Linearly independent vectors each subtracted by a linear combination of them are linearly dependent if coefficients add up to $1$ The task
I've been given the following problem:

Let $v_1, \ldots, v_n$ linearly independent vectors in an $\mathbb{F}$-vector space $V$ and $u = \lambda_1 v_1 + \ldots + \lambda_n v_n$ a linear combination of those vectors.
Prove that
$$v_1 - u, \ldots, v_n - u \text{ are linearly dependent} \Leftrightarrow \lambda_1 + \ldots + \lambda_n = 1 $$

Unfortunately, I have some trouble with this question. What steps would need to be taken in order to prove this? One hint might be the fact that if a set of vectors is linearly independent, it means any linear combination of them has only one clear set of coefficients (and the other way around).
EDIT: After angryavian's answer, I've understood how to prove the first direction. Unfortunately, the other direction still leaves me quite puzzled. A hint would be greatly appreciated.
 A: Mistake in counterexample: $\mu_1$ can be real number, so they are linearly dependent.

Proving the original problem:
$\implies$ direction: If $v_1 - u, \ldots, v_n - u$ are linearly dependent,
there exist $c_1, \ldots, c_d$ not all zero such that
$$0 = c_1 (v_1 - u) + \cdots + c_n (v_n - u)
= (c_1 - (c_1 + \cdots + c_n)\lambda_1) v_1 + \cdots + (c_n - (c_1 + \cdots + c_n)\lambda_n) v_n.$$
By linear independence of $v_1,\ldots, v_n$,
$$c_i - (c_1 + \cdots + c_n) \lambda_ i =0, \text{ for all } i = 1, \ldots, n.$$
Summing over all $i$ yields $$c_1 + \cdots + c_n = (c_1 + \cdots + c_n)(\lambda_1 + \cdots + \lambda_n).$$
If $c_1 + \cdots + c_n \ne 0$, then we are finished.
The other situation $c_1 + \cdots + c_n = 0$ cannot happen, since then $c_1 (v_1 - u) + \cdots + c_n(v_n - u) = c_1 v_1 + \cdots + c_n v_n \ne 0$, by linear independence of $v_1, \ldots, v_n$, which contradicts our premise.
$\impliedby$ direction: 
It is your job to find $c_1, \ldots, c_n$ not all zero such that
$$0 = c_1 (v_1 - u) + \cdots + c_n (v_n - u)
= (c_1 - (c_1 + \cdots + c_n)\lambda_1) v_1 + \cdots + (c_n - (c_1 + \cdots + c_n)\lambda_n) v_n.$$
There is a relatively straightforward choice of $c_1,\ldots, c_n$ (in terms of the $\lambda_1, \ldots, \lambda_n$) that will make this work.

Overkill matrix approach:
WLOG assume $v_1, \ldots, v_n$ span $V$ (else work in the subspace spanned by $v_1, \ldots, v_n$). We consider matrices and vectors using the basis $v_1, \ldots, v_n$.
The columns of the matrix $I - \lambda \mathbf{1}^\top$ are $v_1 - u, \ldots, v_n - u$. Here, $\lambda \mathbf{1}^\top$ is the $n \times n$ matrix consisting of the column $(\lambda_1, \ldots, \lambda_n)$ repeated $n$ times.
It suffices to show that $0$ is an eigenvalue of $I - \lambda \mathbf{1}^\top$ if and only if $\lambda_1 + \cdots + \lambda_n = 1$.
One can show that the eigenvalues of $I - \lambda \mathbf{1}^\top$ are $1$ and $1 - (\lambda_1 + \cdots + \lambda_n)$.
A: If you consider the matrix having as columns the coordinates of the new vectors with respect to the given linearly independent set (everything takes place in the subspace of which $\{v_1,\dots,v_n\}$ is a base):
\begin{bmatrix}
1-\lambda_1 & -\lambda_1 & \dots & -\lambda_1 \\
-\lambda_2 & 1-\lambda_2 & \dots & -\lambda_2 \\
\vdots & \vdots & \ddots & \vdots \\
-\lambda_n & -\lambda_n & \dots & 1-\lambda_n
\end{bmatrix}
you see that this matrix is
$$
I-
\begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n \end{bmatrix}
\begin{bmatrix} 1 & 1 & \dots & 1 \end{bmatrix}
$$
which is not invertible (meaning that $\{v_1-u,\dots,v_n-u\}$ is linearly dependent) if and only if $1$ is an eigenvalue of
$$
A=
\begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n \end{bmatrix}
\begin{bmatrix} 1 & 1 & \dots & 1 \end{bmatrix}
$$
This is a matrix with rank at most one, so it has at most one nonzero eigenvalue, namely
$$
\begin{bmatrix} 1 & 1 & \dots & 1 \end{bmatrix}
\begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n \end{bmatrix}
=\lambda_1+\lambda_2+\dots+\lambda_n
$$
