Eigenvalues of a resriction of a compact operator

Question

Let $H$ be a real separable Hilbert space, let $A$ be a compact operator on $H$ whose spectrum does not contain $1$, i.e., $1-A$ is invertible, and let $P$ be an orthogonal projection on some proper subspace $S$ of $H$. Then is it possible for $PAP$ to have $1$ as one of its eigenvalues?

Answer 1

Yes, this can happen even in finite dimension. Let $H=\mathbb C^2$ and $$ A=\begin{bmatrix}1&1\\1&1\end{bmatrix}. $$ Then the eigenvalues of $A$ are $0$ and $2$. Meanwhile, if you consider $P=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$, the eigenvalues of $$ PAP=\begin{bmatrix} 1&0\\0&0\end{bmatrix} $$ are $0$ and $1$.