# How does one find the image of a linear transformation, given that T is a rotation followed by a projection?

I'm struggling to understand the intuition behind this because I am unsure which coordinates I am mapping to. I think that I understand how to find the kernel (by essentially finding the null space of the matrix after multiplying the rotational matrix with that of the projection), but I think that I recall having to map coordinates from a preimage to form the image under T (this could be wrong)? And when it says that T is a rotation followed by a projection, I am not sure if this means that we were initially looking at a graph under a rotation, or if we were initially looking at some graph 'X' that was transformed by a rotation and a projection.

If someone could clarify how I should be thinking about a problem like this, I'd be very grateful. My apologies if I am completely off the mark / sound like I'm blabbering on about nothing. I think that this is partially augmented by the fact that I'm having trouble understanding the formal purpose of a kernel and an image.

• Let's assume T is a mapping from $R^n$ to $R^n$. First of all, the rotation will be a linear map from $R^n$ onto $R^n$, and then the projection will map $R^n$ onto the subspace W of $R^n$ on which you're projecting; so the image of T will be the subspace W. – user84413 Jul 19 '15 at 23:22

Any rotation $\varphi$ around the origin (say in the plane) is a linear transformation on the vectors: respects addition and multiplication by scalar.
So, - once a basis is fixed - it has a corresponding matrix $M$ that expresses this transformation by matrix multiplication: $M\cdot v=\varphi(v)$ for all $v$.
Similarly, any projection $\psi$ to some subspace is a linear transformation, hence it also has a corresponding matrix, say $P$.
Finally, the composition of transformations corresponds to the multiplication of the corresponding matrices: $$\psi(\varphi(v))=P\cdot M\cdot v$$
So, now your transformation (matrix) $T$ is of the form $$T=P\cdot M$$