Rank of the difference of two projection matrices.

Question

Let $P_1,P_2$ be two $n\times n$ projection-matrices such that the column space of $P_2$ is contained in the column space of $P_1$. Then we have that $P_1-P_2$ is also a projection matrix with the rank of $P_1-P_2$ being $\operatorname{rank}(P_1) - \operatorname{rank}(P_2)$.

I do not see why have have $\operatorname{rank}(P_1-P_2) = \operatorname{rank}(P_1) - \operatorname{rank}(P_2)$ ?

Do you see why this is?, can you please explain it?

EDIT: A "projection-matrix" here means a matrix of an orthogonal projection. In other words, a symmetric idempotent matrix.

Answer 1

Note: I'm assuming orthogonal projections.

It is not hard to check that the two projections commute (that is, the order of application does not matter). That means they can be diagonalized in a common basis. Indeed, it is not hard to write it down in that basis (sort the basis so that the zero diagonal elements occur last). If you do so, you can immediately see why that claim is true.

Answer 2

Hint: Use the fact that if $A$ is idempotent, then $rankA=traceA$. Note that $P_1$, $P_2$ and $P_1-P_2$ are all projections and hence idempotent.