The vectors $v_{1},\dots,v_{n}$ form a linearly independent set. Addition of vector $u$ makes this set linearly dependent. How to prove that $u \in span\{v_{1}\ldots,v_{n}\}$.
I was able to prove the reverse.
The vectors $v_{1},\dots,v_{n}$ form a linearly independent set. Addition of vector $u$ makes this set linearly dependent. How to prove that $u \in span\{v_{1}\ldots,v_{n}\}$.
I was able to prove the reverse.
There are $a_0, \dots, a_n$ not all zero, such that $a_0u + a_1v_1 + \dots a_nv_n = 0.$ If $a_0 = 0$, the set will become independent, which is contradictory. Clearly $a_0 \neq 0$, as $v_1, \dots, v_n$ are linearly independent, and therefore we get
$u = -\frac{a_1}{a_0}u_1 - \dots - \frac{a_n}{a_0}u_n $
Ok since Alexander has already given you the answer, let's look at how just rearranging an equation can tell you so many things:
Problem 1: If $v_1, \cdots v_4$ are linearly independent vectors in $\mathbb{R}^4$, show that $\{v1, v2, v_3\}$ is a linearly independent set of vectors.
Use the contrapositive and solve the problem!
If $v_1, v_2, v_3$ are linearly dependent, then $\exists c_1, c_2, c_3$ such that $\sum c_i v_i$ = $0$. So if $c_4 = 0$ and $c_1, c_2, c_3 \neq 0$, then $c_1 v_1+ c_2 v_2 + c_3 v_3 = 0 \times v_4 = 0$ so $v_1, \cdots v_4$ are linearly dependent and we are done!