Trace identity deduction - linear algebra.

In a lemma building up to the proof of Cartan's criterion for solubility they deduce something that I don't follow. This is likely due to some deficit in my knowledge of linear algebra.

I'll outline the general argument in case the context matters:

$\lambda_1,\ldots,\lambda_n$ are eigenvalues for a linear transformation. $E$ is the $\mathbb{Q}$-vector space spanned by them. Trying to show that every $f$ in the dual of $E$ is zero. $y$ is the diagonal matrix with entries $f(\lambda_1),\ldots,f(\lambda_n)$. Then there are some arguments which conclude with $0=\sum f(\lambda_i)^2$.

They then say that this implies $f(\lambda_i)=0$. I can't see why this is true, can someone point it out?

Thanks.

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Ah. Is this Humphreys? I remember loving this proof – muzzlator Mar 27 '13 at 17:19
I don't have the book at hand, but this is strange at first sight. Every $f$ in the dual being zero implies that all those eigenvalues are zero; is that what is being proved here? In any case the dual would be $L_\Bbb Q(E,\Bbb Q)$, so its elements take rational values, and their squares are positive. – Marc van Leeuwen Mar 27 '13 at 17:37
I've uploaded the proof from the book here: i.imgur.com/egjmNM1.png – muzzlator Mar 27 '13 at 17:43
That version explains some details better, it's from Humphreys? I'm learning from a set of lecture notes at the moment. – Lobstered Mar 27 '13 at 18:11

$f(\lambda_i)^2 \geq 0$ and so $$\sum f(\lambda_i)^2 = 0 \iff f(\lambda_i) = 0$$