linear algebra - If $dim V = n$ and $S$ is a linearly independent set in $V$, then $S$ is a basis for $V$, True or False?

I'm currently taking a linear algebra course, and the topic of the current section is dimensions of vector spaces. I came across the titled question in the practice problems. Online sources that I've seen say the answer is False, because we don't know the number of elements in $S$, therefore we cannot imply that $S$ is a basis for $V$.

I'm sure that I'm missing some key piece of information or I'm implying too much, but I thought it was True with Theorems 11 and 12 backing up my reasoning.

Theorem 11: Let $H$ be a subspace of a finite-dimensional vector space $V$. Any linearly independent set in $H$ can be expanded, if necessary, to a basis for $H$. Also, $H$ is finite-dimensional and $$\dim H \le \dim V$$

Theorem 12 (Basis Theorem): Let $V$ be a p-dimensional vector space, where $p \ge 1$. Any linearly independent set of exactly $p$ elements in $V$ is automatically a basis for $V$. Any set of exactly $p$ elements that spans $V$ is automatically a basis for $V$.

My reasoning behind thinking the answer is True is as follows...

We don't know the number of elements in $S$, but we don't need that information to infer that $S$ is a basis for $V$, due to the aforementioned theorems. Since $S$ is known to be linearly independent, we know $S$ can have at most $n$ elements. By the Basis Theorem, $S$ is automatically a basis for $V$. Before you say it, yes I know the theorem states that $S$ must have exactly $n$ elements, but this is where I feel Theorem 11 comes into play.

If $S$ has less than $n$ elements, it is still linearly independent, so by Theorem 11 we know there exist elements that can be added to $S$ to expand $S$ into a basis for $V$.

I'm curious to know why what I have stated is incorrect, so if anyone can enlighten me I would appreciate it greatly. Thank you!

Answer

You're missing the piece that $dim S \le dim V$ includes the case where $dim S \lt dim V$. By Theorem 11, $S$ can be expanded to a basis for $V$, but this does not mean that $S$ is necessarily already a basis for $V$. Theorem 12 says that $S$ would be a basis for $V$ if $S$ contained exactly $dim V$ elements ($S$ is already linearly independent).