Prove $(a_1, a_2, ... , a_n)$ is basis of V by n-r vectors $(a_{r+1}, ... , a_n)$

Question

Let $V$ be a $K$-vector space, $dim_K = n > r$

and $(a_1, a_2, ... , a_r)$ $∈ V$ is linearly independent set of vectors.

How can I prove that $(a_1, a_2, ... , a_n)$ is basis of $V$ from $n-r$ vectors $(a_{r+1}, ... , a_n)$

Answer 1

As $ r < n $, $ \{a_1, \ldots, a_r\} $ does not span V, there exists an element $ a_{r+1} \in V - \text{span}(\{a_1, \ldots, a_r\}) $. Thus, $ \{a_1, \ldots, a_{r+1}\} $ is a linearly independent set. If we repeat this process $ n - r $ many times, we get a set $ \{a_1, \ldots, a_n\} $ of $ n $ linearly independent vectors in $ V $, which has dimension $ n $, so this set is a basis for $ V $.

I encourage you to fill in the details about why the set $ \{a_1, \ldots, a_{r+1}\} $ is, in fact, linearly independent and why $ |\{a_1, \ldots, a_n\}| = \text{dim}(V) $ implies that $ \{a_1, \ldots, a_n\} $ is a basis for V. These are not too hard to verify, but understanding them is the key to this problem.