Proving "Rotation Matrix is Orthogonal: Necessary & Sufficient

[brotherbobby]

I'd like to prove the fact that - since a rotation of axes is a length-preserving transformation, the rotation matrix must be orthogonal.

By the way, the converse of the statement is true also. Meaning, if a transformation is orthogonal, it must be length preserving, and I have been able to prove it. This is the so called "sufficient" statement.

But how to prove the "necessary" statement above? Any help?

 

[mpresic]

I am sure there are easier ways to do this than this procedure. Take u to be a vector. Take v to be the transformed vector, of the same length. Translate v to have the same origin as u. Now you can find the angle between u and v (with the same origin) by taking the dot product. The cross product between u and v (with the same origin) will provide a vector perpendicular to both u and v, call it w. Vector w will have length sin (angle between u and v). Divide the w by sin angle to make the vector unit length. Now you can form the rotation matrix that will rotate u into v, or vice versa, v into u. You can multiply either by the transpose to verify you come up with the unit matrix.

You might try this as a last resort although I am sure there are easier ways. If you do accomplish this, you will learn a lot about rotations.

 

[PeterDonis]

if a transformation is orthogonal, it must be length preserving, and I have been able to prove it.

How did you prove it? If all the steps in your proof can be reversed, then you have the proof you are looking for.

 

[brotherbobby]

How did you prove it? If all the steps in your proof can be reversed, then you have the proof you are looking for.

No am afraid, it is not like that, the argument cannot be reversed like that. Let me present it so that you know what I mean.

Imagine the rotation matrix to be orthogonal and prove that it will preserve the length of a vector when the coordinate system undergoes rotation.

The "new" length of the vector is $x'_i x'_i = R_{ij}x_j R_{ik} x_k \text{, (summation on repeated indices implied)} = R_{ij} R_{ik} x_j x_k = \delta_{jk} x_j x_k \text{(since the rotation matrix is orthogonal)}=x_k x_k = x_i x_i$,

which shows that the length is invariant.

The argument cannot be reversed. I need to begin from the fact that the length is an invariant, i.e. $x'_i x'_i = x_i x_i$ and then derive that if it is so, the rotation matrix has to be orthogonal, i.e. $R_{ij} R_{ik}=\delta_{jk}$. Any help?

 

[samalkhaiat]

No am afraid, it is not like that, the argument cannot be reversed like that. ... Any help?

Yes, you can reverse your argument. x_{i} x_{i} = \delta_{jk} x_{j}x_{k} , \ \ \ \ \ \ \ (1) \bar{x}_{i} \bar{x}_{i} = R_{ij}R_{ik} x_{j} x_{k} . \ \ \ \ \ (2) So, if the left-hand-sides of (1) and (2) are equal, then R_{ij}R_{ik} = \delta_{jk}. You can also do it by writing R_{ij} = (e^{\omega})_{ij} = \delta_{ij} + \omega_{ij} + O(\omega^{2}) , work to first order in \omega and show that \omega_{ij} = - \omega_{ji}. This means that the matrix R is orthogonal: R^{-1}_{ij} = R_{ji} = R^{T}_{ij} . More generally, given \bar{x} = Rx, where x = (x_{1}, x_{2}, \cdots , x_{n})^{T} and R is a real n \times n, you can easily show that \{ (\bar{x})^{2} = (x)^{2} \} \ \Leftrightarrow \ \{ R^{-1} = R^{T} \} .

 

[Fredrik]

Definition: Let X be an inner product space over $\mathbb R$. For each $x,y\in X-\{0\}$, define the angle between $x$ and $y$, denoted by $\theta(x,y)$, by
$$\cos\theta(x,y)=\frac{\langle x,y\rangle}{\|x\|\|y\|}.$$
Note that the Cauchy-Schwartz inequality ensures that the right-hand side is in the interval [-1,1].

Theorem: Let X be an inner product space over $\mathbb R$. If $R:X\to X$ is surjective onto $X$ and preserves distances and angles, then $R$ is an orthogonal linear transformation (i.e. a linear transformation that preserves norms).

Proof: We will prove that $R$ preserves norms. Let $x\in X$. Since $R$ preserves distances, we have $\|Rx\|=\|R(x-0)\|=\|x-0\|=\|x\|$. Since $x$ is an arbitrary element of $X$, this means that $R$ preserves norms.

We will prove that $R$ preserves inner products. Let $x,y\in X-\{0\}$. Since $R$ preserves angles and norms, we have
$$\frac{\langle x,y\rangle}{\|x\|\|y\|}=\cos\theta(x,y)=\cos\theta(Rx,Ry)=\frac{\langle Rx,Ry\rangle}{\|Rx\|\|Ry\|} =\frac{\langle Rx,Ry\rangle}{\|x\|\|y\|}.$$ This implies that $\langle x,y\rangle=\langle Rx,Ry\rangle$. Since $x,y$ are arbitrary elements of $X$, this means that $R$ preserves inner products.

We will prove that $R$ is linear. Let $a,b\in\mathbb R$. Let $x,y,z\in X$. Let $w\in X$ be such that $Rw=z$.
\begin{align*}
&\langle z,R(ax+by)\rangle =\langle Rw,R(ax+by)\rangle =\langle w,ax+by\rangle =a\langle w,x\rangle+b\langle w,y\rangle =a\langle Rw,Rx\rangle+b\langle Rw,Ry\rangle\\
& =\langle z,aRx+bRy\rangle.
\end{align*} Since $z$ is an arbitrary element of $X$, this implies that $R(ax+by)=aRx+bRy$. Since $a,b$ are arbitrary real numbers and $x,y$ are arbitrary elements of $X$, this means that $R$ is linear.

Comment: You may still be wondering why a norm-preserving linear transformation would be orthogonal in the sense $R^TR=I$. The standard inner product on $\mathbb R^3$ is given by $\langle x,y\rangle =x^Ty$. So for all $x\in\mathbb R^3$, we have
$$x^Tx=\langle x,x\rangle =\|x\|^2=\|Rx\|^2=\langle Rx,Rx\rangle =(Rx)^T(Rx)=x^TR^TRx.$$ This implies that $R^TR=I$. (The idea is to make clever choices of x that I don't have time to explain right now).