# Orthogonal character of rotation matrix

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1. Oct 2, 2015

### 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?

2. Oct 2, 2015

### 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.

3. Oct 2, 2015

### Staff: Mentor

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

4. Oct 12, 2015

### brotherbobby

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?

5. Oct 12, 2015

### samalkhaiat

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 $$\{ (\bar{x})^{2} = (x)^{2} \} \ \Leftrightarrow \ \{ R^{-1} = R^{T} \} .$$

Last edited: Oct 12, 2015
6. Oct 12, 2015

### Fredrik

Staff Emeritus
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).

Last edited: Oct 12, 2015