According to my understanding, the SVD decomposition consists of decomposing a matrix into a rotation, scaling, then another rotation. This suggests to me one can use that to create arbitrary linear transformations. This is what I am attempting to do.
Let $A\in \mathbb{R}^{2 \times 3}$ be a matrix defined from its SVD decomposition as a rotation (with $z$ as axis), then unitary scaling and finally an identity matrix, that is:
$$A=I_{2} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix}R_z^T $$
where $I_2$ is the $2\times 2$ identity matrix, and
$$R_z=\begin{bmatrix} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
That yields: $$A=\begin{bmatrix} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \end{bmatrix}$$
If we consider $Av$, then the $z$ component of $v$ has no influence over $Av$. Intuitively that just seems wrong; I know $z$ won't rotate since that is the axis of rotation, but it is just gone. How to make intuitive sense of this?