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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.

Sergio
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  • What do you mean by “surface arguments”? And which connection do you want made explicit? – Joppy Oct 02 '18 at 23:47
  • 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? – Lubin Oct 03 '18 at 00:01
  • @Joppy Pythagoras' proof was a cut and paste argument, Euclid's was also about areas. – Sergio Oct 03 '18 at 14:29
  • @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. – Sergio Oct 03 '18 at 14:31
  • So "surface" means area? And "scalar product" means inner product? – David K Oct 03 '18 at 15:07
  • @David K Yes. I'm sorry, I didn't notice my wording was somewhat unusual. – Sergio Oct 04 '18 at 22:45
  • You can have an inner product space on $\mathbb R^2$ in which the inner product is different from the usual one, and the norm is not based on the Pythagorean Theorem. I'm not sure about the premise of this question. – David K Oct 19 '18 at 11:40

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