Show that the system $v_1 - u, \dots, v_n - u$ is linearly dependent precisely when $\lambda_1 + \dots + \lambda_n=1$

Hint 1 If $\lambda_1+..+\lambda_n=1$ then $$u = \lambda_1(u-v_1) + \dots + \lambda_n (u-v_n)=(\lambda_1+...+\lambda_n)u- \left( \lambda_1v_1 + \dots + \lambda_n v_n\right)$$

Now use the definition of $u$ and the relation on $\lambda's$

Hint 2 Assume that $$\beta_1(u-v_1) + \dots + \beta_n (u-v_n)=0$$ with not all coefficients 0.

Then $$(\beta_1+...+\beta_n)u-\beta_1 v_1-...-\beta_n v_n=0 \\ (\beta_1+...+\beta_n)(\lambda_1v_1 + \dots + \lambda_n v_n)-\beta_1 v_1-...-\beta_n v_n=0 \\ \left((\beta_1+...+\beta_n)\lambda_1-\beta_1\right)v_1 + \dots +\left((\beta_1+...+\beta_n)\lambda_n-\beta_n\right)v_n=0 \\ $$

Using Linear independence you get $$(\beta_1+...+\beta_n)\lambda_1=\beta_1 \\ (\beta_1+...+\beta_n)\lambda_2=\beta_2 \\ .... \\ (\beta_1+...+\beta_n)\lambda_n=\beta_n $$

Now add the relations together, and prove that $\beta_1+...+\beta_n \neq 0$.

We may assume without loss of generality that $v_1,\dots,v_n$ are the standard basis vectors of $\mathbb K^n$. (Why?)

Hence $u=(\lambda_1,\dots,\lambda_n)^t \in \Bbb K^n$ and to determine linear dependence of the vectors $v_1-u,\dots,v_n-u$ it is enough to calculate the determinant of $$ \begin{pmatrix} \lambda_1 -1 & \lambda_1 & \lambda_1 & \cdots & \lambda_1 \\ \lambda_2 & \lambda_2-1 & \lambda_2 & \cdots & \lambda_2 \\ \lambda_3 & \lambda_3 & \lambda_3-1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \lambda_{n-1} \\ \lambda_n & \lambda_n & \cdots & \lambda_{n-1} & \lambda_n-1 \end{pmatrix}. $$ If we subtract the second column from the first, the third from the second, etc. we get $$ \begin{pmatrix} -1 & 0 & 0 & \cdots & \lambda_1 \\ 1 & -1 & 0 & \cdots & \lambda_2 \\ 0 & 1 & -1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \lambda_{n-1} \\ 0 & 0 & \cdots & 1 & \lambda_n-1 \end{pmatrix}. $$ Now add the first row to the second, then the second to the third, … to obtain $$ \begin{pmatrix} -1 & 0 & 0 & \cdots & \lambda_1 \\ 0 & -1 & 0 & \cdots & \lambda_1+\lambda_2 \\ 0 & 0 & -1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \lambda_1+\cdots+\lambda_{n-1} \\ 0 & 0 & \cdots & 0 & \lambda_1+\cdots+\lambda_n-1 \end{pmatrix}. $$ This is an upper triangular matrix and the determinant is (up to a sign) equal to $\lambda_1+\cdots+\lambda_n-1$ which is zero precisely when $$ \lambda_1+\cdots+\lambda_n = 1. $$