Michael wrote in comment #11:

> More precisely, let \\(\mathbf{V}\\) be a matrix in a vector space and \\(\mathbf{V^\ast}\\) be matrix in its dual space. Then:

> \[\mathbf{V^\ast} \cdot \mathbf{V} = \mathbf{I} = \mathbf{V} \cdot \mathbf{V^\ast}\]

> \[\mathbf{V^\ast} = \mathbf{V}^{-1}\]

I don't know what a "matrix in a vector space" is, or what "matrix in its dual space" is, but I don't like this answer to Puzzle 283 - it makes my hair stand on end. I'm hoping someone corrected you and people eventually figured out some better answer! I have a lot of catching up to do here.

> More precisely, let \\(\mathbf{V}\\) be a matrix in a vector space and \\(\mathbf{V^\ast}\\) be matrix in its dual space. Then:

> \[\mathbf{V^\ast} \cdot \mathbf{V} = \mathbf{I} = \mathbf{V} \cdot \mathbf{V^\ast}\]

> \[\mathbf{V^\ast} = \mathbf{V}^{-1}\]

I don't know what a "matrix in a vector space" is, or what "matrix in its dual space" is, but I don't like this answer to Puzzle 283 - it makes my hair stand on end. I'm hoping someone corrected you and people eventually figured out some better answer! I have a lot of catching up to do here.