True or false? If a diagonalizable operator has only the characteristic values $0$ and $1$, it is a projection.
My attempt: Suppose $T:V\to V$ is diagonalizable with eigenvalue $0$ and $1$. Then $\exists B=\{\alpha_1,…,\alpha_n\}$ basis of $V$ such that $\forall i\in J_n$, $\alpha_i$ is eigenvector of $T$. Let $\alpha_1,…,\alpha_r$ and $\alpha_{r+1},…,\alpha_n$ are eigenvectors corresponding to eigenvalue $1$ and $0$, respectively. Let $\alpha \in V$. Then $\alpha =\sum_{i=1}^n a_i\cdot \alpha_i$. So $T(\alpha)= \sum_{i=1}^n a_i\cdot T(\alpha_i)= \sum_{i=1}^r a_i\cdot \alpha_i$ and $T^2(\alpha)=T(T(\alpha))=T(\sum_{i=1}^r a_i\cdot \alpha_i)= \sum_{i=1}^r a_i\cdot \alpha_i$. Thus $T^2(\alpha)=T(\alpha)$, $\forall \alpha \in V$. Hence $T$ is a projection. Is my proof correct?
Aliter:
Diagonalizable operators have minimal polynomials with distinct linear factors.
Therefore, in this case $m_T(x)=x(x-1)\implies T^2=T.$