Let $v_{1},v_{2},\ldots,v_{k},v$ be vectors in a vector space $V$, and define $W_{1} = \text{span}\{v_{1},v_{2},\ldots,v_{k}\}$ and $W_{2} = \{v_{1},v_{2},\ldots,v_{k},v\}$.
(a) Find necessary and sufficient conditions on $v$ such that $\dim(W_{1}) = \dim(W_{2})$.
(b) State and prove a relationship involving $\dim(W_{1})$ and $\dim(W_{2})$ in the case that $\dim(W_{1})\neq\dim(W_{2})$.
MY ATTEMPT
(a) We have that $\dim(W_{1}) = \dim(W_{2})$ iff $v\in\text{span}\{v_{1},v_{2},\ldots,v_{k}\}$.
Let us prove the implication $(\Leftarrow)$ first.
Indeed, if $v\in\text{span}\{v_{1},v_{2},\ldots,v_{k}\}$, then $W_{1} = \text{span}\{v_{1},v_{2},\ldots,v_{k}\} = \text{span}\{v_{1},v_{2},\ldots,v_{k},v\} = W_{2}$.
On the other hand, let us prove that $(\Rightarrow)$.
Since $W_{1} = \text{span}\{v_{1},v_{2},\ldots,v_{k}\}$, we can extract from $\{v_{1},v_{2},\ldots,v_{k}\}$ a basis $\mathcal{B}_{1} = \{v_{1},v_{2},\ldots,v_{n}\}$ for $W_{1}$, where $n\leq k$.
Thus $W_{1} = \text{span}\{v_{1},v_{2},\ldots,v_{n}\}$ and $\dim(W_{1}) = n$.
If $v\not\in W_{1}$, then $v\not\in\text{span}\{v_{1},v_{2},\ldots,v_{n}\}$. Consequently the set $\{v_{1},v_{2},\ldots,v_{n},v\}$ is linear indenpendent, which is a contradiction.
Indeed, if it was the case, we would have $\dim(W_{2}) = n+1$, which contradicts the hypothesis.
(b) If $\dim(W_{1})\neq\dim(W_{2})$, then $\dim(W_{2}) = \dim(W_{1}) + 1$.
Based on the previous argument, if $\dim(W_{1})\neq\dim(W_{2})$, then $v\not\in\text{span}\{v_{1},v_{2},\ldots,v_{k}\} = W_{1}$, which is the same set as $W_{1} = \text{span}\{v_{1},v_{2},\ldots,v_{n}\}$ (according to the previous notation). Therefore we conclude that $\{v_{1},v_{2},\ldots,v_{n},v\}$ is LI.
More precisely, $\dim(W_{2}) = \dim(W_{1}) + 1$, as previously claimed.
Could someone double-check my arguments or provide another approach to solve it?