Dimension and generated subspace.

Question

Let $V$ be a $n-$dimensional vector space over a field $\mathbb{K}$. If we know that $$\dim \text{Span}\{v_1,\dots, v_s\}=s$$ can we conclude that $v_1,\dots, v_s$ are linearly independent?

My solution.

We consider the case $s=2$. We suppose that $v_1$ and $v_2$ be linearly dependent. Then exist $\lambda\in \mathbb{K}$ such that $$v_1=\lambda v_2,$$ so $v_1\in \text{Span}(v_2)$ and therefore $$\text{Span}(v_1,v_2)=\text{Span}(v_2).$$ Since $v_2\ne 0$ we have that $$\dim\text{Span}(v_2)=1$$ and this is a contradiction. Now it is sufficient to conclude with the induction on s.

Answer 1

If $S$ generates a subspace of dimension $s$, then you can find a subset $T$ of $S$ such that $|T|=s$ and $T$ is linearly independent. Applying to your case, we only have one choice of $T$, which is $S$ itself, and thus this proves its linear independence.

Answer 2

Let $U$ be a (finite-dimensional) subspace with dimension $n = \dim U$. Suppose we have the set $S = \{ v_1, ..., v_n \}$ such that ${\rm Span} S = U$. Then $S$ is linearly indepedent. (That is, $S$ is a basis for $U$)

Proof

Suppose not; that is, suppose $S$ is not linearly independent. Then there exists a non-trivial set of coefficients $\alpha_1, ..., \alpha_n$ such that $\alpha_1 v_1 + ... + \alpha_n v_n = 0$. Because the solution is nontrivial, at least one of these coefficients satisfies $\alpha_k \ne 0$, $1 \le k \le n$. Then we can write $v_k = -\sum_{i \in \{ 1, ..., n \} - \{ k \}} \frac{\alpha_i}{\alpha_k} v_i$; i.e., $v_k$ is a linear combination of the other vectors. As such, we can remove it from the set $\{ v_1, ..., v_n \}$ without changing the span of that set (you should prove this for yourself).

Now, we have a set $\{ v_1, ..., v_n \} - \{ v_k \}$, which has $n - 1$ elements, whose span is equal to $U$. This shows that $\dim U \le n - 1$, which is a contradiction, for we assumed that $n = \dim U$. Therefore, $S$ must be linearly independent.