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Orthogonal character of rotation matrix

  1. Oct 2, 2015 #1
    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. jcsd
  3. Oct 2, 2015 #2
    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.
     
  4. Oct 2, 2015 #3

    PeterDonis

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    How did you prove it? If all the steps in your proof can be reversed, then you have the proof you are looking for.
     
  5. Oct 12, 2015 #4
    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?
     
  6. Oct 12, 2015 #5

    samalkhaiat

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    Yes, you can reverse your argument. [tex]x_{i} x_{i} = \delta_{jk} x_{j}x_{k} , \ \ \ \ \ \ \ (1)[/tex] [tex]\bar{x}_{i} \bar{x}_{i} = R_{ij}R_{ik} x_{j} x_{k} . \ \ \ \ \ (2)[/tex] So, if the left-hand-sides of (1) and (2) are equal, then [itex]R_{ij}R_{ik} = \delta_{jk}[/itex]. You can also do it by writing [tex]R_{ij} = (e^{\omega})_{ij} = \delta_{ij} + \omega_{ij} + O(\omega^{2}) ,[/tex] work to first order in [itex]\omega[/itex] and show that [itex]\omega_{ij} = - \omega_{ji}[/itex]. This means that the matrix [itex]R[/itex] is orthogonal: [tex]R^{-1}_{ij} = R_{ji} = R^{T}_{ij} .[/tex] More generally, given [itex]\bar{x} = Rx[/itex], where [itex]x = (x_{1}, x_{2}, \cdots , x_{n})^{T}[/itex] and [itex]R[/itex] is a real [itex]n \times n[/itex], you can easily show that [tex]\{ (\bar{x})^{2} = (x)^{2} \} \ \Leftrightarrow \ \{ R^{-1} = R^{T} \} .[/tex]
     
    Last edited: Oct 12, 2015
  7. Oct 12, 2015 #6

    Fredrik

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    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
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