Proving the trace equals the dimension of an image

Question

If we set that an operator $P:V \to V$ is called a projector when $P^2 = P$ (and its dimension is less than infinity), how can I show that $\text{Trace}(P) = \dim \text{Image}(P)$?

Answer 1

Since $P^2 = P$ we see that $P$ has as its minimal polynomial $x^2 - x = x*(x-1)$. Since the minimal polynomial has no repeated roots, we know that $P$ must be diagonalizable and all of its eigenvalues must be 0 or 1. In general the number of non-zero eigenvalues (counted with multiplicity) is the dimension of the image of a linear map (this is basically rank nullity theorem). Trace is equal to the sum of the eigenvalues. We see that the sum of eigenvalues is equal to the multiplicity of 1 as an eigenvalue which in turn is equal to the dimension of the image of $P$.

Answer 2

Choose a basis $\mathcal B$ for the image of $P$ and extend this basis to a basis $\mathcal E$ of all of $V$. The diagonal entries of the matrix of $P$ with respect to the basis $\mathcal E$ has $|\mathcal B|$ $1$'s and rest of the diagonal entries are $0$. Thus $\text{trace}(P) = |\mathcal B| = \dim\text{image}(P)$.