Trace Equals Rank – Proof for Idempotent Matrices

Question

How could we prove that the "The trace of an idempotent matrix equals the rank of the matrix"?

This is another property that is used in my module without any proof, could anybody tell me how to prove this one?

Answer 1

Sorry to post solution to this such a old question, but "The trace of an idempotent matrix equals the rank of the matrix" is very basic problem and every answer here is using the solution using eigen values. But there is another way which should be highlighted.

Solution:

Let $A_{n\times n}$ is a idempotent matrix. Using Rank factorization, we can write $A=B_{n\times r}C_{r\times n}$ where $B$ is of full column rank and $C$ is of full row rank, then $B$ has left inverse and $C$ has right inverse.

Now, since $A^2=A$, we have $BCBC=BC$. Note that, $$BCBC=BC\Rightarrow CBC=C\Rightarrow CB=I_{r\times r}$$

Therefore $$\text{trace}(A)=\text{trace}(BC)=\text{trace}(CB)=\text{trace}(I_{r\times r})=r=\text{rank}(A)\space\space\space\blacksquare$$