7

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.

  • It's not an odd question for a mathematician. If you look you can find that vector space inside any other vector space, which does seem to make it the smallest. – Ethan Bolker Feb 27 '16 at 23:49
  • 1
    The only thing smaller would be the empty set. Is that a vector space? – amd Feb 27 '16 at 23:49
  • Most definitions of a vector space require the set to be nonempty. Therefore, the empty set is not a vector space. In this setting, smallest means that if $V$ is any vector space, then there is a subspace of $V$ isomorphic to $\mathcal{V}$. In other words, for any vector space $V$, $\mathcal{V}$ is inside $V$. Therefore $\mathcal{V}$ is a vector space because it satisfies the vector space properties and it is the smallest because every other vector space contains it. – Michael Burr Feb 27 '16 at 23:51
  • Really, for it to be the smallest possible vector space, this means that it is a subspace of every other vector space. With that definition the proof of course becomes trivial. – Rellek Feb 27 '16 at 23:54
  • Is {1} also inside of any Vector space? –  Feb 27 '16 at 23:54
  • 2
    @JosiahBlaisdell How do define 1 in a vector space? It's true that 1 is defined in the underlying field, but then {1} is just a subset of the field of scalars. – user4894 Feb 27 '16 at 23:55
  • @user4894 Hmmm, I don't think you can... I see... –  Feb 27 '16 at 23:57
  • If you look closely, see that ${θ}$ is the smallest, because the smaller structure than ${θ}$ is $φ$, the null set, which doesn't contain a null vector and hence not a vector space. – Anik Bhowmick Aug 01 '18 at 09:49

3 Answers3

2

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 ...)

Neal
  • 32,659
  • yes. thank you. –  Feb 28 '16 at 00:13
  • By contradiction, assuming {0} is not the smallest vector space, there must be a smaller vector space that is in every vector space. But that would mean there's another element in every vector space that can alone form a vector space. That's not possible since the identity element is 0. –  Feb 28 '16 at 00:22
0

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.

Adam Francey
  • 574
0

There are three conditions for a set of vectors V to be a vector subspace. Those three conditions are:

  1. The set V must contain the zero vector.
  2. The set V must be closed under scalar multiplication.

  3. 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.

Enusi
  • 496