# Pythagorean theorem in linear algebra

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.

• What do you mean by “surface arguments”? And which connection do you want made explicit? – Joppy Oct 2 '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 3 '18 at 0:01
• @Joppy Pythagoras' proof was a cut and paste argument, Euclid's was also about areas. – Sergio Oct 3 '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 3 '18 at 14:31
• So "surface" means area? And "scalar product" means inner product? – David K Oct 3 '18 at 15:07