Is it possible that a subspace is not a vector space

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
14 replies · 2K views
tze liu
Messages
57
Reaction score
0
<Mentor's note: moved from general mathematics to homework. Thus no template.>

Prove subspace is only a subset of vector space but not a vector space itself.
Even a subspace follows closed under addition or closed under multiplication,however it is not necessary to follow other 8 axioms in vector space.
if some subspace follow closed under addition or closed under multiplication but don't follow Distributive axioms of c(x+y) = cx + cy(mean they cannot be added in this way),then it is a subspace but not vector space.
thank
right?
 
Physics news on Phys.org
If the subspace is a subspace of a vector space, then it has to inherit the other axioms from the parent space.
Otherwise: subspace of what?
 
  • Like
Likes   Reactions: tze liu
tze liu said:
Prove subspace is only a subset of vector space but not a vector space itself.
then what is a subspace by definition?
tze liu said:
Even a subspace follows closed under addition or closed under multiplication,however it is not necessary to follow other 8 axioms in vector space.
I can imagine such a set but it is not called subspace in accordance to standard terminology.
I mean that even in ##\mathbb{R}## there are subsets those are closed under + but they are not subspaces of the one dimensional vector space ##\mathbb{R}##
 
Last edited:
  • Like
Likes   Reactions: tze liu
zwierz said:
I mean that even in ##\mathbb{R}## there are subsets those are closed under + but they are not subspaces of the one dimensional vector space ##\mathbb{R}##
Well, there are the two trivial ones that are subspaces.
 
##\mathbb{Z}## is not a subspace
 
fresh_42 said:
You said for a subspace ##U## with ##x,y \in U## and ##c \in \mathbb{F}## you have ##c\cdot x\, , \,x+y \in U##.
With this, can you say something about ##cx+cy##? And why is ##c(x+y) \neq cx +cy## impossible?
what is the reason that closed under addition and closed under multiplication lead to the conclusion that c(x+y) = cx + cy?

x+y=y+x closed under addition
c(x+y)=c(y+x) closed under mutiplication
cx+cy is closed under add / muti
but these 3 properties does not give the conclusion that c(x+y)=cx+cy?
 
zwierz said:
then what is a subspace by definition?

I can imagine such a set but it is not called subspace in accordance to standard terminology.
I mean that even in ##\mathbb{R}## there are subsets those are closed under + but they are not subspaces of the one dimensional vector space ##\mathbb{R}##
subspace is a subset of vector space that follows the closed add/muti rule.
 
mfb said:
Who is talking about Z?
it is just an example of a subset that is closed under + but it is not a subspace of the vector space ##\mathbb{R}## (over the field ##\mathbb{R}##)
tze liu said:
subspace is a subset of vector space that follows the closed add/muti rule.
seems that you use not standard definition
 
tze liu said:
what is the reason that closed under addition and closed under multiplication lead to the conclusion that c(x+y) = cx + cy?
You inherit the definition of multiplication and addition from the parent space.@zwierz: Your earlier post, taken literally, claimed that no set closed under addition is a subspace of R. That is wrong.
There are sets closed under addition that are not subspaces, sure, but that is a weaker statement than you made.
 
mfb said:
Your earlier post, taken literally, claimed that no set closed under addition is a subspace of R. That is wrong.
which post do you mean? cite it please
 
fresh_42 said:
Instead it is very likely starting to confuse him as wrong statements already have been made
which wrong statements do you mean cite them please