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Let $V$ be an inner product space and $v_1,\ldots,v_n\in V$ be basis with $(v_i,v_j)\leq 0$ for $i\neq j$.

Suppose that there exists vectors $v_1^*,\ldots,v_n^*\in V$ satisfying $(v_i,v_j^*)=\delta_{ij}$, where $\delta_{ij}$ is a Kronecker delta.

I want to prove that $(v_i^*,v_j^*)\geq 0$ for every $i,j$.

I can prove this assertion when $n=2$. Here is the proof:

Let $v_2^*=a_1v_1+a_2v_2$.

Then, $0=(v_2^*,v_1)=a_1(v_1,v_1)+a_2(v_2,v_1)$, $1=(v_2^*,v_2)=a_1(v_1,v_2)+a_2(v_2,v_2)$.

Hence, we have $ a_1=-\frac{a_2(v_2,v_1)}{(v_1,v_1)}$, hence $a_2=\frac{(v_1,v_1)}{(v_2,v_2)(v_1,v_1)-(v_2,v_1)^2}$.

By Cauchy-Schwarz inequality, $a_2\geq 0$ and since $(v_2,v_1)\leq 0$, we get $a_1\geq 0$. Similar things holds for $v_1^*=b_1v_1+b_2v_2$.

I want to generalize this to $\operatorname{dim}V=n$. But, bilinear calculus is somewhat awkward. Are there any simple proofs?

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But the inner-product (,) is defined only for vectors. How would you define it for covectors/duals? Maybe you are using the fact that V and $V^*$ are isomorphic (both being finite-dimensional) , and then you can pull-back the inner-product? – gary Jun 20 '11 at 15:48
Can you show us your proof for dim V<=3? It is generally helpful in situations like these to consider prevents it from working for $n>3$? – Vladimir Sotirov Jun 20 '11 at 16:01
Then you just use the isomorphism between the two to pull-back the inner-product, and then use bilinearity. – gary Jun 20 '11 at 16:05
@Peter: One usually has a bilinear pairing $V\times V^*\to\mathbf{F}$, but this bilinear pairing is not equal to the inner product in $V$; your using the same notation for both is at best confusing. Likewise for the inner product on $V^*$. But if you are defining the inner product on $V^*$ by pulling back to $V$ along the non-coordinate-free isomorphism $v_i\mapsto v_i^*$, then by definition you would have $(v_i^*,v_j^*)_{V^*} = (v_i,v_j)_{V} \leq 0$. So that cannot be the definition you are using. I think you're going to have to clarify and distinguish between the three bilinear pairings. – Arturo Magidin Jun 20 '11 at 16:41
@Arturo: I think it's pretty clear that the isomorphism is the standard one arising from the scalar product on $V$, not the one from the pairing. But I agree this question conflates too many things together. – Marek Jun 20 '11 at 16:44

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