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?