Exercise 5.A.14

Proof. Suppose $\lambda$ is eigenvalue of $P$. It follows that $u=\lambda (u+w)$ for some $u\in U$ and $w\in W$. Thus $(1-\lambda )u=\mathit{\lambda w}$. Since $U+W$ is direct sum, that is $U\cap W=\{0\}$, if $u\ne 0$, then $w=0$, implying $\lambda =1$. Similarly, if $w\ne 0$, then $u=0$, implying $\lambda =0$. Hence $0$ and $1$ are only eigenvalues and $U$ and $W$, respectively, are the corresponding eigenvectors. □