Common eigenvector of a sequence of compact operators

Question

Let $H$ be a separable, infinite-dimensional Hilbert space and suppose we have a sequence of norm-one compact operators $(A_n)$ on $H$ which all have 1 as an eigenvalue. Can we pass to a subsequence to extract a sequence of $A_n$'s having a common eigenvector?

Answer 1

I think not. Split your Hilbert space into a direct sum of two dimensional spaces. On the $n$th two dimensional space, find a matrix whose eigenvalues are $1/(2n-1)$ and $1/(2n)$, whose eigenvectors are $(\cos(1/n),\sin(1/n))$ and $(-\sin(1/n),\cos(1/n))$. Since the eigenvalues are all distinct, the eigenvectors are uniquely determined. The matrix is Hermitian, hence its norm will be its spectral radius, which is $1$.