Dimension of this subspace of $\mathbb R^4$ generated by $\{v_{i1},v_{i2},v_{i3},v_{i4}\}$ with $\sum_{i=1}^4 v_{ij}=0$ for $i,j\in\{1,2,3,4,\}$ Let $$v_i=(v_{i1},v_{i2},v_{i3},v_{i4}),\ \ for\ \ i=1,2,3,4$$ be four vectors in$\mathbb R^4$ such that $$\sum_{i=1}^4v_{ij}=0\ \ for\ \ each\ \ j=1,2,3,4.$$  $W$ be the subspace  generated by $\{v_1,v_2,v_3,v_4\}.$ Then the dimension $d$  of $W$ over  $\mathbb R$  is 
$A.\ d=1\ \ or\ \ d=4$
$B.\ d\le 3$
$C.\ d\ge 2$
$D.\ d=0\ \ or\ \ d=4$
Now writing them vectors as follows :
$$v_1=\ \ v_{11}\ \ \ v_{12}\ \ \ v_{13}\ \ \ v_{14}\\v_2=\ \ v_{21}\ \ \ v_{22}\ \ \ v_{23}\ \ \ v_{24}\\v_3=\ \ v_{31}\ \ \ v_{32}\ \ \ v_{33}\ \ \ v_{34}\\v_4=\ \ v_{41}\ \ \ v_{42}\ \ \ v_{43}\ \ \ v_{44}$$
$\sum_{i=1}^4 v_{ij}=0$  gives $$v_{i1}+v_{i2}+v_{i3}+v_{i4}=0\ \ for\ \ i=1,2,3,4.$$ i.e. *sum of $i$-th co-ordinate of all the  $4$  vectors is $0$. So,if  we know  any $3$  of  them  then the co-ordinates of the  $4$-th can be found and so dimension can never be $4.$  So , strike off options $A$  and  $D.$
Now , we are left with  the  options  $B$  and $C.$  Now , my question  is  how  do  I know  if  the  dimension can be reduced to $2$  or  not i.e. whether $B$ or $C$  is  correct $?$
Thanks.

Take the four vectors $(1,0,0,0), (0,1,0,0), (0,0,1,0)$ and $(-1,-1,-1,0) .$ These four satisfy the given conditions and the first three are linearly independent . Hence the answer will be $B.$
