How do you find a matrix that maps a rotation of $60º$ around a line $r(t) = 0 + t (1, 1, 1)$

Question

Find a matrix that maps a rotation of $60º$ around a line $r(t) = 0 + t (1, 1, 1)$

I have some problems with these kind of rotations, because you aren't rotating around a $x, y$ or $z$ axis, so you can't directly use the rotational formula i.e.: $$\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\\ & & 1 \end{pmatrix}$$ I have tried to change basis, where the orthonormal basis is $v_1 = \frac{\sqrt 3}{3} (1, 1, 1)$, $v_2 = \frac{\sqrt2}{2} (1, -1, 0)$, $v_3 = \frac{\sqrt 6}{6} (1, 1, -2)$, but then got stuck. I know I should somehow make a matrix that maps from standard basis to my new basis.

Am I even on the right path or is this completely wrong?

Answer 1

Consider the linear map such that its matrix with respect to $B=\{v_1,v_2,v_3\}$ is$$M=\begin{bmatrix}1&0&0\\0&\frac12&-\frac{\sqrt3}2\\0&\frac{\sqrt3}2&\frac12\end{bmatrix}.$$So, if$$P=\begin{bmatrix}\frac1{\sqrt3}&\frac1{\sqrt2}&\frac1{\sqrt6}\\\frac1{\sqrt3}&-\frac1{\sqrt2}&\frac1{\sqrt6}\\\frac1{\sqrt3}&0&-\frac2{\sqrt6}\end{bmatrix}$$and if $T$ is the matrix of your rotation with respect to the standard basis, then $M=P^{-1}TP$ and therefore$$T=PMP^{-1}=\begin{bmatrix}\frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\end{bmatrix}.$$