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Vector space or inner product space  ambiguous! 
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#1
Mar2612, 11:52 PM

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Sometimes the way authors write a book makes you wonder if they are at their wits' end: 'A vector space with an inner product is an inner product space!'
I am not sure if I have gone crazy but to me it is obvious that if you have a vector space wrapped up in a nice gift basket and sent off to your address from Amazon, then the inner product comes with the vector space for free. After all, the inner product is obtained from any two vectors in a vector space, so I just can't understand why a vector space cannot be preequipped with an inner product, that is, why the hell isn't a vector space and an inner product space the same thing? 


#2
Mar2712, 01:08 AM

P: 772

Because the definition of vector space doesn't include an inner product.
An inner product is a function mapping a pair of vectors to an element of the underlying field; until you have defined such a function, you do not have an inner product. The definition of vector space does not include such a function, therefore, a vector space is not necessarily an inner product space. 


#3
Mar2712, 08:12 AM

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Given any (finite dimensional) vector space there are an infinite number of possible "inner products". For example, choose any basis, [itex]\{e_1, e_2, ..., e_n\}[/itex]. We can now write two vectors, [itex]u= a_1e_1+ a_2e_2+ ...+ a_ne_m[/itex] and [itex]v= b_1e_1+ b_2e_2+ ...+ b_ne_n[/itex], written in terms of that basis.
We define the inner product [itex]<u, v>= a_1b_1+ a_2b_2+ ...+ a_nb_n[/itex]. Choosing a different basis will give a different inner product. (And the "theoretical meat" of the GramSchmidt orthogonalization process is that, given any abstractly defined inner product there exist a basis in which that inner product is as given above. And note the "finite dimensional". There exist infinite dimensional vector spaces which do not have any inner product. 


#4
Mar2712, 03:59 PM

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Vector space or inner product space  ambiguous!



#5
Mar2712, 09:16 PM

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#6
Mar2812, 08:23 AM

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#7
Mar2812, 10:08 AM

P: 772




#8
Mar2812, 10:14 AM

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And "almost all" such sums do not converge.



#9
Mar2812, 10:58 AM

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$$ \begin{align} A\big((x_1,x_2),(y_1,y_2)\big) &=(x_1+x_2,y_1+y_2)\\ S\big(a,(x_1,x_2)\big) &=(ax_1,ax_2)\\ I\big((x_1,x_2),(y_1,y_2)\big) &=x_1 x_2+y_1y_2 \end{align} $$ for all ##(x_1,x_2), (y_1,y_2)\in\mathbb R^2## and all ##a\in\mathbb R##, then ##(\mathbb R^2,A,S)## is a vector space over ℝ, and if we denote that space by V, then ##(V,I)## is an inner product space over ℝ. This post may be useful. 


#10
Mar2812, 11:03 AM

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#11
Mar2812, 02:04 PM

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Any uncountable sum with more than countablymany nonzero terms, necessarily diverges. Just partition your uncountable supportset into sets An:={x:x>1/n}; at least one of the sets will have infinitelymany terms. 


#12
Mar2812, 02:06 PM

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#13
Mar2812, 04:25 PM

P: 800

If you want to intersect two sets, there's no point mentioning that the sets are also vector spaces or groups or manifolds or anything else. If you loaded up every definition with all of its derived types, it would be incredibly confusing. A set is such and so. A group is a set with such and so. A Lie group is a group with such and so. You build up complex definitions in terms of simpler ones. If all you care about is the properties of a vector space, why add on additional properties that you don't care about? 


#14
Mar2812, 08:31 PM

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The claim was that we can construct an inner product on any vector space. The argument was: (1) Choose a (Hamel) basis. (2) Every vector is a finite linear combination of vectors from this basis (even if the basis itself is infinite). (3) HallsofIvy's recipe for an inner product still works, regardless of whether the vector space is finite or infinitedimensional. 


#15
Mar2812, 08:40 PM

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Right, my bad, I lost focus and was making a general statement about sums in V.Spaces. The force is back with me now.



#16
Mar2812, 09:26 PM

P: 1,271

The definition isn't a vector space on which there exists an inner product. It's a vector space with some chosen inner product that you have singled out. 


#17
Mar2812, 09:29 PM

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#18
Mar2812, 09:38 PM

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But we're working with a (Hamel) basis for the entire vector space, so (by definition!) every vector is a finite linear combination of elements in the basis!



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