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Can someone give me an idea, why $\forall x: \left<\sum_j \lambda_j \left< x,e_j\right> e_j,x\right>\geq 0$, where the $\lambda_j$'s are fixed, implies that all $\lambda_j$ are $\geq0$,?

(The $x$'s belong to a Hilbert space,the $e_j$'s are an orthonormal basis and the $\lambda_j$'s are real or complex .)

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    Is $x$ fixed? And the $\lambda_j$?2012-06-22
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    The statement needs to be quantified: is the inequality supposed to hold for *all* $x$?2012-06-22
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    @LeonidKovalev Yes2012-06-22

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I understand that this inequality is to hold for all $x$. (Otherwise, $x=(1,1)$ with $\lambda_1=1,\lambda_2=-1$ witnesses that it is false.)

Suppose that there is some $\lambda_i$ not real or not greater than or equal to zero. Then for $x=e_i$ we have $$\langle\sum_j\lambda_j\langle x,e_j\rangle e_j, x\rangle=\langle \lambda_i e_i, e_i\rangle=\lambda_i \not\geq 0$$ so we have a contradiction.

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    But why do you suppose it is not real? I can't see, where that comes into play.2012-06-22
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    We want to show that $\lambda_i$ are all positive, so in particular they are all real. I suppose by contradiction that they are either not real or real and not greater than or equal to zero. (So as to avoid saying that they are not greater than zero when they are in fact incomparable.) It is the source of contradiction, since in the end the expression equals $\lambda_i$ so it cannot be positive (since it is not real or it is smaller than zero).2012-06-22