MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

An eigenvalue of a 2 x 2 matrix satisfies the equation

$$ \left(\begin{array}{cc} a & b \\\\ c & d \end{array} \right)\left( \begin{array}{c} x \\\\ y \end{array}\right) = \lambda \left( \begin{array}{c} x \\\\ y \end{array}\right) $$

Graham Farr multiplies by the identity matrix. It is still defines eigenvalues $(A - \lambda I ) \vec{x} = 0$.

$$ \left(\begin{array}{cc} a & b \\\\ c & d \end{array} \right)\left( \begin{array}{c} x \\\\ y \end{array}\right) = \left(\begin{array}{cc} \lambda & 0 \\\\ 0 & \lambda \end{array} \right)\left( \begin{array}{c} x \\\\ y \end{array}\right) $$

He embeds the complex numbers into the space of 2 x 2 matrices and asks for vectors which are rotated + dilated the matrix.

$$ \left(\begin{array}{cc} a & b \\\\ c & d \end{array} \right)\left( \begin{array}{c} x \\\\ y \end{array}\right) = \left(\begin{array}{cc} \lambda & \mu \\\\ -\mu & \lambda \end{array} \right)\left( \begin{array}{c} x \\\\ y \end{array}\right) $$

The characteristic equation is defines a circle in $\mu,\lambda$. This eigencircle is not really a fixed circle in the plane, but a collection of "characteristic" pairs of values forming a circle of values $(\mu,\nu) \in \mathbb{R}^2$.

\[ \left| \begin{array}{cc} a - \lambda & b - \mu \\\\ c + \mu & d - \lambda \end{array} \right| = 0 \]

Farr made an applet demonstrating these calculations and uses the eigencircles for easy demonstrations of relations between the eigenvectors, sign of the determinant, etc.

Can you have eigencircles in more than two dimensions? I suppose you can ask for two basis vectors $v_1, v_2$ such that $Av_1 = \lambda v_1 + \mu v_2 $ and $Av_2 = - \mu v_1 + \lambda v_2$. It also seems 2 x 2 matrices are singling out not only two directions, but a whole circle's worth of vectors.

Is there a correct higher-dimensional generalization of this? Have these objects been studied under a different name?

share|cite|improve this question
Isn't an "eigencircle" just a way to study complex eigenvectors of a real matrix without extending scalars? – Qiaochu Yuan Apr 26 '12 at 16:14
Multiplication in the complex plane geometrically is rotation and scaling (by a positive constant). Maybe consider $SO(n) \times \mathbb{R}^+$, i.e. positively-scaled, special orthogonal matrices. Farr also mentions the use of quaternions toward the end of the article, though says this still applies to 2×2 matrices. Interesting stuff! – Victor Dods Apr 26 '12 at 17:47
@Qiaochu Does a real matrix have a $SO(2)$ worth of complex eigenvalues? These are 2D invariant subspaces, where the matrix acts as an element of $\mathbb{C} = SO(2) \times \mathbb{R}^+$. – john mangual Apr 26 '12 at 20:49
@John: if $V$ is a 2D invariant subspace which contains no eigenvectors, then extending scalars to $\mathbb{C}$ you can find a pair of complex conjugate eigenvectors with complex eigenvalues which span $V \otimes \mathbb{C}$. In general just apply the structure theorem for f.g. modules over a PID to $\mathbb{R}[x]$. You get simple modules of the form $\mathbb{R}[x]/(x - r)$ (real eigenvectors) and $\mathbb{R}[x]/(x^2 + bx + c)$ for $b^2 - 4c < 0$ (pairs of complex eigenvectors). – Qiaochu Yuan Apr 26 '12 at 22:45
@Qiaochu Maybe these are 1D subspaces on which the action of the matrix is to rotate and dilate (i.e. multiply by $\lambda + \mu i$). A 2 x 2 matrix has exactly 2 complex eigenvalues and the pairs $(\lambda, \mu)$ can take a continuous range of values lying on a circle. So they have different cardinalities. – john mangual Apr 27 '12 at 5:57

They are not exactly what you are looking for, but I think you should take a look at two-parameter eigenvalue problems. For instance check Essentially, they are a system of two equations of the kind you are considering with different matrices but with the same unknowns $\lambda,\mu$. Since you have two equations in typically that setting, you get a finite number of solutions $(\lambda,\mu)$.

share|cite|improve this answer

As I understand the problem it can be generalized. The $\small \lambda,\mu $-matrix can be seen as scaled rotation, and using $ \small r^2 = \lambda^2 + \mu^2 $ we can, because the $\small \lambda,\mu$ - matrix is invertible and is the left factor on the rhs (thus it is not really an eigenvalue analogon in my opinion) we can differently rewrite your original equation as

$\qquad \small \frac1{r^2} \begin{bmatrix} \lambda & -\mu \\\ \mu & \lambda \end{bmatrix} \cdot \begin{bmatrix} a & b \\\ c & d \end{bmatrix} \cdot \begin{bmatrix} x \\\ y \end{bmatrix}=\begin{bmatrix} x \\\ y \end{bmatrix} \cdot 1 $

or $\qquad \small \begin{bmatrix} \cos(\varphi) & -\sin(\varphi) \\\ \sin(\varphi) & \cos(\varphi) \end{bmatrix} \cdot \begin{bmatrix} a & b \\\ c & d \end{bmatrix} \cdot \begin{bmatrix} x \\\ y \end{bmatrix}=\begin{bmatrix} x \\\ y \end{bmatrix} \cdot r $

$\qquad \small (T \cdot A) \cdot X = X \cdot r$

where then the eigen-problem is
$\qquad \small \left\vert T \cdot A - r\cdot I \right\vert = 0 $

The left factor T is simply a rotation-matrix and r the eigenvalue of the rotated matrix $\small T\cdot A$.

If this characterization of the $\small \lambda, \mu$-matrix is the interesting one (which I assume) it can then be generalized to more dimensions in the obvious way and the $\small \lambda ,\mu $'s have to be determined as scalings of the cos/sin-parameters by the eigenvalues of the multidimensionally rotated matrix.

[Update] Here is a picture of the construction according to my interpretation of the problem. I take some example-matrix A $\small \begin{bmatrix} 12&24\\\ 13&83 \end{bmatrix} $ then leftmultiply by the rotationmatrices of angles between 0 and $\small 2 \pi $. Each has then two eigenvalues $\small w_0 , w_1 $. The $\small \lambda,\mu$ -matrix is then $\small T \cdot w_0 $. Note, that the eigenvalues and thus the $\small \lambda,\mu$-matrix may become complex for some rotations. Then I plot the log of the absolute value of the two eigenvalues in different colors. Here is the picture:

alt text

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.