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?


2 Answers 2


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 $i<n$. 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.


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