I wrote a proof for a theorem and would appreciate if someone could check if my proof is sound. Thanks a lot.
$Theorem$:
Every set that contains a linearly dependent set is linearly dependent.
$Proof$:
Let $S=\{v_1, v_2,…,v_k\}$ be a set of vectors. Let $L=\{v_1, v_2, …, v_l\} \subset S$, $l \lt k$, be a linearly dependent subset of $S$.
Since $L$ is linearly dependent, there must exist, without loss of generality, $v_1 \in L$ such that
$$\sum_{i=2}^l \alpha_i v_i = v_1, \ \alpha_i \in \mathbb{R}$$
Then we can write a sum
$$\sum_{i=2}^l \alpha_i v_i + \sum_{i = l+1}^k 0 v_i= v_1 =$$
$$\sum_{i=2}^k \alpha_i v_i = v_1$$
where $\alpha_i=0$ for $l+1 \lt i\lt k$.
Rearranging,
$$\sum_{i=2}^k \alpha_i v_i – v_1 = 0 \iff \sum_{i=1}^k \alpha_i v_i = 0$$
with $\alpha_i \in \mathbb{R}$ not all zero.
Thus, the set $S$ is linearly dependent. This completes the proof.
Best Answer
It is not correct. You actually did not use the fact that $L$ is linearly dependent at all. In the second paragraph of you proof, you should have written that, since $L$ is linearly dependente, there are numbers $\alpha_1,\ldots,\alpha_l\in\mathbb R$ not all of which are $0$ such that $\sum_{j=1}^l\alpha_jv_j=0$.
Besides, you assumed that $S$ is finite, but the statement that you are trying to prove doesn't contain that assumption.