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Vector Space over field of R

by loli12
Tags: field, space, vector
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loli12
#1
Sep4-05, 05:13 AM
P: n/a
Let V = {(a1, a2, ......., an): ai in C for i = 1, 2, ... n}; (C=complex numbers) ; so, V is a vector space over C. Is V a vector space over the field of real numbers with the operaions of coordinatewise addition and multiplication?

I thought the answer to this question is No since after we perform the operations on any two of the elements in C, we are getting some other complex numbers which is not an element in R. But the book says Yes to the question. Can anyone please tell me what's wrong with my concept and what is the correct way to brainstorm this question?

Thanks!
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loli12
#2
Sep4-05, 01:36 PM
P: n/a
does anyone has any idea to the problem? or is there any other missing information that I have to provide? Please Please help!!
HallsofIvy
#3
Sep4-05, 01:54 PM
Math
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Thanks
PF Gold
P: 39,682
Look at the DEFINITION of "vector space over field F".

A vector space over a field is:

A set of objects with addition and scalar multiplication defined such that
1) For all x, y in the set, x+ y is also in the set and all the "group" properties hold.
Okay, if our set is the set of complex numbers the sum is a complex number and it is easy to show that the set of complex numbers, with usual addition, is a group.
In fact, as far as the properties of addition are concerned, the field is irrelevant!

2) For all x in the set, a in the field, ax is in the set, a(x+y)= ax+ay, (a+b)x=ax+ bx (the "distributive laws).
Sure: if x is a complex number, a is a real number, ax is a complex number. The distributive laws hold for all complex numbers and the real numbers are a subset of complex numbers so it doesn't matter if we restrict a to be real!

"I thought the answer to this question is No since after we perform the operations on any two of the elements in C, we are getting some other complex numbers which is not an element in R."

The vectors are in C, not R. It's only the "scalars" we multiply by that have to be in R. Making R the field means we restrict what we can multiply by, not the possible result.


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