# Inner product of dual basis

Assume $V$ is an inner product space over $\mathbb{R}$ with inner product $\left<\cdot,\cdot\right>$. Let $u_1,\cdots,u_n$ be a basis of $V$ and $v_1,\cdots,v_n$ be the dual basis, i.e., $\left<u_i,v_j\right> = \delta_{ij}$. Prove that if $\left<u_i,u_j\right>\leq 0$ for all $1\leq i < j \leq n$, then $\left<v_i,v_j\right>\geq 0$ for all $1\leq i < j \leq n$.

If we let $B=(b_{ij})$ be the matrix from basis $\{v_i\}$ to $\{u_i\}$, that is, $(u_1,\cdots,u_n) = (v_1,\cdots,v_n)B$, then $u_j = \sum_{i=1}^n b_{ij}v_i$, hence $\left< u_j, u_k \right> \leq 0$ and the duality imply $\left< u_j, u_k \right> = \left< \sum_{i=1}^n b_{ij}v_i, u_k \right> = \left< b_{kj}v_k, u_k \right> = b_{kj} \leq 0$ for $k\neq j$.

Now $(v_1,\cdots,v_n) = (u_1,\cdots,u_n) B^{-1}$, if we let $B^{-1} = (c_{ij})$ then $v_j = \sum_{k=1}^n c_{kj}u_k$, so $\left< v_i, v_j \right> = \left< v_i, \sum_{k=1}^n c_{kj}u_k \right> = \left< v_i, c_{ij}u_i \right> = c_{ij}$ for $i\neq j$.

Since $BB^{-1} = I_n$, we have $\sum_{k=1}^n b_{ik}c_{kj} = \delta_{ij}$. But it seems difficult to determine whether $c_{ij}\geq 0$ from this condition. How should I do next?

Not in the same way as you suggested but here is a simple computation based answer. Some trivial calculations are needed. I hope you will be able to figure it out.

Consider $$n=2$$ and without loss of generality let $$u_1,u_2$$ be unit vectors.

Let $$a=(u_1,u_2)$$. Then $$0\geq a>-1$$.

Let $$v_i=a_{i1}u_1+a_{i2}u_2$$ for $$i=1,2$$.

Using $$(u_i,v_j)=\delta_{ij}$$ we get,

$$a_{11}=a_{22}=\frac{1}{1-a^2}, ~~a_{12}=a_{21}=-\frac{a}{1-a^2} \tag{1}$$

Now $$(v_1,v_2)=(a_{11}a_{21}+a_{12}a_{22})+(a_{11}a_{22}+a_{12}a_{21})a \tag{2}$$

Using (1) and (2), we get

$$(v_1,v_2)=\frac{a}{a^2-1}\geq 0$$ since $$0\geq a>-1$$.

Let us consider the general case now.

By induction hypothesis, $$(v_i,v_j)\geq 0$$ for $$i,j\leq n-1$$. It's now sufficient to show that $$(v_n,v_i)\geq 0$$ for $$i. But it's also trivial since $$u_n,u_i$$ generate a two dimensional subspace of $$V$$ and $$v_n,v_i$$ restricted to that subspace form a dual basis of $$u_n,u_i$$. Hence $$(v_n,v_i)\geq 0$$ by $$n=2$$ case.

As @Irfan showed in his answer, you can reduce the problem to $$n=2$$ case. Here is another way proceeding from your approach. We may assume the inner product $$\langle \cdot, \cdot \rangle$$ is the dot product. Recall that $$B = (b_{ij})$$ is given by $$b_{ij} = \langle u_i, u_j \rangle$$, i.e. $$B$$ is a Gram matrix. Thus $$B = X^TX$$ where $$u_1, \cdots, u_n$$ are columns of $$X$$. In particular, $$\det(B) = \det(X)^2 >0$$. Similarly, you can justify that every principal minor of $$B$$ is positive. To sum up, we have the followings:

• $$B$$ has non-positive off diagonals.
• All the principal minors of $$B$$ are positive.

This is one characterization of the nonsingular M-matrices. Since $$B$$ is a nonsingular M-matrix, it is inverse-positive, i.e. $$B^{-1}\geq 0$$ entrywise. The proof of this implication is given in this answer.