Is my linear algebra book's definition of Euclidean space correct?

[quasar987]

My linear algebra book seems to give a different definition than Mathworld.com so I'll state it.

A scalar product over a vectorial space V is a vectorial real function that to every pair of vectors u, v, associates a real number noted (u|v) satisfying the 4 axioms...
1.
2.
3.
4.

A vectorial space of finite dimension with a scalar product is called a euclidean space.


My question is the following: I don't like how that definition sounds. Is it equivalent to: "Let V be a vectorial space of finite dimension. If there exists a scalar product function over V, then V is called a euclidean space." ?


P.S. does anyone knows a good website that teaches about diagonalisation of hermitian matrixes?

 

[HallsofIvy]

I honestly don't understand your question.

You are asking if
"A vectorial space of finite dimension with a scalar product is called a euclidean space."
and "Let V be a vectorial space of finite dimension. If there exists a scalar product function over V, then V is called a euclidean space."
are equivalent. I don't see any difference at all!

 

[quasar987]

I don't see any difference at all!

Thanks for answering my question.

 

[quasar987]

I found the definitious ambigouous because for a scalar product function to exist, you have to define it first. So according to their definition, a vectorial space with no defined scalar product function is not a euclidian space. But as soon as you do define one, it becomes a euclidean space. My definition says: if a vectorial space is such that a scalar product function CAN be defined (i.e. "potentially"), then it is a euclidean space. That's how I saw it.

I'm kind of realising my definition is no better they theirs now... so let me reformulate my question: "What does the definition says?" Does it say that if there is a defined scalar product function over V then V is a euclidean space, or does it say that if a scalar product function over V can be defined, then V is a euclidean space?

 

[matt grime]

You're just coming across common abuse of notations. A euclidean space is one possessing a euclidean norm/inner product. However you do not *have* to use the inner product. R^2 is eulicdean whether or not you use the inner product.

 

[HallsofIvy]

Ah! now I see what the difference is. In a finite dimensional vector space you can show that every possible inner product is equivalent to a "dot product" defined using a particular basis. (That is, given a basis, to find the "dot product" of u and v, write each as a linear combination of the basis vectors:u= a₁e₁+ a₂e₂+... , v= b₁e₁+ b₂e₂+... . The "dot product" is a₁b₁+ a₂b₂+ ... . The "Gram-Schmidt" orthogonalization process essentiall show that, given any inner product, there exist a basis so that inner product is given by the basis.) I.e. all inner products are equivalent so it really doesn't matter which you use. Of course, every finite dimensional vector space can be given an inner product so, in that sense, every finite dimensional vector space is Euclidean!

 

[Palindrom]

every finite dimensional vector space is Euclidean!

Every finite dimensional vector space over R is Euclidean.