A question related to Dimensions & Spanning of square matrix. Let $V$ be a span of square matrix of order $n$. $W$ be a subspace of matrices with entries in each row adding upto zero. Then $dimW =$ ?.
The option given are:


*a. $n$


*b. $\frac{n(n-1)}{2}$


*c. $n-1$


*d. $n (n-1)$

I guess the answer is $n$. Because dimension of space of which $W$ is subspace is 3. So subspace $W$ must have same dimension inspite of any information. Am I right? I am un under graduate student, not a mathematician. Any help is appreciated.
 A: if every row in $W$ sums to zero, then $W v = 0$ where v is a column vector with entries of $[1,1,...,1]$
The kernel of W is non-trivial.
The dimension of $W$ is less then $n.$
It could be considerably less then $n,$ but how much less cannot be determined by the information given.  We only know for sure of one vector in the kernel.
$dim(W)\le n-1$  
A: I have a feeling you really mean that $V$ is the space of square matrices of order $n$. In that case the dimension of $W$ is $n(n-1)$.
Edit: In any row, the first $(n-1)$ elements you can choose arbitrarily, the last one gets determined by them. You do it for each row, hence dimension becomes $n(n-1)$. Explicitly the basis can be given by the matrices $E^{pq}$, for $1 \le p \le n, 1 \le q \le n-1$, where $E^{pq}_{ij} = 1$ if $(i,j) = (p,q)$ and $E^{pq}_{ij} = -1$ if $(i,j) = (p, n)$ and rest are zero.
A: The previous answer is unsatisfactory. In the question, I do not understand the role of $V$. As correctly mentioned in the previous answer, if the matrix w is in W then $w(v)=0$ for $v=(1...1).$ The question says that $W$ is A subspace and not THE subspace $W_1=\{w; w(v)=0\}.$ Therefore  $\dim W\leq \dim W_1=n(n-1).$  If you say THE subspace, the correct answer is d). 
