upsilon t)+b*cos(\upsilon t))$

(There's a proof that we can fold the imaginary part into the $c_2$ solution, since theoretically either $c$ could be complex, but we want only the real part.)

Here's the issue: In the past, I have gotten seemingly different general real solutions from picking either of the eigenvectors--but a solution is supposed to be made up of the fundamental set of solutions, and is also supposed to be unique!

So, the big question: How do we prove that whether we plug $a+ib$ or $a-ib$, we reach the same general real solution?

Or is the answer that we don't? And in fact we reach two different general real solutions with two different fundamental sets, because choosing either of the two eigenvectors gives a different fundamental set?

And if the above hypothesis is correct, then what is the meaning of a general real solution? Shouldn't they be solving the same equation?

  • $\begingroup$ It comes down to the constants of integration being arbitrary. $\endgroup$
    – amd
    Dec 8, 2019 at 22:42

1 Answer 1


If your matrix $A$ is real, then you must have real solutions only and your complex eigenvalues can come only in conjugate pairs. Actually, taking either of the eigenvalues is misleading, because you actually have two complex solutions for two complex conjugate eigenvalues. Each eigenvalue has only one complex solution. And each eigenvalue has only one eigenvector. However, as these two complex solutions are conjugate of each other, at the end while adding these two solutions you get the real part of them. The real part of complex number does not depend whether you take $Re + iIm$ or $Re - iIm$, both of them have real part $Re$. Therefore it does not make any difference, however, note that you get this final solution as addition of the two complex solutions for the two complex conjugate eigenvalues. So the solution goes like this:

$x(t) = (c_1+ic_2)*(a+ib)*e^{(\mu + iv)t} + (c_1-ic_2)*(a-ib)*e^{(\mu - iv)t}$

which is:

$x(t) = 2Re((c_1+ic_2)*(a+ib)*e^{(\mu + iv)t})$

which is also:

$x(t) = 2Re((c_1-ic_2)*(a-ib)*e^{(\mu - iv)t})$.

All simplifies to your expression at the end.

So, although you take any of the two complex conjugate eigenvalues for finding the solution, the solution itself is the addition of the two complex solutions for each of the eigenvalues, and at the end you have two constant coefficients $c_1$ and $c_2$ to be found from initial conditions. The equation is the same, but initial conditions, i.e. $x(0)$, may be different.

  • $\begingroup$ Thank you, this was extremely helpful! $\endgroup$
    – JCK
    Dec 9, 2019 at 16:23
  • $\begingroup$ Nice to hear that. Good luck. $\endgroup$
    – Pasha
    Dec 9, 2019 at 16:27

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