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This question is in the same spirit as this one.

On one hand it is often stated,that the Pythagorean theorem means that Euclidean distance arises from a scalar product. But on the other hand most elementary proofs rely on surface arguments.

Is there a way to make the connection explicit using linear algebra terms ? In other words, is there a version of the Pythagorean theorem of the form "if a norm satisfies this condition regarding determinants, then it is given by a scalar product ?"

Another way of looking at the question is this. $\mathbb{R}^2$ with euclidean norm, determinant, and scalar product, is a model for the Euclidean plane. This somehow means that linear algebra can give us some insight about geometry. Conversely, if the Euclidean plane is a model for a two-dimensional real space with determinants and norm, why must the norm be the one given by a scalar product ? What insight does Euclidean geometry give us about scalar products ?

One could argue that for the Pythagorean theorem angles are as important as surfaces, and that having angles implies a scalar product. But it seems to me the naïve conception of angles is very different from the abstract version of orthogonality in a scalar product space that we are used to.

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  • $\begingroup$ What do you mean by “surface arguments”? And which connection do you want made explicit? $\endgroup$ – Joppy Oct 2 '18 at 23:47
  • $\begingroup$ Well, the Pythagorean Theorem at the very least talks about distances, lengths. What is the distance between two points in the plane? Do you have a definition that is couched in the language of linear algebra? $\endgroup$ – Lubin Oct 3 '18 at 0:01
  • $\begingroup$ @Joppy Pythagoras' proof was a cut and paste argument, Euclid's was also about areas. $\endgroup$ – Sergio Oct 3 '18 at 14:29
  • $\begingroup$ @Lubin a distance is a norm on a vector space. Hence I am asking: what characterises the euclidean norm ? There is a Jordan-Von Neumann-Frechet theorem that says that a norm is euclidean iff it satisfies the parallelogram identity. $\endgroup$ – Sergio Oct 3 '18 at 14:31
  • $\begingroup$ So "surface" means area? And "scalar product" means inner product? $\endgroup$ – David K Oct 3 '18 at 15:07

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