What is so fundamentally different between a rotation and a translation that one can be represented with a single n-vector and the other one needs an n-n matrix?

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    $\begingroup$ I would ask you the reverse: what do you see in common between translations and rotations, beyond the fact that both involve a change of coordinates? $\endgroup$ – Wolphram jonny Jan 2 '15 at 23:24
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    $\begingroup$ I don't think this question deserves either the downvote or vote to close that it received. Understanding translation and rotation are a necessary part of advancing beyond the beginner stage of physics (by beginner phase I mean physics where simple calculus suffices). $\endgroup$ – David Hammen Jan 3 '15 at 1:09
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    $\begingroup$ @DavidHammen : I haven't down-voted, but the reasons for down-voting include "This question does not show any research effort..." The OP knows translations and rotations are represented by vectors and matrices, so he cannot be a beginner. $\endgroup$ – sammy gerbil Aug 23 '16 at 4:41

Both are Euclidean isometries (distance and global-shape preserving maps): the only other kind of Euclidean isometry is a reflexion, although this last one is "discrete" - you can't have a fraction of a reflexion, whereas you can have a fraction $\alpha$ of a translation or rotation: you translate $\alpha$ times as far or rotate through $\alpha$ times a given angle. So the two transformations you mention are the only continuous Euclidean isometries.

What's fundamentally different? Translations commute. Rotations do not. This means that for two translations $T_1,\,T_2$ the order of application does not matter: the resultant is the same whichever way around you choose, i.e. $T_1\,T_2 = T_2\,T_1$. This does not hold for rotations: draw some marks on an orange and check this yourself if you haven't noticed this before. Aside from for rotations about the same axis, $R_1\,R_2 \neq R_2\,R_1$.

All groups (look this word up if you haven't met it) of continuous symmetries that are Abelian ( i.e. groups of symmetries that commute - for which the order does not matter) can be shown to be essentially the same (isomorphic) to either (1) a space of vectors kitted with vector addition or (2) a torus whose surface is labelled orthogonal Cartesian co-ordinates and which behaves essentially the same as a vector space of adding arrows. The only difference in the latter, torus group is that if you translate far enough in a given direction, you can get back to your beginning point. Otherwise the torus group looks exactly like a space of vectors. So you can take this as your fundamental informal reason: any commuting Lie group looks either exactly like a space of vectors, or a "compactified" one that behaves the same aside from one's being able to get back to one's beginning point by a far enough translation in any direction.

Another pithy characterisation is CuriousOne's comment:

A translation affects only one coordinate direction, a rotation affects two. An infinitesimal translation needs one vector, an infinitesimal rotation needs two. Maybe the better way to think about comparing these operations is with their generators than the matrices and vectors? I am sure a theorist can chime in with a better answer regarding the representation theory of Lie-groups.

Some background: in higher dimensions, rotations rotate 2D planes and leave the orthogonal complement of a plane invariant. This is what CuriousOne means by "rotation affects two". The axis concept only works in 3 dimensions: you can define a plane in 3 dimensions as the 2D subspace normal to a vector (in two dimensions you rotate about a point). So, in 3D with a vector's direction standing for the axis and its magnitude standing for the angle, you can indeed represent a rotation by a lone vector. There is even a triangle law for vector addition of rotations, but it is somewhat more involved than addition of translations. Nonetheless, it may surprise you just the same: see the discussion under the heading "Example 1.4: ($2\times 2$ Unitary Group $SU(2)$)" on this page of my website here. Another way of saying this is as David Hammen points out: The axis concept's being workable in 3D is an "accident" of dimension: in $N$ dimensions, a rotation is of the form $e^H$ where $H$ is a real $N\times N$ skew-symmetric ($H=-H^T$: equal to the negative of its transpose) matrix and thus $H$ must have noughts along its leading diagonal and its lower, below leading diagonal triangle is simply the upper, above leading diagonal triangle reflected. There are thus $N\,(N-1)/2$ real parameters needed to specify the rotation, which just happens to be equal to $N$ when $N=3$.

Lastly, one should mention the special relationship between translations and rotations. The translations are special insofar that for any isometry $U$ and translation $T$ we have $U\,T\,U^{-1} = T_1$, where $T_1$ is another translation (not needfully the same as $T$). The technical name for this is that the translations form a normal subgroup of the group of isometries; what this means is that any isometry can be uniquely decomposed into the form $T\,R$, where $T$ is a translation and $R$ a rotation: we say that the isometry group $E(N)$ is the semidirect product (written $E(N)=T(N)\rtimes R(N)$) of the group $T(N)$ of translations and $R(N)$ of rotations).

I should add that, in higher dimensions, I use the word "rotation" more loosely than many authors: I simply mean any homogeneous transformation whose matrix is of the form $e^H$ with $H$ skew symmetric. Many authors split these up into further different classes.


Remember that a point in N-dimensional space is given by an N-tuple, which is (in its essence) a vector.

Basically, when you translate a point or set of points in space you are picking a direction and moving those points along a straight line in that direction. In other words, you are adding two vectors. (Granted, you could perform a complicated translation along a paramaterized path, but the net result in terms of the coordinates would be a simple displacement from the starting point to the ending point.)

On the other hand, a rotation is a fundamentally two-dimensional transformation (i.e. a rotation always defines a plane). This is why the we have the right-hand rule: your thumb is the normal vector to the plane of rotation. And a plane (at least in three dimensions) is uniquely defined by its normal vector.

Now, it should start to be clear why we need only a vector for translations while we need a matrix for rotations. Since translations define a line (a 1-dimensional object), you only need a 1st-order tensor. On the other hand, since rotations define a plane (a 2-dimensional object), you need a 2nd-order tensor.

Long story short, translations simply change the components of a vector such that the change in each component is independent of the change in the other components, whereas rotations mix the components of a vector (i.e. the way that one component changes relies on the values of the other components).


A rotation is a linear transformation on the vector space $\mathbb{R}^n$, and hence a member of the group of linear transformations $\mathrm{GL}_\mathbb{R}(n)$, which is well-known to be the group of $n\times n$ matrices.

A translation is not a linear transformation - it is the addition of a constant vector to every vector of the space, and thus, in particular, it doesn't map the zero vector to the zero vector. Since every matrix is a linear transformation and vice versa (at least in the finite dimensional case), a translation cannot be a matrix, since it is not linear.

There's much more to be said about the structure of the non-abelian rotations in contrast to the abelian translations, but this this the one, definite reason - one is a linear transformation, the other is not, hence one is a matrix, and the other is not.


The fundamental difference between translation and rotation is that the former (when we speak of translation of a whole system) affects all the vectors in the same way, while a rotation affects each base-vector in a different way. This is why we need a matrix, (and this was the question - why a matrix), because in order to pass to a rotated base, we need to specify each vector (function) of the old base in terms of the new base. This is the origin of the $n$x$n$.

Let me exemplify in the 3D space. Assume that you defined a system of coordinates $(x, y, z)$. Along these axes you have the unit vectors $\vec i, \vec j, \vec k$. Now, any point $P$ in space can be described by its position $\vec r = x\vec i + y\vec j + z\vec k$.

If you decide to displace the point $P$ to a new location $P'$ the position of $P'$ will be affected by a vector $\vec r'= \vec a + \vec r$ where $\vec a = \vec {PP'}$. Or, writing in detail,

$\vec r'= (a_x + x)\vec i + (a_y + y)\vec j + (a_z + z)\vec k$.

This is a translation of a point. A translation can be done also in another way, we can translate the origin of our system of coordinates, e.g. by the vector $\vec a$. In this case, any vector $\vec r'$ will be affected in the same way,

$\vec r'= (x - a_x)\vec i + (y - a_y)\vec j + (z - a_z)\vec k$.

To the difference, a rotation leaves the origin in place, and rotates the axes, the unit vectors. So, we need a matrix, not a vector because we need to express each one of the new vectors is affected differently, so we need to express each one in terms of the old unit vectors. In our case of 3D space

$\vec i' = (\vec i' \vec i) \vec i + (\vec i' \vec j) \vec j + (\vec i' \vec k) \vec k$,

$\vec j' = (\vec j' \vec i) \vec i + (\vec j' \vec j) \vec j + (\vec j' \vec k) \vec k$,

$\vec j' = (\vec k' \vec i) \vec i + (\vec k' \vec j) \vec j + (\vec k' \vec k) \vec k$.

The nine inner products given above form the 9 matrix-elements for passing from the old coordinates $x, y, z$ to the new coordinates $x', y', z'$.

A rotation can be not only in the ordinary space, but also in a space of functions. The things go similarly, we have a matrix because we have $n$ vectors to express in terms of $n$ new vectors.


By n, I think you mean the dimension of the space. Take ordinary R3, Euclidean 3-space.

I would imagine that a rotation can be very nicely represented by a 3-vector (x,y,z) where the direction of the vector gives the AXIS and the length of the vector gives the angle of rotation.

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    $\begingroup$ True, but then you couldn't represent a translation in the same way as a vector anymore, so what's the difference. $\endgroup$ – Syd Kerckhove Jan 3 '15 at 2:49
  • $\begingroup$ there is no difference in the amount of information needed to represent them. In either case, 3 numbers will do, either to express a rotation or a translation in R3. $\endgroup$ – leaveswater02 Jan 3 '15 at 3:02
  • $\begingroup$ I agree, but the three numbers are not of the same kind. One has to do with euler angles and the other has to do with positional coordinates. $\endgroup$ – Syd Kerckhove Jan 3 '15 at 3:03
  • $\begingroup$ @SydKerckhove -- This answer uses the axis-angle representation of 3D rotation, but as a "vector". (I used vector in quotes because this is not a vector.) This representation of rotations in three dimensions is distinct from Euler angles. $\endgroup$ – David Hammen Aug 23 '16 at 9:27
  • $\begingroup$ Denoting this "vector" as $\vec\theta$, a rotation in 3D space can be expressed as $\vec v_{\text{rot}} = \vec v + \frac{\sin|\vec\theta|}{|\vec \theta|}\,\vec\theta\times \vec v + \frac{1-\cos|\vec\theta|}{|\vec \theta|^2} \, \vec\theta\times(\vec\theta\times \vec v)$. This is one way to write Rodrigues' rotation formula. Whether this qualifies as "nicely" is a matter of interpretation. $\endgroup$ – David Hammen Aug 23 '16 at 9:34

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