I have studied that given an operator f that can be either a symmetric, or a skew-symmetric, or an orthogonal operator in a Euclidean space $V$ and $U ⊂ V$ be its invariant subspace. Then $U^⊥$ is also an invariant for f.
But I came across a question that asks to provide a counterexample for this i.e. give an example of an operator f on some Euclidean (or Hermitian) vector space such that it has an invariant subspace $U$ and $f(U^⊥)$ does not belong to $U^⊥$
I do not understand this we proved that given any general operator it will have an invariant subspace and we are asking to provide a counterexample to that. It's confusing.
`Define $T: \mathbb R^{2} \to \mathbb R^{2}$ by $T(x,y)=(x+y,0)$. Verify that $U=\{(x,y): y=0\}$ is an invariant subspace but $U^{\perp} = \{(x,y): x=0\}$ is not invaraint.