I was wondering if it was possible to prove existence of a unitary operator $A$ such that:
$\langle Au,u\rangle=0$ for all $u$.
In 2-dimensions it clearly is (just a 90 degrees rotation) and similarly in other finite dimensions. However is it possible to prove this for infinite dimensions?
For all $u\in L^{2}$, say?



Existence of an "Orthogonalizing" Operator
To find such an operator in separable real Hilbert space just take a countable direct sum of rotations of $\mathbb{R}^2$. More formally:


Let $(e\_n)\_{n \in \mathbb{N}}$ be a countable basis for Hilbert space and define an operator $A$ by $Ae\_{2n}=e\_{2n+1}$ and $Ae\_{2n+1}=-e\_{2n}$. If $u=\sum\_{k=1}^\infty a\_ke\_k$ where the $a\_k$'s are real then we have $Au=\sum\_{n=1}^\infty (a\_{2n}e\_{2n+1}-a\_{2n+1}e\_{2n})$ and so $\langle Au, u\rangle = \sum\_{n=1}^\infty (a\_{2n+1}a\_{2n}-a\_{2n}a\_{2n+1})=0$. If the coefficients are allowed to be complex then the minus becomes a plus in the imaginary component and we will typically get a nonzero answer due to the effect of complex conjugation in the definition of inner product.


As I mentioned in the comments, for real matrices of odd finite dimension the characteristic polynomial of $A$ will have a real root, which implies the existence of an invariant subspace. If the corresponding eigenvalue is nonzero then the relation $\langle Au, u\rangle=0$ is obviously impossible when $u$ belongs to this eigenspace, so in odd dimensions this property is impossible for an invertible matrix $A$. This same observation means that the desired property also does not hold for any invertible complex matrix.


