My teacher gave us this proof today, but I don't know if I understood it entirely:
Theorem:
Suppose $V$ a vector space (finitely generated) over the reals.
$$B = \{v_1,\cdots, v_n\}$$
where $B$ is a basis for $V$.
Suppose, also, a set $S\subset V$ L.I., then $S$ is finite and has at most $n$ elements.
Proof:
Let $S = \{y_1,\cdots,y_m\}$, with $m>n$. Suppose $S$ is L.I. Then:
$$\alpha_1y_1+\cdots+\alpha_my_m = 0\tag{1}$$ But since $S$ is greater
than $B$, and $B$ is a basis, $S$ can be spanned by $B$. So any
$y_i\in S$ can be written as a linear combination of $B$, in this way:
$$\\y_1 = \beta_{11}v_1+\cdots+\beta_{n1}v_n\\y_2 =
\beta_{12}v_1+\cdots+\beta_{n2}v_n\\\cdots\\y_m =
\beta_{1m}v_1+\cdots+\beta_{nm}v_n$$ Substituting the system above in
$(1)$, we have:
$$\alpha_1(\beta_{11}v_1+\cdots+\beta_{n1}v_n)+\cdots+\alpha_m(\beta_{1m}v_1+\cdots+\beta_{nm}v_n)
= 0\implies\\ (\beta_{11}\alpha_1+\beta_{12}\alpha_2+\cdots+\beta_{1m}\alpha_m)v_1+\cdots+(\beta_{n1}\alpha_1+\beta_{n2}\alpha_2+\cdots+\beta_{nm}\alpha_m)v_m
= 0$$ But since $B$ is L.I., then the linear combination above implies the coefficients are all $0$. Therefore we end up with the system:$$\begin{cases}\beta_{11}\alpha_1+\beta_{12}\alpha_2+\cdots+\beta_{1m}\alpha_m = 0\\
\cdots \\
\beta_{n1}\alpha_1+\beta_{n2}\alpha_2+\cdots+\beta_{nm}\alpha_m = 0\end{cases}$$
that has $n$ equations, and $m$ variables, therefore is compatible and
indeterminated, that implies it has infinitely many solutions. One of
them is not trivial, so $S$ is L.D.
I think wikipedia has a similar proof, but there, it proves directly that any basis has the same number of elements (could you tell me why a vector $v_i$ suddenly appears in the middle of the proof?. In this proof, the conclusion is that $S$ has at most $n$ elements.
Is my proof correct? I understood it correctly? (I didn't copy, this is my understanding of the proof).
Best Answer
It seems nothing wrong in your understanding. Probably what makes you confused is that the proof on wikipedia is a generalization of what you learnt in your class: your dimension theorem is a version only on finite dimensional vector spaces and that on wikipedia is the full version on both finite or infinite dimensional vector spaces.
In general, in a vector space $V$, the span of a set $S \subseteq V$ is define as all finite linear combinations of elements in $S$. A set $S$ is linearly independent if, a finite linear combination of elements in $S$ is the zero vector implies the coefficients are all zero. The standard way of representing a finite combination is by $\sum_{i \in I}$ where $|I| < \infty$ as in the proof from wikipedia.