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

Theorem 1.7

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 s p a n (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{array}{r} a_{1} u_{1} + a_{2} u_{2} + . . . + a_{n} u_{n} = \bar{0} . \end{array}

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{aligned} \bar{0} & = a_{1} v + a_{2} u_{2} + . . . + a_{n} u_{n} \\ v & = \frac{1}{a_{1}} (- a_{2} u_{2} - . . . - a_{n} u_{n}) \\ v & = - \frac{a_{2}}{a_{1}} u_{2} - . . . - \frac{a_{n}}{a_{1}} u_{n} . \end{aligned}

This tells us that $v$ is a linear combination of the elements of $u_{2}, . . ., u_{n}$ in $S$ . Therefore, by definition, $v \in s p a n (S)$ .

$\Leftarrow$ : Let $v \in s p a n (S)$ , then there exists vectors $w_{1}, w_{2}, . . ., w_{m}$ in $S$ and scalars $b_{1}, . . ., b_{n}$ such that

\begin{array}{r} v = b_{1} w_{1} + b_{2} w_{2} + . . . + b_{n} w_{m} . \end{array}

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

\begin{array}{r} b_{1} w_{1} + b_{2} w_{2} + . . . + b_{n} w_{m} + (- 1) . v = \bar{0} . \end{array}

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. $◼$ .

References:

Linear Algebra 5th Edition