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Somewhat prompted by the discussions of Qiaochu Yuan and Aryabhata in this question, I realized that my understanding of linear/affine transformations thus far had been built on a convoluted series of circular arguments. I will now be asking a question in order to patch the gaps in my knowledge.

Due to my innate tendency to view things geometrically, I had always taken for granted the fact that rotations, reflections, and translations, among other affine transformations, are isometric (That is to say, no matter how you move, spin, or reflect an object, all the lengths and angles do not change at all).

To put it another way, I considered these transformations being isometries as postulates.

My question, then, is if there is a rigorous (so probably non-geometric) way of justifying that these transformations are isometries.

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After mulling over the first part of Jason's answer and suddenly remembering some lessons in CG... I suppose I should be replacing the adjective "linear" with "affine"? – J. M. Aug 25 '10 at 5:43
up vote 6 down vote accepted

The first point I'd make is that translations are not linear transformations because every linear transformation must take 0 to 0, and the only translation that does this is the identity map.

However, here's a general way to see exactly which linear transformations are isometries (including rotations and reflections).

One way is to start with the Euclidean inner product. In the "standard basis", it looks like $\langle v,w\rangle = \sum_i v_i w_i$. Once we have this, the length of a vector $x$ is defined to be $\sqrt{\langle x,x\rangle}$. The notation for this is $|x|$. The angle between two vectors $x$ and $y$ is defined by $cos\theta = \frac{\langle x,y\rangle}{|x||y|}$

We'll say a function "preserves the inner product" if $\langle x , y\rangle = \langle f(x), f(y)\rangle$ for all $x$ and $y$.

Claim 1. A linear function preserves the inner product iff the linear function preserves all lengths.

The proof one way is trivial. Assuming the linear function preserves all lengths, notice that

$|x|^2 + 2\langle x,y\rangle + |y|^2 = \langle x+y,x+y\rangle = \langle f(x) + f(y), f(x) + f(y)\rangle $

$= |f(x)|^2 + 2\langle f(x),f(y)\rangle + |f(y)|^2$.

Now, using the fact that $|x|^2 = |f(x)|^2$, we learn that $\langle x,y\rangle = \langle f(x), f(y)\rangle$. (Note that, as an added bonus, if a linear transformation preserves length, it preserves the inner product and hence it automatically preserves angles as well.)

The next thing worth noting is that you can move a linear transformation around in an inner product. That is, if $A$ is any matrix, then there is a unique matrix $B$ such that $\langle Ax, y\rangle = \langle x,By\rangle$ for all $x$ and $y$. In fact, in an orthonormal basis, $B$ is simply given as the transpose of $A$ - that is, $B = A^t$. The proof is simple: let $e_i$ be an orthonormal basis. Then $A_{ij} = \langle Ae_i, e_j\rangle = \langle e_i, Be_j\rangle = B_{ji}$.

Finally, we come to exactly how to recognize isometries.

Claim 2: the matrix $A$ is an isometry iff $A^t A = Id$. It turns out, this will automatically imply $AA^t = Id$.

Here's the proof.

Assume $A$ is an isometry. Then $\langle x,y\rangle = \langle Ax, Ay\rangle = \langle x, A^t A y\rangle$. Thus, we have $\langle x,y\rangle = \langle x, A^tA y\rangle$ for all $x$ and $y$. If we set $y = x$, then we get $|x|^2 = |x||A^tA x|cos\theta$. But since $A$ is an isometry, so is $A^t$ and hence $|A^t A x| = |x|$. Thus, we get $|x|^2 = |x|^2 cos\theta$ which tells us $\theta = 0$, i.e, that $A^tA x = x$.

Conversely, if $A^tA = Id$, then $\langle x,y\rangle = \langle x, A^t A y\rangle = \langle Ax, Ay\rangle$, showing that $A$ is an isometry.

Finally, one just verifies that every rotation matrix and reflection matrix satisfies $A^tA = Id$.

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I also wanted to add the following theorem, which I don't remember how to prove at the moment (but I remember the proof involves drawing triangles a looking at their image): Suppose f is a map from R^n to R^n which preserves lengths and satisfies f(0) = 0. Then f is automatically a linear transformation. – Jason DeVito Aug 25 '10 at 5:16
"Finally, one just verifies that every rotation matrix and reflection matrix satisfies..." - well, I deal in numerical linear algebra so this is now my point of familiarity. Also, your point regarding translations is duly noted. Thanks. – J. M. Aug 25 '10 at 5:20
Jason, I'm not sure that the theorem you claim is true. Consider, for example, the map ff which rotates the vector $x$ by $||x||_2 {\rm mod} 2 \pi$ radians. Certainly, the norm of $f(x)$ is the same as the norm of $x$, $f(0)=0$, but $f$ is not a linear transformation. We even have that this $f$ is continuous. – alexx Aug 25 '10 at 6:31
Alex, You're right. What I should have said is that if f(0) = 0 and f preserves all distances (not just lengths of vectors), then f will be a linear transformation. Thanks for catching that! – Jason DeVito Aug 25 '10 at 14:54

You implicitly define rotations, translations, and reflections as certain kinds of affine transformations. Thus it suffices to check algebraically that they preserve distances. (There is no need to check that they preserve angles: that follows immediately by polarization.)

That's not terribly deep. More insight is afforded through the usual geometric demonstration that reflections and translations are composed of a finite number of reflections. That a reflection is an isometry is not only geometrically obvious but also easy to show algebraically.

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As in my comment on the question to which you refer, exactly how you define things is critical. Suppose we consider transformations of the plane where the plane is defined to be $\mathbb{C}$ and a transformation of the plane is a 1-1 function $f:\mathbb{C}\to\mathbb{C}$. An isometry is a transformation that preserves distances, so a transformation $f$ is an isometry if and only if $|f(a)-f(b)|=|a-b|$ for all $a,b\in\mathbb{C}$.

Theorem: Any function of the form $f(z)=cz+d$ (direct isometry) or $f(z)=c\bar{z}+d$ (opposite isometry) where $c,d\in\mathbb{C}$ and $|c|=1$ is an isometry.

proof: (direct isometry) $|f(a)-f(b)|$ $=|ca+d-(cb+d)|$ $=|ca-cb|$ $=|c(a-b)|$ $=|c|\cdot|a-b|$ $=1\cdot|a-b|$ $=|a-b|$. (opposite isometry) $|f(a)-f(b)|$ $=|c\bar{a}+d-(c\bar{b}+d)|$ $=|c\bar{a}-c\bar{b}|$ $=|c(\bar{a}-\bar{b})|$ $=|c|\cdot|\overline{a-b}|$ $=1\cdot|a-b|$ $=|a-b|$.

Also note that if $f$ and $g$ are isometries, so is $f\circ g$.

For our purposes and for simplicity, define a translation by $a+bi$ to be the function $T_{a+bi}(z)=z+a+bi$, a rotation about 0 by $\theta$ to be the function $R_\theta(z)=(\cos\theta+i\sin\theta)z$, and reflection over the real axis to be $r_\mathbb{R}(z)=\bar{z}$. (It should be relatively easy to show that these definitions align with any geometric definitions.) Each of these is clearly an isometry, based on the above theorem.

For a rotation about some point $p$ rather than about 0, apply $T_p\circ R_\theta\circ T_{-p}$. For a reflection over a line parallel to the real axis and passing through $bi$, apply $T_{bi}\circ r_\mathbb{R}\circ T_{-bi}$. For a reflection over a line not parallel to the real axis, intersecting the real axis at $a$ at an angle $\theta$, apply $T_a\circ R_\theta\circ r_\mathbb{R}\circ R_{-\theta}\circ T_{-a}$. Each of these is also an isometry because it is the composition of isometries.

Thus, any rotation, reflection, or translation of the plane is an isometry.

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I'm already familiar with manipulations in the complex plane, but never really thought to just start from the definitions of the operations themselves (as I said, I took them for granted). Thanks! – J. M. Aug 25 '10 at 5:00
It may be worth noting that the theorem I stated and proved is half of a theorem called the Complex Number Isometry Theorem in the 3rd edition of UCSMP Precalculus and Discrete Mathematics (which is stated there without proof). The other half is that all isometries of the complex plane can be expressed in one or the other of those two forms (I don't know how to prove that part, though). – Isaac Aug 25 '10 at 5:16
the converse follows from the argument in Jason DeVito's answer. – Qiaochu Yuan Aug 25 '10 at 16:46

Ok, this one is pretty easy.

Check out the theory of orthogonal matricies on wiki. The basic idea is that the matrix corresponding to the isometric linear transformation must be orthogonal. The proof is really pretty, but basically the idea is to show that an inner product of two vectors

(X,Y) is equal to (AX,AY) for a matrix A, if and only if the transpose of A is the same as it's inverse. i.e. A is orthogonal. The next step is to show that the rotation matrix is orthogonal, and that's a rigorous proof.

This idea is easily generalized to non-euclidean metrics. Although it might be at too high a level, I absolutely love matrix lie groups by brian C hall. The first chapter is on google books, and contains some simple proofs. This book only gets super hard in chapter 3;lie groups/algebras is one of my favorite subjects and is certainly where this line of questioning will lead you.

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Let's consider $\mathbb{R}^3$ as a vector space on $\mathbb{R}$, endowed with the standard scalar product $\langle,\rangle$ (also know as $\mathbb{E}^3$, the Euclidean space). We define a function $f: \mathbb{R}^3 \to \mathbb{R}^3$ to be an isometry if $||f(u)-f(w)|| = ||u-w||$ for all $v,w \in \mathbb{R}^3$, i. e. if $f$ preserve distances.

Transaltions are isometries

We define a translation by $z$ the map $\tau_{z}: \mathbb{R}^3 \to \mathbb{R}^3$ such that $\tau_{z}(u) = u+z$ for all $u \in \mathbb{R}^3$. Then we have $$||\tau_{z}(u)-\tau_{z}(w)|| = ||(u+z) - (w+z)|| = ||u-w||$$ for all $u,w \in \mathbb{R}^3$ so we are done.

Now we consider reflections and rotations. They are both real unitary linear maps (also known as orthogonal, or members of $O(3)$). The following are equivalent propieties of these transformations of the space:

  • $\langle L(u),L(w)\rangle = \langle u,w\rangle$ for all $L \in O(3)$ $u,w \in \mathbb{R}^3$ or they preserve scalar product
  • $||L(u)|| = ||u||$ for all $L \in O(3)$ $u \in \mathbb{R}^3$ or they preserve lenghts
  • $||L(u)|| = ||u||$ for all $L \in O(3)$ and $u \in \mathbb{R}^3$ with $||u|| = 1$ or the unit sphere is $L$-invariant

Now, we'll see that all real unitary transformation of the space are isometries, proving then that reflections and rotations are of this type we have finished.

Real unitary transformations are isometries

If $L \in O(3)$ then $$||L(u)-L(w)|| = \sqrt{\langle L(u)-L(w),L(u)-L(w)\rangle} = \sqrt{\langle L(u-w),L(u-w)\rangle}$$ by definition of norm and using the linearity of L, and $$\sqrt{\langle L(u-w),L(u-w)\rangle} = \sqrt{\langle u-w,u-w\rangle} = ||u-w||$$ by using one of the three proprieties above.

Reflections are unitary transformations (and thus isometries)

If $S\subseteq\mathbb{R}^3$ is a subspace, the we define reflection in respect to $S$ the linear map $\rho_{S}: S \oplus S^{\bot} \to S \oplus S^{\bot}$ such that if $u = u_{S}+u_{S^{\bot}}$ then $\rho_{S}(u) = u_{S}-u_{S^{\bot}}$. So we have $$||u_{S}±u_{S^{\bot}}|| = \sqrt{||u_{S}||^2 + ||u_{S^{\bot}}||^2}$$ and then $||\rho_{S}(u)|| = ||u||$ for all $u \in \mathbb{R}^3$, that is one of the three equivalent proprieties above.

Rotations are unitary transformations (and thus isometries)

This is a little weird. As you can read here, perhaps the best way to define rotations is to consider them as paricular members of $O(3)$, so there is nothing to prove.

Note: what has been said remains true for an arbitrary space $V$ of dimension $n$ on $\mathbb{R}$ endowed with a positive product $\langle ,\rangle$ because of the isomorphism with $\mathbb{E}^n$, this is the power of linear algebra!

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