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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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    If we choose a basis relative to which the inner product $(-,-)$ is a dot product, and if we let $P$ be the matrix whose columns is the new basis $\{v_i\}$, then $(P^{-1})^T$ will have columns $\{v_i^*\}$. $M=P^TP$ will be a real symmetric matrix with $m_{ij}=(v_i,v_j)$ as will $M^{-1}$ with $m^{-1}_{ij}=(v_i^*,v_j^*)$. So unless I'm mistaken the problem is equivalent to showing that if $M$ is a real symmetric matrix with positive eigenvalues (and positive diagonal?), then $M$ has non-negative non-diagonal entries if and only if $M^{-1}$ has non-positive non-diagonal entries.2011-06-21

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