$\operatorname{tr}(A^\text{T}A) = \operatorname{tr}(A)$ if $A$ is idempotent.

Question

I know that $ \operatorname{tr}(A)=\operatorname{rank}(A)$ if $A$ is idempotent. Now I need to calculate $\operatorname{tr}(A^\text{T}A)$. I have done some simulation which shows $\operatorname{tr}(A^\text{T}A)=\operatorname{tr}(A)$ but I don’t know how to prove it.

Sorry if I’m making a wrong claim.

Answer 1

It doesn't hold. Take eg. the matrix $$A= \left[\begin{array}{cc} \frac{2}{3} & -\frac{2}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{array} \right].$$ We see that $\operatorname{tr} A = 1,$ but the trace of $$A^TA = \left[\begin{array}{cc} \frac{2}{3} & -\frac{1}{3} \\ -\frac{2}{3} & \frac{1}{3} \end{array} \right] \left[\begin{array}{cc} \frac{2}{3} & -\frac{2}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{array} \right] =\left[\begin{array}{cc} \frac{5}{9} & -\frac{5}{9} \\ -\frac{5}{9} & \frac{5}{9} \end{array} \right]$$ is $\frac{10}{9}.$

Answer 2

Take a simple counterexample $$A= \left[\begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array} \right]$$ $\operatorname{tr} A = 1,$ $$A^TA= \left[\begin{array}{cc} 2 & 0 \\ 0 & 0 \end{array} \right]$$

$\operatorname{tr} A^TA = 2$.