$$T(x) = \left[ \begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right] \cdot \left[ \begin{array}{c} x \\ y \\ 1 \end{array} \right] $$
You have to solve it by definition: $$T(w_{1}) + T(w_{2}) = T(w_{1} + w_{2}),$$ and $$\alpha T(x) = T(\alpha x).$$
I am new here, I do not know the format, I’m sorry for any inconvenience that this may cause. Just wondering if the exercise above is a linear transformation or not.
$ L : \mathbb{R}^n \rightarrow \mathbb{R}^m $ is a linear transformation if and only if there is a matrix, $ A $, such that $ L( z ) = A z $... What you have here is sort of like that... BUT...
But here $L$ is a bit tricky: it is a function of two variables. Another thing you know about a linear transformation is that it should map the zero vector to the zero vector. What happens if you pick $ x = y = 0 $?