Linear Dependence of Vectors and Matrices Let $W=\left \{ w_{1},...,w_{m} \right \}$ and $V=\left \{ v_{1},...,v_{n} \right \}$ for $m>n$ where W and V are both independent sets of vectors. ($W$ being an independent set of vectors means $w_1,\cdots,w_m$ are independent. Similarly for $V$)
Why the following cannot be true?

$$\begin{bmatrix}
w_{1}\\ 
w_{2}\\
\vdots \\
w_{m} \end{bmatrix}\neq\begin{bmatrix}
a_{11} & &a_{1n} \\ 
\vdots &\ddots &\vdots\\ 
 a_{m1}&  & a_{mn}\\
\end{bmatrix}
\begin{bmatrix}
v_{1}\\ 
v_{2}\\
\vdots \\
v_{n} \end{bmatrix}$$

 A: Suppose otherwise one has the equality. Then one can write that 
$$
B=AC
$$
where $rank(B)=m$ and $rank(C)=n$. Now, by a property of matrix ranks:
$$
m=rank(B)\leq\min\{rank(A),rank(C)\}=\min\{rank(A),n\}
$$
which is impossible because by the assumption $m>n$.  

[Added later.] I like Martin's answer better than mine: if the equality is true, then it follows that $span(W)\subset span(V)$, which is impossible since $m>n$.
A: The rows of $Av $ belong to the span of $v_1,\ldots,v_n $, which has dimension $n $. An $n $-dimensional space cannot contain  $m>n $ linearly independent vectors.

So, here is an argument that shows that if $v_1,\ldots,v_n$ and $w_1,\ldots,w_m$ linearly independent and they span the same space, then $n=m$. We can write 
$$
w_k=\sum_{j=1}^n\alpha_{kj}v_j
$$
for certain coefficients $\{\alpha_{kj}\}$. Similarly, we can write 
$$
v_j=\sum_{\ell=1}^m\beta_{j\ell}w_\ell
$$
for certain coefficients $\{\beta_{j\ell}\}$. Combining the two equalities, we get 
$$
w_k=\sum_{j=1}^n\sum_{\ell=1}^m\alpha_{kj}\beta_{j\ell}\,w_\ell=\sum_{\ell=1}^m\left(\sum_{j=1}^n\alpha_{kj}\beta_{j\ell}\right)\,w_\ell.
$$
By the linear independence of $w_1,\ldots,w_m$ we obtain
$$
\sum_{j=1}^n\alpha_{kj}\beta_{j\ell}=\delta_{k\ell}.
$$
In particular, $$\tag{1}\sum_{k=1}^m\sum_{j=1}^n\alpha_{kj}\beta_{jk}=\sum_{k=1}^m\delta_{kk}=m.$$
Now we can repeat all the above with the roles of $v_j$ and $w_k$ exchanged, and we obtain 
$$
v_j=\sum_{\ell=1}^m\sum_{h=1}^n\beta_{j\ell}\alpha_{\ell h}\,v_h
=\sum_{h=1}^n\left(\sum_{\ell=1}^m\beta_{j\ell}\alpha_{\ell h}\right)\,v_h,
$$
and so 
$$
\sum_{\ell=1}^m\beta_{j\ell}\alpha_{\ell h}=\delta_{jh}.
$$
Thus,
$$\tag{2}
\sum_{j=1}^n\sum_{\ell=1}^m\beta_{j\ell }\alpha_{\ell j}=\sum_{j=1}^n\delta_{jj}=n.
$$
But the sums in $(1)$ and $(2)$ are the same, and so $m=n$.
