# Prove that 3d rotation is linear

In a 2d space, a transformation is linear if $f(v+w) = f(v) + f(w)$ and $f(kv) = k*f(v)$, and rotation preserves addition so it is linear. In a 3d space, similar rules apply: $(x, y, z) + (l, j, k) = (x + t, y + j, z + k)$, and $k(x, y, z) = (kx, ky, kz)$.

It follows that applying a given rotation matrix in 3d space

(for example $M = \begin{bmatrix} 1 & 0 & 0 \\ 0 & cos\theta & -sin\theta \\ 0 & sin\theta & cos\theta\end{bmatrix}$ )

should also be linear. The above is just a rotation along the x-axis so it seems trivial. How can I prove that all 3d rotations are linear, given a rotation matrix M?

Edit

How can I prove that all 3d rotations are linear using the definition of linearity, ie using multiplication and addition? Can I use the fact that rotation itself is multiplication, and it preserves addition?

• I think this is probably related to the parallelogram rule for adding vectors – Akiva Weinberger Apr 27 '16 at 23:38
• @AkivaWeinberger I think it is too, just not sure how to show that. – jackwise Apr 27 '16 at 23:42
• IMHO, this question either doesn't make sense or is a tautology. If your rotation is given as a matrix, then since matrix multiplication is linear, the rotation is linear. I think you should pick a definition of rotation that doesn't use matrix at all. e.g. any transform which 1) fix the origin and 2) preserve euclidean distances between any two points and then show that transform has to be a linear one. – achille hui Apr 28 '16 at 3:08

Any rotation is a rotation around some axis. You can write it as a change of basis matrix times a standard rotation matrix (similar to $R_x$, but just being around another non standard axis) in the new basis, then back to the standard basis. So it is just matrix multiplication, and matrix multiplication is linear.

$M$ is a rotation matrix and generates a linear transformation $T$. It operates on vectors $v$ by $T(v)=Mv$. Matrix multiplication is linear, so $T$ is a linear transformation.

• I'm not that great at words & math - could you please give an example? – jackwise Apr 27 '16 at 23:45
• It may be clearer to note that any rotation matrix can be written as a product $M=R_xR_yR_z$. But all we need is that a rotation is a matrix, and that matrix multiplication is linear. – jdods Apr 27 '16 at 23:57
• Is it necessary to change the basis at all? It seems that if the rotation matrix multiplication is linear, that should be enough to begin with. – theREALyumdub Apr 27 '16 at 23:57
• No, but since op knew a standard rotation matrix was linear, I figured it would help since any rotation is rotation around an axis. I.E. just a standard rotation in some basis. – jdods Apr 28 '16 at 0:00
• @jdods I clarified my question, not sure if your answer still applies. – jackwise Apr 28 '16 at 0:30

I think this question has nothing to do with bases and matrix representations. This question is an excellent example of treating linear transformations on their own terms rather than analyzing their component space representations.

A transformation is linear if it doesn't matter if you add two vectors first and then transform the sum or if you transform each vector first and then add together the results + and the same for multiplication by a scalar. Here's a video that explains this: https://www.lem.ma/17

And here's a video that shows that rotations are linear: https://www.lem.ma/gs The video is for rotations in the plane, but there is no difference as far as linearity is concerned.