True or false? If a diagonalizable operator has only the characteristic values 0 and 1, it is a projection.
Solution
The assertion is \(\boxed{\text{true}}\).
Since \(T\) is diagonalizable, there is a basis of eigenvectors \(v_1,\ldots,v_n\) with \[ Tv_j=\lambda_jv_j, \qquad \lambda_j\in\{0,1\} \] for every \(j\) [@HK71]. Therefore \[ T^2v_j=\lambda_j^2v_j=\lambda_jv_j=Tv_j, \] because \(0^2=0\) and \(1^2=1\). Thus \(T^2\) and \(T\) agree on a basis, hence \[ T^2=T. \] An idempotent linear operator is a projection, so \(T\) is a projection.