Proofs from Linear independence

If the coefficients in the displayed representation of $v$ are unique, suppose a non-zero relation

exists. Then

is another representation of $v$. Since some $c_j\neq0$, the two representations are different, a contradiction. Thus the set is linearly independent. Conversely, suppose

Subtracting gives

Independence forces $\lambda_j-\alpha_j=0$ for every $j$, so the two representations are the same.

Write $S=\{u_1,\ldots,u_n\}$. Suppose

If $\mu\neq0$, then $v$ is a linear combination of the vectors in $S$, contradicting $v\notin\operatorname{span}(S)$. Hence $\mu=0$ and the independence of $S$ forces $\lambda_1=\cdots=\lambda_n=0$. Therefore $S\cup\{v\}$ is linearly independent.