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If $A:H \rightarrow H$ is Linear, then A is uniquely given by the values { $\langle Ah,h \rangle : h \in H$}.

My aim to prove is that given a $x$ I can obtain the value $Ax$ with the values given, then I though that { $Ax :x \in H$} is a subspace of $H$ then it would be nice that this set is a basis of this subspace. But I can not see how to prove that.

energy
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    This depends on the scalars being complex. Please make explicit if you are assuming complex scalars. The first answer gives a real counterexample. – Jonas Meyer Jan 19 '17 at 00:57
  • http://math.stackexchange.com/questions/57350/for-complex-matrices-if-langle-ax-x-rangle-langle-bx-x-rangle-for-all-x, http://math.stackexchange.com/questions/524970/if-the-expectation-langle-v-mv-rangle-of-an-operator-is-0-for-all-v-is-t?noredirect=1&lq=1, http://math.stackexchange.com/questions/1373644/prove-that-the-linear-transformations-are-the-same?noredirect=1&lq=1, http://math.stackexchange.com/questions/315942/prove-that-there-is-no-non-zero-linear-operator-on-c2-such-that-alpha?noredirect=1&lq=1, http://math.stackexchange.com/q/298353/1424 – Jonas Meyer Jan 19 '17 at 01:01

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This is not true over real scalars. consider the two-dimensional real Hilbert space with the operator defined by the matrix $$ \begin{pmatrix}0 & -1 \\ 1 & 0 \end{pmatrix} $$ For this operator, $\langle Ah, h\rangle =\langle (-h_2, h_1), (h_1, h_2)\rangle = 0$ but the operator is not zero.

Same example applies in infinite dimensional real spaces like $\ell^2$. Using the standard orthonormal basis $(e_n)$, the operator is defined by $e_{2n-1} \mapsto e_{2n}$ and $e_{2n}\mapsto -e_{2n-1}$.

  • If I recall correctly it applies only for a self-adjoint continuous linear operator. – DanielWainfleet Jan 18 '17 at 23:38
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    Works for normal operators, too. Interestingly, an attempt to find a reference hit another question that claimed the result for all linear operators. :$ –  Jan 18 '17 at 23:43
  • It works for all operators on complex inner product spaces, not for real ones of dimension greater than 1. – Jonas Meyer Jan 19 '17 at 00:56
  • @JonasMeyer How do you show that for non-normal operators on a complex Hilbert space? – Lukas Betz Jan 19 '17 at 01:04
  • @LeBtz: See the links in the second comment on the question (or to the right if you can see a "Linked" section. – Jonas Meyer Jan 19 '17 at 01:04
  • In zaq's link (and perhaps here) there may be an implicit assumption of complex scalars, but the statement in the linked question is wrong for asserting the numerical radius equals the operator norm in general. – Jonas Meyer Jan 19 '17 at 01:09
  • @JonasMeyer Ah, ok. I somehow thought that we need to be able to calculate the norm by looking at $\langle Ax, x\rangle$ but I see now that what we actually need is weaker. Thanks for the other links. I didn't see the argument in the second comment as there was a clear mistake in the proof (taking an infimum of a complex set). – Lukas Betz Jan 19 '17 at 01:16
  • The exercises was put like I wrote, but then is wrong, thanks, what could I add about scalars, with complex scalars this result would be true? Thanks – energy Jan 19 '17 at 19:52