Need clarification:Definition of Euclidean n-space and Inner Product Space

Can anyone kindly clarify my concern about the definition of Euclidean n-space and Inner Product Space?

I've the definition of these as:

Euclidean n-space:

When is combined with the standard operations of vector addition, scalar multiplication, vector length and the dot product the resulting vector space is called Euclidean n-space.

Inner Product Space:

A Vector space with an inner product is called an inner product space.



My question is what does it mean when you say a with .... vector length, dot product the resulting vector space is called Euclidean n-space or a vector space with an inner product is called Inner Product Space?


But doesn't the operation of finding vector length, dot product or inner product be applied to every vector?


So what does it mean actually(it's not clear to me) when you add the clause " with vector length, dot product or a vector space with an inner product"?

Don't worry I found the answer at this website:

Pauls Online Notes : Linear Algebra - Inner Product Spaces

The clause simply means the vectors follow dot product/inner product axioms.

Thanks.

Then you probably learned that there are many more vector spaces than just .

For example, the set of all polynomials, p(x), of degree n or less (for fixed n), is a vector space. We can define an "inner product" to be .

More generally, the vector space, , of functions, f(x), defined on [a, b] such that the (Lebesque) intgral exists, is an infinite dimensional vector space. And we can define the inner product as [itex]<f, g>= \int_a^b f(x)g(x)dx[/tex]. The "length" of such a vector would be

, the set of all functions, f(x), defined on [a, b], such that exists also forms a vector space but one on which we cannot define an inner product on this space. We can define the length of a vector to be .

Quote:

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Originally Posted by HallsofIvy

Then you probably learned that there are many more vector spaces than just .

For example, the set of all polynomials, p(x), of degree n or less (for fixed n), is a vector space. We can define an "inner product" to be .

More generally, the vector space, , of functions, f(x), defined on [a, b] such that the (Lebesque) intgral exists, is an infinite dimensional vector space. And we can define the inner product as [itex]<f, g>= \int_a^b f(x)g(x)dx[/tex]. The "length" of such a vector would be

, the set of all functions, f(x), defined on [a, b], such that exists also forms a vector space but one on which we cannot define an inner product on this space. We can define the length of a vector to be .

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First of all thanks HallsofIvy for replying. I understand what you are saying about different vector spaces.

But sorry for not understanding a part of what you said. For can't there be inner product like this(because it follows 4 properties
of inner product):
?

So why this won't be an inner product? Because you said:

"And we can define the inner product as "

Why can we do this but not with absolute value sign?

Quote:

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Originally Posted by HallsofIvy

, the set of all functions, f(x), defined on [a, b], such that exists also forms a vector space but one on which we cannot define an inner product on this space. We can define the length of a vector to be .

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Why can't we define inner product in this vector space where , defined on , such that ?


Does anyone know?

After a hard thinking I finally understood. Thanks HallsofIvy.