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?
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$\begingroup$ What other conditions are you putting on A? for instance, is it allowed to have non-trivial kernel? (I assume not but you should make your assumptions/conditions more clear) $\endgroup$– Yemon ChoiCommented Sep 10, 2013 at 13:48
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$\begingroup$ Yes sorry I realized this after I posted. A non-trivial kernel indeed. In fact it would be nice to know that this is possible for a unitary operator A. $\endgroup$– PlogCommented Sep 10, 2013 at 13:50
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1$\begingroup$ Are you working in real or complex space? If the former, in dimension three the characteristic polynomial of $A$ is odd and therefore has a real root, which implies the existence of a one-dimensional invariant subspace. If the latter then the existence of a one-dimensional invariant subspace also holds even in dimension two. The desired property clearly cannot hold in the presence of a one-dimensional invariant subspace which does not lie in the kernel. $\endgroup$– Ian MorrisCommented Sep 10, 2013 at 14:27
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$\begingroup$ Ah yes, should have mentioned I'm working in real space. $\endgroup$– PlogCommented Sep 10, 2013 at 14:29
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1$\begingroup$ What about taking an orthogonal splitting $H=H_1\oplus H_2$, with $H_i$ both isomorphic to $H$, an isometry $B:H_1\to H_2$, and set $A=[0\; -B^*;B\; 0]$ ($2\times 2$ block matrix) ? $\endgroup$– BS.Commented Sep 10, 2013 at 14:47
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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.
answered Sep 10, 2013 at 14:44