I am asked to prove following
If the $n$ vectors $\bf {u_1, u_2, u_3,..., u_n}$ span $V$ then $\bf {u_1, u_2, u_3,..., u_m}$ where $m < n$ does not span V.
If I am not missing anything, the proposition above is not correct.
Let's consider an example:
Let $\bf w$ be arbitrary vector. Suppose it can be rewritten as the following linear combination
$k_{1}\mathbf{u}_{1} + k_{2}\mathbf{u}_{2} + \cdots + k_{n}\mathbf{u}_{n} = \bf w$
Where $k_{n} = 0$. Since $k_{n}\mathbf{v}_n = \bf O$, we have
$k_{1}\mathbf{u}_{1} + k_{2}\mathbf{u}_{2} + \cdots + k_{n-1}\mathbf{u}_{n-1} + \mathbf O = \bf w \implies$
$k_{1}\mathbf{u}_{1} + k_{2}\mathbf{u}_{2} + \cdots + k_{n-1}\mathbf{u}_{n-1} = \bf w $
So if we let $m=n-1$, we can say that both vectors $\bf {u_1, u_2, u_3,..., u_n}$ and $\bf {u_1, u_2, u_3,..., u_m}$ span $V$ and $m < n$.
The definition of the "span" (in the book I'm reading) only says that, if some vectors, say $\mathbf{v}_{1},\mathbf{v}_{2},\cdots,\mathbf{v}_{n}$, span vector space, say $V$, then any vector in $V$ can be written as the linear combination of $\mathbf{v}_{1},\mathbf{v}_{2},\cdots,\mathbf{v}_{n}$. There is no mentioning that scalars must be non-zero.
So my question is: am I misunderstanding the definition of the span, or for the proposition to be correct, author should say "form a basis for $V$" instead of "span $V$"?
As $n$ vectors span the full space, $V$ is finite dimensional and you can select a basis among the vectors $\newcommand\vu{\mathbf{u}}(\vu_1,\vu_2,...,\vu_n)$. Wlog. the basis are the first $m$ vectors, then $(\vu_1,\vu_2,...,\vu_m)$ also spans $V$.
Of course, if the dimension of $V$ is $n$, then the original system is already a basis, a minimal spanning system at the same time as a maximal linearly independent system.