Linear dependency - Alternative way to prove this statement

Question

Just wondered if there is an alternative way to prove the following statement, and if my proof is correct.

Statement:

If $v_1,...,v_n$ are vectors in $R^n$, and for every $v_i$ , $(i=1,...,n)$, the sum of the coordinates in $v_i$ is $0$, then $v_1,....v_n$ are linearly dependent.

Proof (sorry if not written very formally):

Define a matrix $A$ with the vectors in it's rows:

$ A=\begin{pmatrix}-v_1-\\...\\-v_n- \end{pmatrix}_{n*n} $

The sum of every row in $A$ is $0$, then $0$ is an eigenvalue of A with eigenvector:

$u=\begin{pmatrix}1\\...\\1 \end{pmatrix}_{n*1}$

Hence, $A$ is not invertible, and therefore it's rows are linearly dependent,

and so the vectors $v_1,...,v_n$ are linearly dependent.

Answer 1

Your proof is correct.

I think that it is simpler to say that all those vectors belong to the space$$\left\{(x_1,\ldots,x_n)\in\mathbb{R}^n\,\middle|\,\sum_{k=1}^nx_k=0\right\},$$whose dimension is $n-1$. And $n$ vectors in a $n-1$-dimensional space are always linearly dependent.