Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was trying to figure out, how many degrees of freedoms a $n\times n$-orthogonal matrix posses.The easiest way to determine that seems to be the fact that the matrix exponential of an antisymmetric matrix yields an orthogonal matrix:

$M^T=-M, c=\exp(M) \Rightarrow c^T=c^{-1}$

A antisymmetric matrix possesses $\frac{n(n-1)}{2}$ degrees of freedom.

BUT: When I also thought about how to parametrize these freedoms explicitly (without the exponential) I remembered, that rotations in $\mathbb{R}^n$ can be parametrized using $n-1$ angles or cosines.

I dont' understand, where the remaining parameters are hidden?

My guess is, that a orthoganl transformation in $n>3$ can be more complicated than a rotation or that there are different types of rotation (containing reflectiong or such things) and that all the combination of these different types accounts for the rest of the parameters.

Thanks you for your help!

share|cite|improve this question

3 Answers 3

The result you remembered is wrong. I am actually not sure what you are referring to. Maybe you are thinking about Euler angles in $3$ dimensions, which are terribly misleading. Each Euler angle should not be thought of as a rotation about a different axis, but as a rotation in a different $2$-dimensional subspace spanned by two axes. The fact that these are the same thing in $3$ dimensions is a consequence of the anomalous identity ${3 \choose 1} = {3 \choose 2}$ (equivalently, the existence of the cross product) and does not hold in higher dimensions.

For example, in four dimensions there are four coordinate axes, say $x, y, z, w$, and ${4 \choose 2} = 6$ Euler angles: $xy$-rotation, $xz$-rotation, $xw$-rotation, $yz$-rotation, $yw$-rotation, $zw$-rotation. So the correct version of this result really does give the correct number of parameters.

share|cite|improve this answer
In any case, what is true is that an orthogonal transformation in dimensions greater than $3$ can be more complicated than a rotation of one of the above types; in particular, its fixed subspace can have any even codimension. Whether these are "rotations" are not depends on your definition of "rotation": my definition is an orthogonal transformation of determinant $1$, so... – Qiaochu Yuan Mar 20 '11 at 22:48
This is it! There are n(n-1)/2 Euler angles. This is exactly the number of freedoms of a orthogonal matrix! (Turns out, the (n-1) angles I was taking about are the number of angles needed to specify a direction in n-dimensional space!) – JSG Mar 21 '11 at 0:08

Although every rotation will have an matrix which is orthogonal, not every orthogonal matrix is the matrix of a rotation. And you need more than just the angles in the general case to specify a rotation.

In fact, every orthogonal matrix is similar to a block-diagonal matrix, each block $2\times 2$ and at most a $1\times 1$ block; the $2\times 2$ blocks correspond to rotations, and the $1\times 1$ block corresponds either to $1$ or to $-1$ (an identity or a reflection about a codimension $1$ subspace).

Each rotation can be determined by a single angle, (and the $1\times 1$ block gives you a furhter degree of freedom, but this does not actually give you a full fledged "dimension"). But you need more than the angles to specify the rotations, you need to specify the $2$-dimensional subspaces in which they act.

In $\mathbb{R}^3$, you can get away with just three angles because you can determine a single vector with two angles (think spherical coordinates); this gives you the normal vector to the plane of rotation; then specify the angle of rotation, and this gives you the $3 = \frac{3(2)}{2}$ parameters.

But in $\mathbb{R}^n$ with $n\gt 3$, you need to specify the $2$-dimensional subspace and the angle to specify the rotation (and do this enough times). This cannot be determined by a single vector plus an angle any more the way it does in $\mathbb{R}^3$, because these planes are no longer codimension $1$ (they cannot be given as "the orthogonal complement to this one-dimensional subspace).

share|cite|improve this answer
@Theo; Thanks. My connection flaked and I couldn't deal with it as quickly as I would have liked. – Arturo Magidin Mar 20 '11 at 22:47
No problem, no harm ensued :) I removed my comment as it is no longer relevant. – t.b. Mar 20 '11 at 23:00
@user4514: Any particular reason for yelling, though? (All caps are interpreted as shouting; you can use * or _ for italics, if what you want is emphasis). – Arturo Magidin Mar 21 '11 at 2:09
@Arturo: When I say degrees of freedom, I really mean "dimension of space of parameters", so maybe that is where the communication is breaking down. Specifying the plane of rotation is two degrees of freedom because it's specified by the normal vector, which can be thought of as a point on $S^2$ (or, really $\mathbb{R}P^2$), which is 2 dimensional. The "did I reflect?" part is a binary yes or no value and not really a degree of freedom (it contributes components to the Lie group, not dimension). – Jason DeVito Mar 21 '11 at 4:00
@Jason: No, no; I'm the one screwing this up, which actually just woke me up enough that I couldn't get back to sleep without first trying to fix this. – Arturo Magidin Mar 21 '11 at 8:42
up vote 0 down vote accepted

To describe a rotation in $\mathbb{R}^n$ on needs $n(n-1)/2$ parameters, which equals the number of freedoms that a orthogonal matrix possesses.

(To describe a direction in $\mathbb{R}^n$ one needs $n-1$ angles. See

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.