I've been working on this for almost half hour, can someone answer this question perhaps? Thanks.
Let {$\vec u_1, \vec u_2, \ldots , \vec u_k$} be a linearly independent set of vectors in $\mathbb{R}^n$, and let $\vec v$ be a vector in $\mathbb{R}^n$ such that
$$ \vec v = c_1 \vec u_1 + \cdots + c_k \vec u_k$$
for some scalars $c_1, c_2, \ldots , c_k$, with $c_1 \ne 0$. Prove that {$\vec v, \vec u_2, \ldots , \vec u_k$} is linearly independent.
suppose the converse:
if they are linearly dependent,we have scalars $a_1,\ldots,a_{(k-1)}$ that $v=a_1u_2+\cdots+a_{(k-1)}u_k=c_1u_1+\cdots+c_ku_k$ so ....
u can complete the proof!try a little!
\begin{align} & b_1 \vec v + b_2 \vec u_2 + \cdots + b_n \vec u_k \\[8pt] = {} & b_1(c_1 \vec u_1 + \cdots + c_k \vec u_k) + b_2 \vec u_2 + \cdots + b_n \vec u_k \\[8pt] = {} & (b_1 c_1) \vec u_1 + (b_1 c_2 + b_2 ) \vec u_2 + \cdots + (b_1 c_k + b_k) \vec u_k \tag a \end{align} If this adds up to $\vec 0$ then, since $\vec u_1, \ldots, \vec u_k$ are linearly independent, all of the coefficients in line $(\text{a})$ must be $0$. That implies $b_1 c_1=0$. But it is given that $c_1\ne0$; therefore $b_1=0$. Thus the rest of the coefficients in line $(\text{a})$ reduce to $b_2, \ldots, b_k$, and they must all be $0$.
hint $(\vec{u}_k)$ linearly independent means $\sum_{i=1}^k a_i\vec{u}_i = 0$ has a unique solution $a_i \equiv 0$.
Assume $v, \vec{u}_2, \ldots$ is not linearly independent. That means there is a non-zero solution to $a_1v + a_2 \vec{u}_2 + \ldots + a_k \vec{u}_k = \vec{0}$. Substitute the definition of $\vec{v}$ and use the linear independence of the $u$'s to get a contradiction.
