# How do you formally prove that rotation is a linear transformation?

The fact that rotation about an angle is a linear transformation is both important (for example, this is used to prove the sine/cosine angle addition formulas; see http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry) and somewhat intuitive geometrically. However, even if this fact seems fairly obvious (at least from a diagram), how does one turn the picture proof into a formal proof? On a related note, it seems likely that many formal proofs using a diagram will end up relying on Euclidean geometry (using angle/side congruence properties), but isn't one of the points of linear algebra to avoid using Euclidean geometry explicitly?

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With what definitions of "rotation" and "linear transformation" are you working? –  Isaac Aug 24 '10 at 23:37
A sketch: matrices represent linear transformations, and rotations can be represented by matrices. A bit handwave-y though. –  Guess who it is. Aug 24 '10 at 23:41
@J. Mangaldan: this is just restating the question. –  Qiaochu Yuan Aug 25 '10 at 0:00
As I said... handwave-y. Looking at it again, circular too. :) –  Guess who it is. Aug 25 '10 at 0:27
With these definition's guest's question becomes: if $f:\mathbb R^2 \to \mathbb R^2$ is an isometry of metric spaces such that $f(0)=0$, $f(1)=1$ and $f(i)=i$, then $f=Id_{\mathbb R^2}$. And proving reduces quickly to the issue that straight lines are the unique continuous curve that minimizes length between points. A standard (and non-trivial) thing to prove. –  Ryan Budney Aug 25 '10 at 17:42

The problem comes in defining rotation so that this whole thing doesn't wind up circular, doing so fully would undoubtedly tie us all in knots, but one can prove the linearity of rotations by observing a few choice properties:

1. Rotations preserve angles
2. They fix the origin
3. And they preserve lengths

Now just draw a parallelogram $OACB$ (spanned by vectors $\vec{OB}, \vec{OA}$- with $\vec{OC}=\vec{OB}+\vec{OA}$), rotate and chase the angles and lengths to observe that $R(\vec{OB})+R(\vec{OA})= R(\vec{OC})= R(\vec{OB}+ \vec{OA})$.

It's sort of Euclidianish but it's the only decent way to play it (especially if we want higher dimensions) without the 'lucky guess' approach championed by Qiaochu.

If you feel pushed, you can turn the angle and length chase into norms and inner products but that would be a tragic waste of an isomorphism- here we are dealing with a geometric object in the language of vectors and to treat it non-geometrically would put us at an unnecessary disadvantage.

A good deal of higher mathematics depends on finding 'the best place' to solve a given problem, but the best place need not be the most abstract one: translating something about PDE into hilbert space theory may help find a solution, but it may also rob us of our geometric techniques and intuition.

A great example is something called the h-cobordism theorem, which can (and does- Milnor's take is a gem) take up a small book, using formal techniques (morse theory) to prove it, but can be done in half the time with some geometric (handlebody) jazz.

Vector spaces are another string to our bow, a new place to do math, and at times a more rigorous place- but they are not (at least from my perspective) the only place.

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There's a pun in the first sentence somewhere. Rotation... circular... knots... –  Qiaochu Yuan Aug 25 '10 at 3:13
Why khovanov course! Knot arf glad mu made the link Qiaochu! :p –  Tom Boardman Aug 25 '10 at 13:54

Maybe my first answer was not enough "formal", since I didn't say explicitely what I was assuming a "rotation" is -nor "angle". (Exercise: (1) find in which places I was using those unsaid definitions of "rotation" and "angle" and (2) which definitions are they. :-) )

Well, it was a sort of "high-school proof". The problem about the definition of "rotation", as some answers have already pointed out, or implied, is that, if you really want to be formal, you end saying that a rotation is some kind of linear transformation by definition. Full stop.

Nevertheless, I think I can produce an "undergraduate" proof of this fact, avoiding circular arguments. It has the merit that is valid in any dimension. For which I need the following

Convention. Whatever "rotation" means,

1. It is a map of some vector space $V$,
2. Which has a way of measuring "lengths" and "angles" of its vectors, and
3. "Rotations" preserve those "lengths" and "angles".

Now, a fancy way to have "lengths" and "angles" in a vector space is to have a dot product in it. So let's assume that $V$ is an euclidian vector space; for instance, a real vector space such as $\mathbb{R}^n$ with the standard dot product.

Then, in such a $V$, "length" means "norm", which is $\| v \| = +\sqrt{v\cdot v}$.

We have to be more careful with "angles" because the standard definition already involves rotations. To avoid a circular argument, we define the (non-oriented) angle determined by two vectors $v, w$ as the unique real number $\theta = \widehat{vw} \in [0, \pi ]$ such that

$$\cos \theta = \frac{v\cdot w}{\|v\| \|w\|} \ .$$

Notice that this definition makes sense since, because of the Cauchy-Schwarz inequality, we always have $-1 \leq \frac{v\cdot w}{\|v\| \|w\|} \leq 1$ and $\cos : [0, \pi] \longrightarrow [-1,1]$ is bijective.

So, "rotation" is some kind of map $f: V \longrightarrow V$ which preserves norms and angles. Since the dot product can be expressed in terms of norms and (cosines of) angles,

$$v\cdot w = \|v\| \|w\| \cos\widehat{vw}$$

rotations preserve dot products:

$$f(v) \cdot f(w) = v\cdot w \ .$$

Now, let's show that a map that preserves the dot product is necessarily linear:

\begin{align} \| f(v+w) - f(v) -f(w) \|^2 &= \left( f(v+w) -f(v) -f(w) \right) \cdot \left( f(v+w) -f(v) -f(w) \right) \\\ &= \left( f(v+w)\cdot f(v+w) - f(v+w) \cdot f(v) - f(v+w)\cdot f(w) \right) \\\ &{} \qquad + \left( -f(v)\cdot f(v+w) + f(v) \cdot f(v) + f(v)\cdot f(w) \right) \\\ &{} \qquad + \left( -f(w)\cdot f(v+w) + f(w)\cdot f(v) + f(w)\cdot f(w) \right) \\\ &= \left( (v+w)\cdot (v+w) - (v+w) \cdot v - (v+w)\cdot w \right) \\\ &{} \qquad + \left( -v\cdot (v+w) + v \cdot v + v\cdot w \right) \\\ &{} \qquad + \left( -w\cdot (v+w) + w\cdot v + w\cdot w \right) \\\ &= \| v+w - v -w \|^2 = 0 \ . \end{align}

So

$$f(v+w) = f(v) + f(w) \ .$$

A similar computation shows that

$$\|f(\lambda v ) - \lambda f(v) \|^2 = \|\lambda v - \lambda v\|^2 = 0 \ .$$

Thus

$$f(\lambda v) = \lambda f(v) \ .$$

Hence, our rotation $f$ is a linear transformation.

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+1 for your "understandable" solution by anyone that knows some vector calculus, which is my case. –  Américo Tavares Aug 25 '10 at 22:39
Thank you, Américo. –  a.r. Aug 25 '10 at 23:48

If you want to be formal and avoid Euclidean geometry, it seems to me that the right way to go is the other way around: first define certain linear transformations (the ones which preserve the dot product and have determinant $1$) and then prove that they have the properties you expect them to have. More precisely, the group $\text{SO}(2)$ is abelian with Lie algebra $\mathbb{R}$, and the parameterization $\theta \mapsto \left[ \begin{array}{cc} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{array} \right]$ is the exponential map from the Lie algebra to the group. The methods for doing this should be covered in any book on Lie theory.

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Maybe you don't even have to say there is a $\theta$; just say you have a quantity c and another quantity s that are related by $c^2+s^2=1$. At least one can't get preconceived notions about angles. –  Guess who it is. Aug 25 '10 at 0:36
What 'properties' do you expect the transforms to have? How do you prove that those properties uniquely define rotation? Is rotation already defined (which is missing in your answer), or is this your way of defining rotation (in which case isn't it, in your own words, "restating the question"?). Please elaborate. –  Aryabhata Aug 25 '10 at 0:46
@Moron: the essential property is that SO(2) acts simply transitively on the unit circle. If you accept the existence of the circle, this is not hard to show. Is the existence of the circle what's in question here? (For this just follow the standard procedure for showing that square roots of non-negative real numbers exist.) –  Qiaochu Yuan Aug 25 '10 at 2:15
@Qiaochu: The property was part of the question, but the main issue was, if you are defining what rotation is, you haven't really answered the question ("restating the question" in your words). If you are not, what is the definition of rotation you are using? btw, I didn't downvote. –  Aryabhata Aug 25 '10 at 2:30
@Moron: the approach I am advocating is, as Tom says, a "lucky guess" approach. We define an abstract object and then show it has the properties that we "intuitively know" that rotation has, so we might as well call the abstract object rotation. (This is what we do, for example, when we formally construct R as a model for the Euclidean line.) The issue I had with your answer is that it did not address the second half of the approach. I am also trying to address the OP's desire to ignore Euclidean geometry; certainly it should in principle be possible to prove that Euclid's axioms are... –  Qiaochu Yuan Aug 25 '10 at 3:17

This depends on your definitions of "rotation" and "angle", but, if you are willing to accept the sum and difference formulas of your link http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry , then you may follow the reasoning in Quiochu Yuan's answer there the other way around: given a vector $(x,y) \in \mathbb{R}^2$, let $\alpha$ be the angle determined by $(x,y)$ and the $x$-axis and $r= \sqrt{x^2 + y^2}$ its length. Then, of course,

\begin{eqnarray*} x &=& r \cos \alpha \\ y &=& r \sin \alpha \end{eqnarray*}

If you rotate $(x,y)$ an angle $\theta$, you'll obtain the vector $(x',y')$

\begin{eqnarray*} x' &=& r \cos (\alpha + \theta ) \\ y' &=& r \sin (\alpha + \theta ) \ . \end{eqnarray*}

Now you apply those sum and difference formulas and get

\begin{eqnarray*} x' &=& r \cos\alpha\cos\theta - r \sin\alpha\sin\theta &=& \cos\theta \cdot x - \sin\theta \cdot y\\ y' &=& r \sin\alpha\cos\theta + r \cos\alpha\sin\theta &=& \sin\theta \cdot x + \cos\theta\cdot y \ . \end{eqnarray*}

Which is the same as saying that $(x',y')$ is obtained from $(x,y)$ multiplying with the matrix

$$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

Since the rotation $(x,y) \mapsto (x',y')$ is the same as multiplication by a matrix, it is a linear transformation.

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As an exercise in trigonometry.

The transformation of rectangular co-ordinates $(x,y)$ of $P$ into the rectangular co-ordinates $(x',y')$ of $P$ by a rotation of the axes is given by:

$x=x^{\prime }\cos \phi -y^{\prime }\sin \phi$

$y=x^{\prime }\sin \phi +y^{\prime }\cos \phi$.

This transformation is invertible.

Edit: This way of proving the transformation of rectangular co-ordinates does not "avoid using Euclidean geometry explicitly" (in this case trigonometry), as writen in the question.

However the same method can be used to derive the transformation of rectangular into spherical co-ordinates, which is a non-linear transformation.

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First how do you non-geometrically define rotation?

Perhaps in the 2D case, you can define it as multiplication by a complex number (which is in fact true!)

Once you do that, use the definition of a Linear Transformation.

Show that the map T satisfies:

1) $T(a + b) = T(a) + T(b)$ for any vectors $a,b$ (or complex numbers if you chose to go that route)
2) $T(\alpha b) = \alpha T(b)$ for any scalar $\alpha$ and any vectors $a,b$.

Note that, you can define complex numbers as vectors in the abstract without having to rely on any Euclidean Axioms.

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This is, again, just restating the question. (How do we know that multiplication by complex numbers is rotation?) –  Qiaochu Yuan Aug 25 '10 at 0:03
@Qiaochu: I said, define rotation to be multiplication by a complex number. What is wrong with that? It matches with the geometric intuition... Given the OP never gave a formal definition, I took the liberty :-) –  Aryabhata Aug 25 '10 at 0:24

You can prove this by looking at the appropriate diagram and using trig.

However, what is interesting is that it is not too difficult to show that any isometry of $\mathbb R^n$ (with its usual inner product) is necessarily linear. So if you accept that rotations are isometries (which to me at least is much more self-evident than them being linear) then you get linearity immediately.

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An isometry is not necessarily (quite) linear since it need not fix the origin. Slightly more fancily: the isometry group of $\mathbb{R}^n$ is the semidirect product of $\mathbb{R}^n$ (acting on itself by translation) by the orthogonal group $O(n)$. –  Pete L. Clark Aug 25 '10 at 7:18