Say I am interested in a linear operator $A: \mathbb{R}^2 \to \mathbb{R}^2$ who's matrix in a standard orthonormal basis is $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. $A$ acts on the real plane as a rigid rotation by 90 degrees counter clockwise. The two pairs of eigenvalues and eigenvectors are $\left(i, \begin{bmatrix} 1 \\ -i \end{bmatrix} \right)$ and $\left(-i, \begin{bmatrix} 1 \\ i \end{bmatrix} \right)$. I can pick either eigenvector and take the real and imaginary parts to define a basis for $\mathbb{R}^2$. Now, the eigenvalue $i$ represents rotation by $90$ degrees counterclockwise, but $-i$ represents rotation by $90$ degrees clockwise.

So how am I supposed to determine, given two eigenvalue-eigenvector pairs, what direction the rotation is in for a real operator on a real vector space?

Thank you.

  • $\begingroup$ The operator $A$ is not diagonalizable in the real setting. $\endgroup$
    – Fabian
    Commented Mar 19, 2017 at 21:30
  • $\begingroup$ I understand. $A$ does not preserve any 1-dimensional subspaces in $\mathbb{R}^2$. There is only a 2-dimensional subspace that is $A$-invariant, and thats $\mathbb{R}^2$ itself. $A$ rotates every vector of $\mathbb{R}^2$. My question is: given the two complex conjugate eigenvalues, how do I determine what direction the rotation is in in $\mathbb{R}^2$? The act of conjugation of an eigenvalue reverses the sense of rotation. $\endgroup$
    – Evan
    Commented Mar 19, 2017 at 21:32

1 Answer 1


In general, you can’t.

The notion of direction of a rotation is intimately connected to that of orientation. Eigenvectors, however, don’t carry any orientation information: If $v$ is an eigenvector, then so is $-v$; mix and match at will. Complex eigenvalues of a real matrix come in conjugate pairs, so they’re no help, either. The best you can do is to recover the cosine of the rotation angle (which you can get from the trace without explicitly computing eigenvalues), which of course leaves you with a sign ambiguity.

It might be a bit hard to see why this is so in two dimensions, but consider a rotation in $\mathbb R^3$ instead, which can be described by its axis and angle. The sign of that angle depends on your choice of orientation for the axis, though. For example, a 90-degree counterclockwise rotation about the positive $z$-axis looks like a clockwise rotation when viewed from the opposite direction. The direction of rotation depends entirely on your choice of basis.

Update: Having said all that, it does seem possible to disambiguate the angle by looking at how the eigenvalues match up with the eigenvectors, at least in an oriented orthogonal basis for $\mathbb R^2$. If we measure angles consistently with that orientation, that is, a positive angle takes the first basis vector into the second, the matrix of a rotation expressed relative to such a basis has the familiar form $$R=\pmatrix{\cos\theta&-\sin\theta\\\sin\theta&\cos\theta}.$$ The eigenvalues of this matrix are $\cos\theta\pm i\sin\theta$, and for $\theta\ne n\pi$, $(\pm i,1)^T$ are independent eigenvectors. If we normalize $\theta$ to $[0,2\pi)$, we find that the eigenvalue and eigenvector pairing differs for $0\lt\theta\lt\pi$ and $\pi\lt\theta\lt2\pi$. In the former case, the eigenvalue $\lambda$ with $\Im(\lambda)>0$ has $(i,1)$ for its eigenvalue, while in the latter it’s the eigenvector with $\Im(\lambda)\lt0$. (This is just a slightly roundabout way of examining the sign of $\sin\theta$.) So, by examining this pairing it is possible to resolve the sign ambiguity, but ascribing a direction to the rotation is still more a matter of your point of view: rotating clockwise by $\theta$ is indistinguishable from rotating counterclockwise by $2\pi-\theta$.


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