Exercise 6.5.16

This example show the finiteness in the previous exercise is important. Let $V$ be the space of sequence defined in Exercise 6.2.23. Also use the notation $e_i$ in the same exercise. Now we define a unitary operator $U$ by

It can be check that $\parallel U(x)\parallel =\parallel x\parallel$ and $U$ is surjective. So $U$ is an unitary operator. We denote $W$ to be the subspace

and so we have

Now, $W$ is a $U$-invariant subspace by definition. However, we have $e_2\notin U(W)$ and $W^{\text{⊥}}$ is not $U$-invariant since $U(e_1)=e_2\notin W^{\text{⊥}}$.