Question about zero element in vector spaces

randommacuser
Messages
23
Reaction score
0
Suppose I have a set involving trigonometric functions, with addition defined as multiplication of two vectors. If this is a vector space, the zero vector has to be unique. If cos (0) works as the zero vector, then cos (2*pi), etc. also work. Does this mean the set is not a vector space, because the zero element is not unique? Or is it still a vector space (all other axioms check out) because cos (0) = cos (2*pi) = 1 ?
 
Physics news on Phys.org
I'm a little confused... what's your underlying field? How is multiplication between a field element and vector element defined?

Steven
 
As long as it satifies the axioms of a vector space it is a vector space. The set of vectors may be a partition of another set defined by an equiavlence relation (which I think is what you're getting at).
 
The zero vector in a vector space of functions is the ZERO function if f(x)=0 for all x. Is that the kind of thing you're after?
 
I understand now. You guys were a lot of help!
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
Back
Top