Prove that $\text{span}\left\{\underline{w}_1,...,\underline{w}_n,\underline{u}\right\}=\text{span}\left\{\underline{w}_1,...,\underline{w}_n\right\}$

Question

Suppose that $\underline{u},\underline{w}_1,...,\underline{w}_n\in\mathbb{R^n}$

$\text{span}\left\{\underline{w}_1,...,\underline{w}_n,\underline{u}\right\}=\text{span}\left\{\underline{w}_1,...,\underline{w}_n\right\}$ iff $\underline{u}$ is a linear combinations of others.

My suggested proof:

Let $\bf{v}\in$ $\underline{u},\underline{w}_1,...,\underline{w}_n$

So there are scalars $c_1,...c_n$ such that

$\bf{v}$$=c_1\underline{w}_1+...+c_n\underline{w}_n+0\underline{u}=c_1\underline{w}_1+...+c_n\underline{w}_n$

So $\text{span}\left\{\underline{w}_1,...,\underline{w}_n,\underline{u}\right\}=\text{span}\left\{\underline{w}_1,...,\underline{w}_n\right\}$?

I'm not sure if the proof is correct.

Answer 1

If $V=$ Span$\{u,w_1,\ldots,w_n\}=$ Span$\{w_1,\ldots,w_n\}$ then for any $v\in V$ there exists $s,c_i,d_i$ such that \begin{align} v &=su+c_1w_1+\cdots+c_nw_n\\ &=d_1w_1+\ldots+d_nw_n. \end{align} We can accept without loss of generality $s\neq 0$ so $u=\sum_{i=1}^n \frac{1}{s}(d_i-c_i)w_i$. Other implication is obvious so we are done.

EDIT $\frac{1}{s}$ added to the expression of $u$. Since Span$\{u,w_1,\ldots,w_n\}=$ Span$\{w_1,\ldots,w_n\}$, the set $\{u,w_1,\ldots,w_n\}$ is linearly dependent so the expansion of $v$ is not unique in terms of $\{u,w_1,\ldots,w_n\}$. When $s=0$, the $c_i$'s must be $d_i$ and when $s\neq 0$ there are different $c_i$ and $d_i$.

Actually I noticed shortest proof of the proposition while typing the edit. If $V=$ Span$\{u,w_1,\ldots,w_n\}=$ Span$\{w_1,\ldots,w_n\}$ then $u\in V=$ Span$\{w_1,\ldots,w_n\}$. Thus $u$ is a linear combination of $w_1,\ldots,w_n$

Answer 2

No, the proof is incorrect.

Saying that $v\in\operatorname{span}\{w_1,\dots,w_n,u\}$ means you can write $$ v=\alpha_1w_1+\dots+\alpha_nw_n+\beta u $$ and there's no reason why $\beta$ should be $0$.

Trace of a proof.

First of all, show that $$ \operatorname{span}\{w_1,\dots,w_n\}\subseteq \operatorname{span}\{w_1,\dots,w_n,u\} $$

Thus equality is the same as the reverse inclusion.

Now, suppose $$ \operatorname{span}\{w_1,\dots,w_n\}\supseteq \operatorname{span}\{w_1,\dots,w_n,u\} $$ Then $u\in\operatorname{span}\{w_1,\dots,w_n\}$, so…

Conversely, suppose $u=\gamma_1w_1+\dots+\gamma_nw_n$ and consider $v\in\operatorname{span}\{w_1,\dots,w_n,u\}$. Then you can write $$ v=\alpha_1w_1+\dots+\alpha_nw_n+\beta u =\alpha_1w_1+\dots+\alpha_nw_n+\beta(\gamma_1w_1+\dots+\gamma_nw_n) $$ so…

Answer 3

$$span\{w_1,...,w_n,u\}=span\{w_1,...,w_n\} \iff u = \sum_{i=1}^{n}\alpha_iw_i$$

($\implies$)

$$\sum_{i=1}^{n}\alpha_iw_i+\beta u=\sum_{i=1}^{n}\gamma_iw_i$$ $$\beta u=\sum_{i=1}^{n}\gamma_iw_i - \sum_{i=1}^{n}\alpha_iw_i$$ $$\beta u=\sum_{i=1}^{n}(\gamma_i-\alpha_i)w_i$$

($\impliedby$)

$$u = \sum_{i=1}^{n}\alpha_iw_i \rightarrow u \in span\{w_1,...,w_n\}$$