Exercise 1.2.5

Proof. Take $v_{r+1}$ that can not be represented as ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^r{\alpha }_k{v}_k$. It is possible because $v_1,v_2,...,v_r$ are not generating. Now we need to show $v_1,v_2,...,v_r,$ $v_{r+1}$ are linearly independent. Suppose that $v_1,v_2,...,v_r,v_{r+1}$ are linearly dependent, i.e.,

and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^{r+1}\left|{\alpha }_k\right|\ne 0$. If ${\alpha }_{r+1}=0$, then

and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^r\left|{\alpha }_k\right|\ne 0$. This contradicts that $v_1,v_2,...,v_r$ are linearly independent. So ${\alpha }_{r+1}\ne 0$. Thus $v_{r+1}$ can be represented as

This contradicts the premise that $v_{r+1}$ can not be represented by $v_1,v_2,...,v_r$. Thus, the system $v_1,v_2,...,v_r,v_{r+1}$ is linearly independent. □