1. ## Visualization of rotation

I am trying to visualize the following rotation of R^3, but it is very difficult. I want to get the answer by intuition, and not by using the Rodrigues rotation formula or conjugation of matrices, etc.

Problem statement: Determine the matrix that represents the following rotation of R^3: an angle of pi/2 about the fixed axis containing the vector (1,1,0)^t

Here is what I have tried in my diagram:

![Coordinate axes][1]

Should I find a 3x3 rotation matrix A such that A(1,1,0)^t=(-1,1,0)^t?

[1]: http://i.stack.imgur.com/88MKh.png

2. ## Re: Visualization of rotation

Consider a rotation by an angle of pi/2 about the fixed axis containing the vector $e_1$. That would be a rotation that would take $e_2\mapsto e_3$ and $e_3 \mapsto -e_2$. So, find a basis for $\mathbb{R}^3$ containing (1,1,0)[/tex]. Then, you want a 3x3 rotation matrix that will take (1,1,0) to itself, the second basis vector to the third, and the third basis vector to negative the first.

3. ## Re: Visualization of rotation

Would the basis {[1,0,0],[0,1,0]} work? For the rotation matrix, could you show me an example?

4. ## Re: Visualization of rotation

Oh, I should have mentioned that you want an orthonormal basis. For example: (1,1,0), (-1,1,0), (0,0,1) are a set of three orthogonal vectors. So, an orthonormal basis would be: $\dfrac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, \dfrac{1}{\sqrt{2}}\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$. Also, you can always use the Gram-Schmidt process if you cannot find a simple orthogonal basis to normalize.

So, you want:

$A\begin{pmatrix}\dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{2}} & 0 \\ \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}\dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} & 0 & -\dfrac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}$

That means $A = \begin{pmatrix}\dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} & 0 & -\dfrac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}\dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{2}} & 0 \\ \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix}^{-1} = \dfrac{1}{2}\begin{pmatrix}1 & 1 & \sqrt{2} \\ 1 & 1 & -\sqrt{2} \\ -\sqrt{2} & \sqrt{2} & 0\end{pmatrix}$