I was reading Sheldon Axler's Linear Algebra Done Right and in Section 2.A, there is a lemma 2.23,
In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.
Although there is a proof given in the book, I sought to think of other proofs. Then, I managed to come up with a proof that rely on the rank of a matrix. But I am not entirely sure if it is correct, so I would appreciate if anyone can verify my proof.
$\textit{Proof:}$ Assume for contradiction that there exist a linearly independent list of vectors, $u_1, u_2,\cdots , u_m$ in a vector space $V$ but it is not a span of $V$ and a spanning list of vectors $w_1, w_2, \cdots , w_n$, where $m>n.$ There exist $a_{i,j}\in \mathbb{F}$ ($V$ is a vector space over $\mathbb{F}$) such that $$\begin{align*} u_i = \sum_{j=1}^n a_{j,i}w_j \end{align*}$$ as $w_1, w_2,\cdots ,w_n$ spans $V$. Let $v_i= \begin{pmatrix} a_{1,i} \\ a_{2,i} \\ \vdots \\ a_{n,i}\end{pmatrix}$, $A= \begin{pmatrix} v_1 & v_2 &\cdots &v_m\end{pmatrix}$ where $A$ is the matrix containing $v_i$ in the $i$th column.
Claim: $v_1, v_2 , \cdots , v_m$ are linearly independent and $\textrm{Rank($A$)}= n$.
$\textit{Proof:}$ Assume for contradiction that $v_1, v_2 , \cdots , v_m$ are not linearly independent, then there exist $k \in \{1,2, \cdots , m\}$ such that $v_k$ can be represented by the other $v_i$'s by $$v_k = b_1v_1+ \cdots + b_mv_m$$ where $b_1, b_2 , \cdots, b_m \in \mathbb{F}$ and $b_k=0$. So, $$\begin{align*} \begin{pmatrix} a_{1,k} \\ a_{2,k} \\ \vdots \\ a_{n,k}\end{pmatrix} &= b_1\begin{pmatrix} a_{1,1} \\ a_{2,1} \\ \vdots \\ a_{n,1}\end{pmatrix} + \cdots + b_m\begin{pmatrix} a_{1,m} \\ a_{2,m} \\ \vdots \\ a_{n,m}\end{pmatrix} \\ &= \begin{pmatrix} b_1a_{1,1}+ \cdots +b_m a_{1,m}\\ b_1a_{2,1} +\cdots + b_ma_{2,m} \\ \vdots \\ b_1a_{n,1}+ \cdots + b_ma_{n,m}\end{pmatrix}\end{align*}$$ and we see that $b_1u_1 + \cdots +b_m u_m = a_{1,k}w_1+ \cdots + a_{n,k}w_n=u_k$, contradicting that $u_1,u_2,\cdots , u_m$ are linearly independent. Since $v_1, v_2, \cdots , v_m$ are linearly independent, the only solution to $A\mathbf{x}=\mathbf{0}$ is the zero vector. This means $\textrm{Rank($A$)}=n$. $\quad \square$
If $v=c_1w_1 + c_2 w_2+ \cdots + c_nw_n$ is a vector in $V$, where $c_1, c_2, \cdots , c_n \in \mathbb{F}$, by our claim, there is a solution vector, $t=\begin{pmatrix} t_1 \\ t_2 \\ \vdots \\ t_m\end{pmatrix}$ such that $$\begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,m} \\ a_{2,1} & a_{2,2} & \cdots &a_{2,m} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,m}\end{pmatrix}\begin{pmatrix} t_1 \\ t_2 \\ \vdots \\ t_m\end{pmatrix}= \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_m\end{pmatrix}$$ or equivalently, $$\sum_{j=1}^m a_{i,j}t_j=c_i.$$ This gives $$\begin{align*} v&= \sum_{i=1}^n c_iw_i\\ &= \sum_{i=1}^n \sum_{j=1}^m a_{i,j}t_jw_i \\ &= \sum_{j=1}^mt_j\sum_{i=1}^n (a_{i,j}w_i)\\ &= \sum_{j=1}^m t_ju_j.\end{align*}$$ Hence, $u_1, u_2, \cdots , u_m$ spans $V$, a contradiction. $\quad \blacksquare$
Edit: The thing about this proof is that I feel that something is not quite right, in particular, the proof of the claim where I made arguments about linear independence and the rank of $A$.
This cannot be true. You have that $m>n$. Then if Rank$(A)=n$, the maximal number of linearly independant column vectors is also $n$ (since row-rank = column-rank), therefore there cannot be $m$ linearly independent column vectors.
I don't see where this comes from. Can you please elaborate?