# Is an orthogonal operator with determinant equal to $1$ or $-1$ always a rotation or a reflection?

I know that in $R^{2}$ an orthogonal operator with determinant $1$ or $-1$ is either a rotation or a reflection. What I was wondering was whether this result holds in vector spaces with higher dimensions. In a vector space like $R^{4}$ are orthogonal operators also just rotations and reflections when their determinant is $1$ or $-1$ or is this not the case. If it doesn't hold in higher dimension would someone mind explaining why it doesn't hold?

• It’s not even true in $\mathbb R^3$, where a determinant of $-1$ could mean that you have a combination of rotation and reflection. Things get even more interesting in higher dimensions, as Ted Shifrin hints in his answer. The eigenvalues of the operator can tell you what’s going on, as in, for example this question. – amd Apr 22 '16 at 23:45

First of all, an orthogonal map (in finite dimensions) always has determinant $\pm 1$. But in $\Bbb R^4$, you can have a rotation in the $x_1x_2$-plane along with a rotation in the $x_3x_4$-plane, for example, and I would not call that a rotation.
• Using your logic would it be correct to assume that rotations and reflections don't exist in $R^{n}$ where $n > 3$ – Shah Kamal Apr 22 '16 at 21:40
• Of course they exist. But a general orthogonal map need not be one. If your rotation is in the $x_1x_2$-plane and the $x_3x_4$-plane is fixed, then this is generally considered a rotation. I suppose you need to give me a definition if you're not happy. – Ted Shifrin Apr 22 '16 at 21:45
• @ShahKamal : but you can generate the orthogonal group of $\mathbb{R}^n$ only from reflections of 1D subspaces and rotations of 2D subspaces – reuns Apr 22 '16 at 22:22
Orthogonal operators, or matrices $A$, are usually defined by the property that $A^TA = AA^T = I$ which automatically forces $\det A = \pm 1$.