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I was thinking about how the dual basis (w.r.t. to the standard scalar product) in $\mathbb{R^n}$ looks like if the original basis had obtuse angles between all its vectors.

Therefore I want to look at the inverse of the matrix $A$ which contains the scalar products of the basis vectors. Assuming the vectors are normed, $A$ just has 1‘s on its diagonal, is symmetric and has nonpositive numbers between 0 and -1 everywhere else.

Now my hypothesis is, after looking at some examples, that $A^{-1}$ only has nonnegative entries, which would mean the dual basis would have only acute angles.

Is this true? I couldn’t find any linear algebra argument supporting this.

return true
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  • It's not clear to me what you mean by the "dual basis" in this context. In any case, it's probably helpful to note that the angle between two vectors in $\Bbb R^n$ is obtuse iff their scalar product is negative. – Ben Grossmann Jan 23 '22 at 18:50
  • That‘s why $A$ only has nonpositive entries except on the diagonal. I‘m wondering about how $A^{-1}$ looks. – return true Jan 23 '22 at 19:28

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Yes, it is true.

If $A$ is the matrix of inner products of basis vectors, then it is an invertible Gram matrix and therefore positive definite. If its off-diagonal entries are non-positive, then $A$ is an $M$-matrix. As I show in my answer here, the inverse of such a matrix necessarily has non-negative entries.

Ben Grossmann
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