Theory of Matrices proof for basis, spanning set, linear independence

Question

Let $V$ be a finite dimensional space, $\dim V=n$, and let the set $s= \{v_1, v_2,..., v_n\}$ in $V$.

1) if $s$ is linearly independent then $s$ is a basis of $V$

2) if $s$ is spanning then $s$ is a basis of $V$

The question is asking to prove this. I am able to show the reverse of the above but I keep getting stuck. I appreciate any help.

Answer 1

For part 1. You can show that if $\{s_1,\ldots,s_n\}$ doesn't generate $V$, then there exist a vector $w\in V $ such that $w$ is not a linear combination of $\{s_1,\ldots,s_n\}$. Then you must show that the set

$$\{s_1,\ldots,s_n,w\}$$ is linearly independent (check this please using that $\{s_1,\ldots,s_n\}$ is l.i. by part 1 hypothesis) and this is a contradiction with $\dim(V)=n$. Then $\{s_1,\ldots,s_n\}$ must generate $V$.

In general. If $k\leq\dim (V)$ and $\{v_1,\ldots,v_k\}$ is a linearly independent set of vectors in $V$. Then the set ontained adding a vector $w\in V$ that is not a linear combination of $\{v_1,\ldots,v_k\}$ is always linearly independent.

Also. You can show the following: If a set of $m$ vectors $\{u_1,\ldots,u_m\}$ generate a vector space $V$, then any linearly independent set of vectors in $V$ doesn't contain more than $m$ elements. You can use this for part 2.