How to find explicit formula for the following isometry?

Question

I want to transform smoothly (and if possible by rotation) $(2,0,0,0)$ to $(1,1,1,1)$ in Euclidean space isometrically. How to find explicit formula for this transformation?

Thanks.

Answer 1

Let $b=\{e_1,e_2,e_3,e_4\}$ be the standard basis of $\mathbb{R}^4$. Let

• $u_1=(1,1,1,1)$;
• $u_2=(1,-1,0,0)$;
• $u_3=(0,1,-1,0)$;
• $u_4=(0,0,1,-1)$.

Apply Gram-Schmidt to $\{u_1,u_2,u_3,u_4\}$, thereby obtaining an orthonormal basis $b'=\{f_1,f_2,f_3,f_4\}$ of $\mathbb{R}^4$. Of course, $f_1=\frac12(1,1,1,1)$. Finally, consider the map $T\colon\mathbb{R}^4\longrightarrow\mathbb{R}^4$ such that $(\forall j\in\{1,2,3,4\}):f(e_j)=f_j$. Then $T$ is an isometry and $T(2,0,0,0)=(1,1,1,1)$.

Answer 2

There are many such rotations. Let's first make an orthonormal basis so that one of the rotations only touches the first two coordinates. One such basis is $$ (1,0,0,0)\\ \frac{1}{\sqrt3}(0,1,1,1)\\ \frac{1}{\sqrt2}(0,1,-1,0)\\ \frac{1}{\sqrt6}(0,1,1,-2) $$ In this basis, your vectors become $(2,0,0,0)$ and $\left(1, \sqrt3,0,0\right)$. That's a rotation through $\pi/3$ radians. So in this basis, the rotation has matrix $$ \begin{bmatrix} \cos(\pi/3) & -\sin(\pi/3)&0&0\\ \sin(\pi/3) & \cos(\pi/3)&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix} $$ (If you want the whole rotation as an actual movement, not just an instantaneous transformation, just insert a $t$ next to all the $\pi$'s and let it vary from $0$ to $1$.) Finally, to get your rotation in the standard basis, we have to transform this matrix to your basis. That means that the final result is $$ \begin{bmatrix} 1&0&0&\vphantom{\frac1{\sqrt2}}0\\ 0&\frac{1}{\sqrt3}&\frac{1}{\sqrt2}&\frac1{\sqrt6}\\ 0&\frac{1}{\sqrt3}&\frac{-1}{\sqrt2}&\frac1{\sqrt6}\\ 0&\frac{1}{\sqrt3}&0&\frac{-2}{\sqrt6} \end{bmatrix} \times \begin{bmatrix} \cos(\pi/3) & -\sin(\pi/3)&\vphantom{\frac1{\sqrt2}}0&0\\ \sin(\pi/3) & \cos(\pi/3)&0\vphantom{\frac1{\sqrt2}}&0\\ 0&\vphantom{\frac1{\sqrt2}}0&1&0\\ 0&0\vphantom{\frac1{\sqrt2}}&0&1 \end{bmatrix} \times \begin{bmatrix} 1&\vphantom{\frac1{\sqrt2}}0&0&0\\ 0&\frac{1}{\sqrt3}&\frac{1}{\sqrt3}&\frac1{\sqrt3}\\ 0&\frac{1}{\sqrt2}&\frac{-1}{\sqrt2}&0\\ 0&\frac{1}{\sqrt6}&\frac{1}{\sqrt6}&\frac{-2}{\sqrt6} \end{bmatrix} $$ where I've used that the inverse of an othogonal matrix is its transpose.

Note on how I came up with the basis: I wanted the first basis vector to be a unit vector in the direction of $(2,0,0,0)$, because that's easy and means that as long as all other vectors have $0$ as first component they will be orthogonal to that one. I wanted the second unit vector to be one that makes the $(1,1,1,1)$ be in the span of the first two basis vectors. So those two pretty much found themselves.

I just picked the third one because it was a vector obviously orthogonal to the previous two. By then, the last one was fixed: I wanted it to be orthogonal to the third basis vector, so the second and third component had to be equal, and I wanted it to be orthogonal to the second basis vector, so that forced the final component. Finally, I scaled them to be unit length.