Why is the trivial vector space the smallest vector space? My book (Elementary Linear Algebra by Andrilli) says:

The set $\mathcal{{V}}$ = {${\mathbb {0}}$} is a vector space AND is
  the smallest vector space.

Then the book asks why $\mathcal{{V}}$ is the smallest vector space. I have no idea where to even start to explain why $\mathcal{{V}}$ is the smallest space. It seems like an odd question to ask.
 A: I suspect part of your confusion is, 

what does "smallest" mean? 

It seems to imply a partial ordering somehow, so here are two possible definitions:


*

*$V$ is smaller than $W$ provided there is an injective linear map $V\to W$.

*$V$ is smaller than $W$ provided $|V| \leq |W|$.


(Bonus questions: Is there any relationship between these definitions? Can one be proven from the other, and vice versa?)
Now, given either definition, say $V$ is the smallest vector space provided $V$ is smaller than $W$ for any vector space $W$.
From this definition, can you prove that $\{0\}$ is the smallest vector space? (Hint: Every vector space must have a $0$ element, so ...)
A: A vector space cannot be empty because it must have an identity element, $0$, so the smallest vector space is not empty.
Any one element set $V = \{v\}$ where $v \neq 0$ cannot be a vector space because it would be missing the identity element.
So if $V = \{0\}$ is a vector space, it must be the smallest because there are no others with only one element.
A: There are three conditions for a set of vectors V to be a vector subspace. Those three conditions are:


*

*The set V must contain the zero vector.

*The set V must be closed under scalar multiplication.

*The set V must be closed under addition of the vectors.


*

*(2) and (3) can (and often are) combined to say that the set must be closed under linear combination of the vectors in the set. 

*If a vector contains ONLY the vector {${\vec0}$}, it happens to satisfy the other two conditions as well. Therefore, the trivial set V contains only the zero vector. 
