How can I prove a set is linearly dependent

Question

The only thing given to me is $\operatorname{span}(S \cup\{u\}) = \operatorname{span}(S)$ and that $S$ is linearly independent. I have to prove that $S \cup \{u\}$ is linearly dependent. I know that to prove it as linearly dependent I have to prove that an element of the set can be expressed as a linear combination of the other elements. I just don't know how to do that with the given information. I would appreciate your help.

Edit: I can write that since $\operatorname{span}(S \cup\{u\}) = \operatorname{span}(S)$, x ε span(S υ {u}) and x ε span(S) implies λ1s1 +...+ λnsn + λu = λ1s1+...+ λnsn Can you help me proceed further? Thank you for the answers. Sorry for the notations...

Answer 1

A hint: $$ u\in\operatorname{span}(S\cup\{u\}) $$

Answer 2

From $\operatorname{span}(S \cup\{u\}) = \operatorname{span}(S)$ we get $u \in \operatorname{span}(S)$, hence there are scalars $t_1,...,t_n$ and $s_1,...,s_n \in S$ such that

$u=t_1s_1+....t_ns_n$ or

$u-t_1s_1-....-t_ns_n=0$.

Can you proceed ?

Answer 3

For any $x \in span \left( S\cup \{u\}\right)$, $x\in span (S)$ according to your assumption. That means, $u$ can be written as $u = \sum_i^n a_i s_i$ for some $a_i$. This implies that $u$ and $s_1,\cdots,s_n$ are not linearly independent.

Answer 4

Since $u \neq 0$ and $u \in span(S)$. There $\lambda_1,...,\lambda_2$ and $x_1,...,x_n \in S$ such that $u=\lambda_1x_1+..,\lambda_n x_n$.

Now since $u \neq 0$ you have that there is at least one $1\leq i \leq n$ such that $\lambda_i \neq 0$. So you have that $\lambda_1x_1+...+\lambda_nx_n +(-u)=0$.

So $0$ is a non trivial linear combination elemements that belong to $S\cup \{u\}.$ Hence $S\cup \{u\}$ is not linear indepented.

Answer 5

Hints. The inclusion $\operatorname{span}(S)\subseteq\operatorname{span}(S\cup\{u\})$ is obvious, so it’s the converse inclusion that’s interesting. Use the fact that $u\in \operatorname{span}(S\cup\{u\})$, so also $u\in \operatorname{span}(S)$.

Answer 6

If $span(S\cup \{u\})=span(S)$ consider $u$... It is in $span(S\cup \{u\})$, hence in $span(S)$. That completes the proof...