Find the matrix AA representing rotation around the direction (1,1,1)(1, 1, 1) by π/2\pi/2 radians counter-clockwise

Consider the linear transformation T:R3â†′R3T : \Bbb R^3 → \Bbb R^3 given by rotation around the direction (1,1,1)(1, 1, 1) by \pi/2\pi/2 radians counter-clockwise (according to the right hand rule).

Find the matrix AA representing TT.

Got no idea how to start this problem. Can someone give me a starting point?

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Find an orthonormal basis in which the matrix has a simple form. Probably one of its vectors will be a normalized (1,1,1)(1,1,1) which will be mapped to itself, and the two other vectors v,wv,w will be orthogonal to it, being mapped to w,-vw,-v.
– Peter Franek
2 days ago

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2 Answers
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Rodrigues’ rotation formula states if \;\vec v\;\;\vec v\; is rotated about unit vector \;\vec k\;\;\vec k\; through an angle \;\theta\;\;\theta\; anticlockwise the rotated vector becomes

\vec v_{\text{rot}}=\vec v\cos\theta+(\vec k\times\vec v)\sin\theta+\vec k\cdot(\vec k\cdot\vec v)(1-\cos\theta)

\vec v_{\text{rot}}=\vec v\cos\theta+(\vec k\times\vec v)\sin\theta+\vec k\cdot(\vec k\cdot\vec v)(1-\cos\theta)

In our case \vec k={1\over\sqrt 3}(1,1,1)^\top,\quad\theta={\pi\over2}\vec k={1\over\sqrt 3}(1,1,1)^\top,\quad\theta={\pi\over2}
If RR is the coveted matrix,
R\vec v=\vec v_{\text{rot}}R\vec v=\vec v_{\text{rot}}
Taking \;\vec v=(1,0,0)^\top\;\vec v=(1,0,0)^\top we get

R\vec v=\text{1st column of }R=(\vec k\times\vec v)+\vec k\cdot(\vec k\cdot\vec v)=\frac{1}{3}(1,1+\sqrt3,1-\sqrt3)^\top

R\vec v=\text{1st column of }R=(\vec k\times\vec v)+\vec k\cdot(\vec k\cdot\vec v)=\frac{1}{3}(1,1+\sqrt3,1-\sqrt3)^\top

Similarly other columns of RR can be computed. In fact,

R=\frac{1}{3}\begin{bmatrix}
1 & 1-\sqrt3 & 1+\sqrt3\\
1+\sqrt3 & 1 & 1-\sqrt3\\
1-\sqrt3 & 1+\sqrt3 & 1\\
\end{bmatrix}

R=\frac{1}{3}\begin{bmatrix}
1 & 1-\sqrt3 & 1+\sqrt3\\
1+\sqrt3 & 1 & 1-\sqrt3\\
1-\sqrt3 & 1+\sqrt3 & 1\\
\end{bmatrix}

  

 

Thank you sooooooo much ! This is actually my first time asking questions on Stack Exchange. Never thought the answer would be explained so clearly!
– Winter Meng
2 days ago

Another solution:

Consider the following direct orthonormal basis:

U=\pmatrix{b\\b\\b},V=\pmatrix{ \ 0\\ \ a\\-a}, W=U \times V=\pmatrix{-2c\\ \ c\\ \ c}U=\pmatrix{b\\b\\b},V=\pmatrix{ \ 0\\ \ a\\-a}, W=U \times V=\pmatrix{-2c\\ \ c\\ \ c}

with a:=1/\sqrt{2}, b:=1/\sqrt{3}, c:=ab=1/\sqrt{6}.a:=1/\sqrt{2}, b:=1/\sqrt{3}, c:=ab=1/\sqrt{6}.

The rotation matrix RR we are looking for is a direct (counterclockwise) \pi/2\pi/2 rotation leaving UU fixed,

Thus RU=U, \ \ RV=W, \ \ RW=-U,RU=U, \ \ RV=W, \ \ RW=-U, the last two equations expressing that the restriction of RR to the plane orthogonal to UU is a “pure \pi/2\pi/2 2D” rotation.

The 3 preceeding relationships can be gathered into a single one, which is:

R.[U,V,W]=[U,W,-V].R.[U,V,W]=[U,W,-V].

Consequently:

R=[U,W,-V].[U,V,W]^{-1}=[U,W,-V].[U,V,W]^{T}R=[U,W,-V].[U,V,W]^{-1}=[U,W,-V].[U,V,W]^{T} (because [U,V,W][U,V,W] is an orthogonal matrix).

giving, with the help of Mathematica:

R=\frac{1}{3}\pmatrix{1&1-\sqrt{3}&1+\sqrt{3}\\1+\sqrt{3}&1&1-\sqrt{3}\\1-\sqrt{3}&1+\sqrt{3}&1}.R=\frac{1}{3}\pmatrix{1&1-\sqrt{3}&1+\sqrt{3}\\1+\sqrt{3}&1&1-\sqrt{3}\\1-\sqrt{3}&1+\sqrt{3}&1}.

which is the same as the solution of Jack’s wasted life.

  

 

Thank you! This is actually my first time asking questions on Stack Exchange. Never thought the answer would be explained so clearly!
– Winter Meng
2 days ago