I'm trying to solve this problem from linear algebra:
Consider a $120^{\circ}$ rotation around the axis $x = y = z$.
$(a)$ Show that the vector $i = (1,0,0)$ is rotated to the vector $j = (0, 1, 0)$. (Similarly $j$ is rotated to $k = (0, 0, 1)$ and $k$ is rotated to $i$.) How is $j − i$ related to the vector $(1, 1, 1)$ along the axis?
$(b)$ Find the matrix $A$ that produces this rotation. Explain why $A^3 = I$. What are the eigenvalues of $A$?
I haven't been taught about rotation matrices (apart from the $2D$ case), and I'm not sure how to think about rotation from an arbitrary axis. This is what I've tried so far:
$(a)$ If $A$ is the rotation matrix, then $Ai = j$ is what I need to prove. But I have no idea on how to come up with $A$. I can see that $j - i = (-1, 1, 0)$ is orthogonal to the vector $(1, 1, 1)$.
$(b)$ $A^3 = I$ because three subsequent transformations by $A$ will rotate a vector by $360^{\circ}$ and so bring it back to its original position. Again, not sure how to come up with eigenvalues since I can't construct $A$.
The matrix for a linear transformation can be found as the columns will be the vectors that your basis vectors end up at.
The columns of $A$ will be were $\hat{i}$, $\hat{j}$, and $\hat{k}$ land respenctively after the rotation. $\hat{i}$ ends up at $\hat{j}$, $\hat{j}$ ends up at $\hat{k}$, and $\hat{k}$ end up at $\hat{i}$.
So $A = \begin{bmatrix} \hat{j} & \hat{k} & \hat{i} \end{bmatrix}=\begin{bmatrix} 0 & 0 &1 \\ 1& 0& 0\\ 0& 1& 0 \end{bmatrix}$
You should be able to do the rest.