In below image author has proved the theorem which states that,
"Let $V$ ve a finite dimensional vector space and $\{v_1,v_2,...,v_n\}$ is basis for $V$ then, "If a set in $V$ has fewer than $n$ vectors then it does not span $V$ "
How author has derived the linear system indicated by an arrow in the picture?
My attempt: using equation (4), equation (5) implies,
$k_1(a_{11}w_1+a_{21}w_2+...+a_{m1}w_m)+k_2(a_{12}w_1+a_{22}w_2+...+a_{m2}w_m)+...+...+k_n(a_{1n}w_1+a_{2n}w_2+...+a_{mn}w_m=0$
Which implies,
$(a_{11}k_1+a_{12}k_2+...+a_{1n}k_n)w_1+(a_{21}k_1+a_{22}k_2+...+a_{2n}k_n)w_2+...+(a_{m1}k_1+a_{m2}k_2+...+a_{mn}k_n)w_m=0$
From this to conclude homogeneous linear system indicated by arrow (in pic) we must have $w_1,...,w_m$ must be linearly independent. But, here we doesn't know whether $w_i$'s are linearly independent or not! So how author has derived homogeneous linear system in equation indicated by arrow (in pic)
Please help..
Let's solve the following linear system of equations for $x_i'$s, $1\le i\le m$:
$\begin{align} &a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n=0\\ &a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n=0\\ &\cdots\quad\cdots\quad \cdots\quad \cdots \\ &a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n=0 \end{align}\tag A$
What property do these $x_i$'s have? It can be shown that they satisfy $x_1v_1+x_2v_2+\cdots+x_nv_n=0\tag B$
Since $m<n$, it follows that $(A)$ has a non-trivial solution i.e., in particular there exist $x_1,x_2,...,x_n$ not all zero. These $x_i$'s must satisfy $(A)$.
But since $v_i$'s are linearly independent, it follows from $(B)$ that $x_i=0$ for all $1\le i\le n$. This contradicts conclusion in last para.
It follows that the assumption that $w_i$'s span $V$ is not correct.