What's the relation between the trace and the range of a projector?

Question

Knowing that, if $P$ is a projector it'll be: $$P=P^2, ~~~P=P^\dagger$$ I was wandering if we can identify a relation between the dimention and the trace of a projector matrix.

Answer 1

I assume we're talking about finite dimensional vector spaces here. The eigenvalues of $P$ are all $1$ or $0$ because it's a projection. The trace remains the same if you diagonalise $P$, so it will be equal to the multiplicity of eigenvalue $1$, that is the rank of $P$.

Answer 2

The trace of $P\in B(H)$ agrees with the dimension of $PH$, even if $H$ is infinite dimensional, and for $PH$ finite or infinite-dimensional. Since $P=P^\dagger P$ is positive, the trace of $P$ is always defined (whether finite or infinite).

So, if $\{e_n\}$ is an orthonormal basis of $PH$, we may complete it to an orthonormal basis of $H$ $\{e_n\}\cup\{f_m\}$. Then $$ \operatorname{Tr}\,(P)=\sum_n\langle e_n|P|e_n\rangle+\sum_m\langle f_m|P|f_m\rangle=\sum_n\langle e_n|P|e_n\rangle=\sum_n\langle e_n|e_n\rangle=\dim(PH). $$