Is W1 a Vector Subspace with Only the Zero Vector?

[andrey21]

Is the following a vector subspace

W₁ {(x₁,x₂,x₃): x₁=x₂=x₃=0}


I usually begin my attempt by finding two members of the set then check which axioms are valid.However I can only think of 1:

(0,0,0)


Any help would be great thank you

 

[henry_m]

Yes, you're right that there is only one element in that set.

Normally, to show a set is a subspace, you must show that certain properties hold for all elements of the set. So you take two arbitrary elements, and check the properties. But here there is only one choice for an arbitrary element so you only have to check this one case: 'for all v in W' means 'for v=0'.

Hope that makes some sense...

 

[andrey21]

Ok so when checking if set is closed under addition should I just do the following:

(0,0,0)+(0,0,0)=(0,0,0)

and scalar multiplication:

2.(0,0,0) = (0,0,0)

 

[henry_m]

Exactly that!

In fact, what you doing generalises to any vector space at all (which is an abstract structure in which you can add and multiply by scalars in case you haven't met the general definition; check wiki for more). In any vector space there are two subspaces, called the trivial subspaces. One of them is the whole space, and one of them is the space containing only the zero vector, which is the one you're looking at (the zero vector is defined by v+0=v for all vectors v).