# Question about a spanning set

Let $V$ be an $n$-dimensional vector space. Then any linearly independent set of vectors $\{v_1, v_2, \ldots, v_n\}$ is a basis for $V$

Proof:

Let $w \in V$. Then $\{v_1, v_2,\ldots, v_n, w \}$ is linearly dependent. Then exists at least one $c_i \ne 0$ s.t. $\displaystyle{c_0w +\sum_{i = 1}^nc_iv_i} = \vec 0.$ If $c_0 = 0,$ then there's at least one $c_i \ne 0$ in the sum $\displaystyle{\sum_{i = 1}^nc_iv_i} = \vec 0.$ Contradiction as $\{v_1, \ldots, v_n\}$ are linearly independent. Hence $c_0 \ne 0$ meaning we can solve for $w: \ w= \displaystyle{\sum_{i = 1}^n\left(-\frac{c_i}{c_0}\right)v_i}.$ Thus $v_i$ span $V$.

We know $\{v_1, v_2,\ldots, v_n, w \}$ is linearly dependent because $\dim(V) = n$ and $v_i \in V$ meaning $v_i$ span $V$. The proof above shows that $\text{span}(v_1,\ldots,v_n) = V.$ Are we proving something we already know from the definition of basis? Is there subtlety here that I am missing?

The definition of basis is a set of vectors that is linearly indepedent and spans the space. This theorem states that in a finite-dimensional vector space, as long as your set has $n$-many vectors and it's linearly indepedent, then it also spans the space. Therefore it is a basis.
Also, we know $B=\{v_1, v_2,\ldots, v_n, w \}$ is linearly depedent because the length of any linearly indepedent list must be less than or equal to the length of a spanning set. Since $\dim V=n$, the spanning sets of $V$ have length $n$, thus since $|B| \not \leq n$, it must be linearly depedent.