In $\mathbb{R}^3$, I'm given the basis $B=\left\{\begin{pmatrix} 0 \\ 5 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 5 \\ 0 \\ 0 \end{pmatrix}\right\}$, and $T$ is a transformation that roates vectors 30 degrees counterclockwise about the $z$-axis.
I'm asked to find $[T]_B^B$ without computations.
By using computations, I found that $[T]_B^B=\left[\left[T\begin{pmatrix} 0 \\ 5 \\ 0 \end{pmatrix}\right]_B \left[T\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\right]_B \left[T\begin{pmatrix} 5 \\ 0 \\ 0 \end{pmatrix}\right]_B\right]=\left[\left[\begin{pmatrix} -5\sin{30} \\ 5\cos{30} \\ 0 \end{pmatrix}\right]_B\left[\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\right]_B\left[\begin{pmatrix} 5\cos{30} \\ 5\sin{30} \\ 0 \end{pmatrix}\right]_B\right] = \begin{pmatrix} \cos{30} & 0 & \sin{30} \\ 0 & 1 & 0 \\ -\sin{30} & 0 & \cos{30} \end{pmatrix}$
I want to understand this geometrically so I can generalize how $[T]_B^B$ will work for this type of transformation, but I don't really understand the result I obtained. The result I obtained looks like a 30 degrees counterclockwise rotation in the $y$-axis rather than the $z$-axis. Why is that? Did I make a mistake in my computations?
Or am I misinterpreting something about the fact that T is being written in terms of the basis $B$? I'm honestly still having a bit of trouble wrapping my head around the meaning of expressing linear transformations in different bases.
It's important also to understand this in a non-computational sense because the second part of the question asks me to repeat this exercise of finding $[T]_B^B$ for the 5 other bases that can be formed by reordering the vectors in basis $B$.
*Side note: Some of my LaTeX formatting doesn't look too great in this problem. I would appreciate pointers on how to improve that as well.
The matrix gives a rotation about the "second" axis. And if you look at your basis, the second axis is in fact the $z$-axis.