# Intuition for complex eigenvalues

The eigenvalues of a rotation matrix are complex numbers. I understand that they cannot be real numbers because when you rotate something no direction stays the same.

My question
What is the intuition that the eigenvalues are complex? Why do they exist at all for rotation matrices? I mean it is not so that every time a calculation is not possible the result is complex (dividing by 0 is not possible at all - the result is not real, but it is not complex either!). The complex numbers seem to cover some middle ground here, but I don't understand how and why they come into play - there don't seem to be any square roots taken from negative numbers...)

• Did you mean "The eigenvalues of a rotation matrix..."? In any event, if you'll look at the two-by-two case, you'll note that the eigenvalues of a rotation matrix are always complex unless the rotation angle is an integer multiple of $\pi$, so there is no reason to expect that higher-dimensional rotation matrices would behave differently. – J. M. ain't a mathematician Dec 23 '10 at 8:44
• @J.M.: Thank you - fixed it! – vonjd Dec 23 '10 at 8:49
• With regards to the eigenvectors: if the matrix is real, and the eigenvalue is complex, you should expect the corresponding eigenvector to have complex components. – J. M. ain't a mathematician Dec 23 '10 at 8:58
• Complex numbers are not just real numbers (real axis) and square roots of negative real numbers (imaginary axis). Most complex eigenvalues of a matrix will be a+bi where a and b are both nonzero. The intuition you need is the Fundamental Theorem of Algebra: any nonconstant polynomial with complex (e.g., real) coefficients has a complex root. – KCd Dec 23 '10 at 13:49

Heuristically, I suppose we could see this being because the standard action of the complex numbers on $V = \mathbb R^{2n}$ is by rotation. That is, $(e_1, \ldots, e_{2n})$ is a basis for $V$, then we define multiplication by $i$ as
$i e_{2k-1} = e_{2k}, \quad i e_{2k} = - e_{2k-1},$ for $k = 1, \ldots, n$,
so multiplying a vector $v$ by a complex number $\lambda$ will correspond to a scaling by a real and a rotation.
Now, if we have a rotation $A$ on the space $V$ and we want to find a line $l$ "invariant" under $A$, then we can try to look for a complex number $\lambda$ such that rotation of $l$ under $A$ is equivalent to the action of $\lambda$ on $l$. Thus, we can try and look for complex eigenvalues $\lambda$ of $A$.