Let $S$ be a subset of an $n$-dimensional vector space $V$, and suppose $S$ contains fewer than $n$ vectors. Explain why $S$ cannot span $V$

Let $S$ be a subset of an $n$-dimensional vector space $V$, and suppose $S$ contains fewer than $n$ vectors. Explain why $S$ cannot span $V$

Proof:

Suppose $S$ is a subset of an $n$-dimensional vector space $V$. Let $S$ be an $n-1$ dimensional vector space. Then $\dim(S)\leq \dim(V)$ and by the isomorphism of $V$ to $\mathbb{R}^n$ and $S$ to $\mathbb{R}^{n-1}$, it follows that $S$ can at most span $\mathbb{R}^{n-1}$, and thus does not span $\mathbb{R}^n$.

Does this proof work?

• No. An $n-1$ dimensional vector space contains more than $n$ vectors. – Matt Samuel Jun 26 '15 at 20:31
• How did you define the dimension of a vector space? Usually it is defined as the cardinality of a basis, but in that case what you are trying to prove would be trivially true: if $S$ spans $V$ then it must contain a basis of $V$, hence it must have more than $\dim(V) = n$ elements. – A.P. Jun 26 '15 at 20:34
• You need to be more cautious to distinguish "subspace" and "subset". A "subspace" contains infinitely many vectors. I guess you planned to say $S$ is an $n - 1$ dimensional subspace. – Zhanxiong Jun 26 '15 at 20:37

If you are dealing with vector spaces over the real (or complex) numbers, you should know that a nonzero subspace always has infinitely many elements. Indeed, if $v\in V$ and $v\ne0$, the map $\mathbb{R}\to V$ defined by $\alpha\mapsto \alpha v$ is injective and its image is contained in every subspace of $V$ which $v$ belongs to.
So you can't “suppose $S$ is a subspace of $V$”, in general.
What you should note is that from any spanning set you can extract a basis, so the subspace spanned by $S$ has a basis consisting of less elements than $n$ (because it's a subset of $S$). Hence the dimension of the subspace $U$ spanned by $S$ is less than $\dim V$; therefore $U\ne V$.