Show that $\mathrm{span}(v_1,\dots,v_k) = \mathrm{span}(v_1,\dots,v_k,v)$ Show that $\mathrm{span}(v_1,\dots,v_k) = \mathrm{span}(v_1,\dots,v_k,v)$ if and only if $v$ is in $\mathrm{span}(v_1,\dots,v_k)$
I am thinking that if $v$ is in $\mathrm{span}(v_1,\dots,v_k$) it must be one of the elements and can be written $\mathrm{span}(v_1,\dots,v_k,v)$ but I'm not sure how to show this mathematically.
 A: Let $V = \text{span}(v_1,v_2,\ldots, v_k)$ and $W = \text{span}(v_1,v_2,\ldots, v_k, v)$.


*

*Suppose $V=W$, then $v\in V$, so that proves one direction trivially.

*Suppose $v\in V$, then you want to prove that $W=V$. Since $V\subset W$, it suffices to show that $W\subset V$. Now, first choose $\alpha_1, \alpha_2,\ldots, \alpha_k \in F$ (the underlying field) so that
$$
v = \sum_{i=1}^k \alpha_iv_i \qquad (\dagger)
$$
Then if $w\in W$, $\exists \beta_1,\beta_2,\ldots, \beta_k, \beta_0$ such that
$$
w = \sum_{j=1}^k \beta_jv_j + \beta_0v
$$
Now replace $v$ in this expression by $(\dagger)$ to see that
$$
w = \sum_{j=1}^k (\beta_j + \beta_0\alpha_j) v_j
$$
Hence, $w\in V$. This is true for any $w\in W$, so $W\subset V$ as required.
A: Prove the implication two ways.
First prove the forward direction $\implies $:
Let $V=\text{span}(v_1, \ldots, v_k)$ and $W = \text{span}(v_1, \ldots, v_k, v)$ and assume $V=W$. We will prove that $v \in V$. This is trivial since $v \in W = V$ and thus $v \in V$.

Now we prove the backwards implication $\impliedby$:
Assume $v \in V$. We need to prove that $V = W$. To do this, we pick any arbitrary vector $w \in W$ and prove that it is in $V$ and we pick any arbitrary vector $x \in V$ and prove it is in $W$, essentially proving that $V \subseteq W$ and $W \subseteq V$.
Let $x$ be any vector in $V$. Then $x = a_1v_1 + a_2v_2 + \ldots + a_k v_k$ for some coefficients $a_i$. It is evident that $x \in W$ since we can write $x$ as a linear combination of $W$ as such: $x = a_1v_1 + a_2v_2 + \ldots + a_k v_k + 0v$. Thus $V \subseteq W$.
Now let $w$ be any vector in $W$. Then $w = b_1v_1 + b_2v_2 + \ldots + b_kv_k + bv$ and since $v \in V$, it can be written as a linear combination of $(v_1, \ldots, v_k)$ i.e. $ v = c_1v_1 + c_2v_2 + \ldots c_kv_k$ for some coefficients $c_i$. We can then substitute this into the linear combination of $w$:
$$
\begin{align}
w &= b_1v_1 + b_2v_2 + \ldots + b_kv_k + bv \\
  &= b_1v_1 + b_2v_2 + \ldots + b_kv_k + b(c_1v_1 + c_2v_2 + \ldots c_kv_k) \\
  &= (b_1 + bc_1)v_1 + (b_2 + bc_2)v_2 + \ldots + (b_k + bc_k)v_k 
\end{align}
$$
and thus $w$ can be written as a linear combination of $(v_1, \ldots, v_k)$ and so $w \in V$. Since our choice of $w$ was arbitrary, $W \subseteq V$. 
Since we proved that $V \subseteq W$ and $W \subseteq V$, it follows that $V=W$ as required. $\tag*{$\blacksquare$}$
Note: I would normally use summation notation but pedagogically i've found writing the combinations out explicitly can sometimes be easier to grasp initially for students.
