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

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 $s\
eq 0$ 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 $s\
eq 0$ 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 $w_1,\ldots,w_n$