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Given a unit vector $\vec a$ $ \epsilon R $, prove the following transformation represents a rotation matrix:

$$ T\vec x=(\vec a \cdot \vec x)\vec a + (\vec a\times \vec x ).$$

If we take a look what this transformation does to the basis vectors $S =(i,j,k)$ in $R^3$ we get the matrix $$T_{SS}\begin{bmatrix}a_1^2 & -a_3 & a_2 \\a_3 & a_2^2 & -a_1 \\ -a_2 & a_1 & a_3^2\end{bmatrix},$$ which is an Skew-symmetric matrix which means the eigenvalues are purely imaginary. The determinant is obviously non negative. And we know the matrix can represents a rotation mirroring or both,

How do I prove that T is really a represents a rotation, and how do I find the axis of rotation?.

Note: I havent used the fact that $\mid \vec a \mid= 1$ .

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  • $\begingroup$ $T_{SS}$ is not skew-symmetric as its diagonal elements are not all $0$. Moreover, every real $3\times3$ matrix has at least one real eigenvalue. Then again, neither is $T_{SS}$ the matrix that corresponds to $T$, so that’s all moot. Check your calculations carefully. $\endgroup$ – amd Jun 1 at 23:41
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Hints!

  • $T\vec{a} = \vec{a}$
  • If $\vec{x} \perp \vec{a}$, then $T\vec{x} = \vec{x} \times \vec{a}$, which is perpendicular to $\vec{x}$ and $\vec{a}$.
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  • $\begingroup$ The first hint tells me $\vec a$ stays fixed which means its the axis of rotation, the second hint should probably tell me why $T$ is a rotation, but I don't quite get it. $\endgroup$ – Nejc.Z Jun 1 at 14:17
  • $\begingroup$ @Nejc.Z That's right for the first one. For the second hint, I'm applying $T$ to the plane of vectors that are orthogonal to this proposed axis of rotation, and obtaining a vector perpendicular to the vector $\vec{a}$ (hence is still in the plane) and is perpendicular to $\vec{x}$, meaning that it has rotated $\pi/2$ radians. So, it is a rotation by $\pi/2$ around $\vec{a}$, with the orientation depending on the direction of $\vec{a}$ (rotating the other way comes from replacing $\vec{a}$ with $-\vec{a}$. $\endgroup$ – Theo Bendit Jun 1 at 14:28
  • $\begingroup$ I get it now, thank you $\endgroup$ – Nejc.Z Jun 1 at 17:09
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Hint.

$$ T = \vec a^{\dagger}\cdot \vec a + \vec a\times $$

then

$$ T^{\dagger}\cdot T = (\vec a^{\dagger}\cdot \vec a + \vec a\times)^{\dagger}\cdot(\vec a^{\dagger}\cdot \vec a + \vec a\times) = (\vec a^{\dagger}\cdot \vec a)^2 -(\vec a\times) ^2 $$

because $(\vec a\times)^{\dagger}=-\vec a\times$

hence

$$ T^{\dagger}\cdot T =\left( \begin{array}{ccc} a_1^4+\left(a_2^2+a_3^2\right) a_1^2+a_2^2+a_3^2 & a_1 a_2 \left(a_1^2+a_2^2+a_3^2-1\right) & a_1 a_3 \left(a_1^2+a_2^2+a_3^2-1\right) \\ a_1 a_2 \left(a_1^2+a_2^2+a_3^2-1\right) & a_2^4+\left(a_2^2+1\right) a_3^2+a_1^2 \left(a_2^2+1\right) & a_2 a_3 \left(a_1^2+a_2^2+a_3^2-1\right) \\ a_1 a_3 \left(a_1^2+a_2^2+a_3^2-1\right) & a_2 a_3 \left(a_1^2+a_2^2+a_3^2-1\right) & a_3^4+a_1^2 \left(a_3^2+1\right)+a_2^2 \left(a_3^2+1\right) \\ \end{array} \right) $$

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