Prove or disprove: If $v_1,v_2,...,v_n$ are vectors in $\mathbb{R}^n$...

Question

and for every $v_i$ , $i=1,..,n$ the sum of coordinates of $v_i$ is $0$, then $v_1,...,v_n$ and linearly dependent.

What I tried to do: This questions was given after I learnt about matrix similarity and eigen vectors and eigen values, so I tried to think in that direction, I couldn't really get the question saying "the sum of coordinates of $v_i$ is $0$), because I always saw coordinates come with basis so I can use them? (would be appreciated if someone explains in what way should I understand that sentence).
But I proceed to think in that direction and reached something: I chose matrix $A$ to be all ones, ($a_{ij}=1\space \forall i,j) $, and let $T(v)=Av$ a linear transformation, and here I got stuck but my idea was to try to say by contradiction that they're linearly independent and reach contradiction if I find that $A$, is similiar to the zero matrix (if I use the sum of coordinates), and then explain that they can't be similar with different ranks, and prove the statement. But I have no idea how to continue and my weak understanding of the question scares me if I'm in the right direction so I came here to ask.
Appreciate all the help, Thanks in advance to everyone.

Answer 1

When talking about the Euclidean space $\mathbb{R}^n$, the default basis is the standard basis.

Let $A$ be a matrix where the column vectors are $v_1,\cdots,v_n$. Then if you consider the row vector $$ w=(\underbrace{1,\cdots,1}_{\textrm{$n$ terms}}) $$ it follows from the matrix multiplication that $wA=0$, which implies that $A$ is not of full rank and hence the column vectors are linearly dependent.

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Notes.

If you write $$ v_k=(a_{1k},\cdots,a_{nk})^T $$ then the matrix $A$ is

\begin{pmatrix} a_{11}&\cdots& a_{1n}\\ \vdots&\vdots&\vdots\\ a_{n1}&\cdots &a_{nn} \end{pmatrix}

Note that the $k$-th column of $A$ is the vector $v_k$.

Answer 2

There's a far simpler proof. Note that if the vectors were linearly independent, they would span $\mathbb R^n$.

I claim that $\bar 1$ is not in the span of these vectors.

To see why, note that $\bar1\cdot\bar1=n$, but $$\bar1\cdot (a_1\cdot v_1+\ldots +a_n\cdot v_n)=a_1\cdot(\bar1\cdot v_1)+\ldots+a_n\cdot(\bar1\cdot v_n)=0+\ldots+0=0$$ for any choice of coefficients $a_1,\ldots,a_n$.

Answer 3

HINT-A system of $n$ vectors $v_1,v_2,.....,v_n$ is linearly dependent if $\sum_{i=1}^nc_iv_i=0$ holds for some non zero vector $(c_1,c_2,.....,c_n)$.

In your case, indeed this non zero vector is $(1,1,.....,1)$.