This is the proof for theorem 1.7 from the book which I don’t think we covered in lecture.

Theorem 1.7

If $S$ is a linearly independent subset of $V$ and $v$ is a vector in $V$ but not $S$. Then $S \cup \{v\}$ is linearly dependent if and if only if $v \in span(S)$.

Proof:

$\Rightarrow$: Since $S \cup \{v\}$ is linearly dependent, then there are vectors $u_1,u_2,...,u_n$ in $S \cup \{v\}$ and not all zero scalars $a_1,...,a_n$ such that

$$ \begin{align*} a_1u_1 + a_2u_2 + ... + a_nu_n = \bar{0}. \end{align*} $$

But we know that $S$ is linearly independent so this means one of the vectors in $\{u_1,u_2,...,u_n\}$ must be $v$. without the loss of generality let $u_1 = v$ and so

$$ \begin{align*} \bar{0} &= a_1v + a_2u_2 + ... + a_nu_n \\ v &= \frac{1}{a_1}(-a_2u_2 - ... - a_nu_n) \\ v &= -\frac{a_2}{a_1}u_2 - ... - \frac{a_n}{a_1}u_n. \\ \end{align*} $$

This tells us that $v$ is a linear combination of the elements of $u_2,...,u_n$ in $S$. Therefore, by definition, $v \in span(S)$.

$\Leftarrow$: Let $v \in span(S)$, then there exists vectors $w_1,w_2,...,w_m$ in $S$ and scalars $b_1,...,b_n$ such that

$$ \begin{align*} v = b_1w_1 + b_2w_2 + ... + b_nw_m. \end{align*} $$

Moving $v$ to the other side, we see that,

$$ \begin{align*} b_1w_1 + b_2w_2 + ... + b_nw_m + (-1).v = \bar{0}. \end{align*} $$

We know here that $v$ is not in $S$ and so none of the vectors $w_1,...,w_m$ are equal to $v$. Moreover, the coefficient of $v$ is none zero. Therefore, by the definition of linear dependence, $S \cup \{v\}$ is linearly dependent. $\blacksquare$.

References:

Linear Algebra 5th Edition