Dimension of vector space of matrices with zero row and column sum.

Let $K$ be a field. For $A \in \mathrm{M}(m \times n, K)$ let $$ R_i(A) = \sum_{j=1}^n A_{ij} \quad \text{for every $1 \leq i \leq m$} $$ and $$ C_j(A) = \sum_{i=1}^m A_{ij} \quad \text{for every $1 \leq j \leq n$}, $$ and set $$ V_{m,n}(K) = \{A \in \mathrm{M}(m \times n, K) \mid R_1(A) = \dotsb R_m(A) = C_1(A) = \dotsb = C_n(A)\}. $$

We show that the map \begin{align*} \Phi \colon V_{m,n}(K) &\to \mathrm{M}((m-1) \times (n-1), K), \\ \quad (a_{ij})_{1 \leq i \leq n, 1 \leq j \leq m} &\mapsto (a_{ij})_{1 \leq i \leq n-1, 1 \leq j \leq m-1} \end{align*} is an isomorphism; it is clearly linear.

First surjectivity: Let $A = (a_{ij})_{1 \leq i \leq n-1, 1 \leq j \leq m-1} \in \mathrm{M}((m-1) \times (n-1), K)$. For all $1 \leq i \leq m-1$ let $a_{in} = -R_i(A)$ and for all $1 \leq j \leq n-1$ let $a_{mj} = -C_j(A)$. Also let $$ a_{mn} = \sum_{\substack{1 \leq i \leq m-1 \\ 1 \leq j \leq n-1}} a_{ij}. $$ For $\hat{A} = (a_{ij})_{1 \leq i \leq n, 1 \leq j \leq m} \in \mathrm{M}(m \times n, K)$ we have that $$ R_i(\hat{A}) = \sum_{j=1}^n a_{ij} = R_i(A) + a_{in} = 0 \quad \text{for every $1 \leq i \leq m-1$} $$ as well as \begin{align*} R_m(\hat{A}) &= \sum_{j=1}^n a_{mj} = \sum_{j=1}^{n-1} a_{mj} + a_{mn} \\ &= -\sum_{j=1}^{n-1} C_j(A) + a_{mn} = -\sum_{j=1}^{n-1} \sum_{i=1}^{m-1} a_{ij} + \sum_{\substack{1 \leq i \leq m-1 \\ 1 \leq j \leq n-1}} a_{ij} = 0. \end{align*} So all row sums of $\hat{A}$ are zero. Simililarly we find that all column sums of $\hat{A}$ are zero. So $\hat{A} \in V_{m,n}(K)$. Because $\Phi(\hat{A}) = A$ this shows the surjectivity of $\Phi$.

For the injecitvity we argue the other way around: For every $A \in V_{m,n}(K)$ we have $A_{in} = -R_i(\Phi(A))$ for every $1 \leq i \leq m-1$ and $A_{mj} = -C_j(\Phi(A))$ for every $1 \leq j \leq n-1$, as well as $$ A_{mn} = -\sum_{j=1}^{n-1} A_{mj} = \sum_{j=1}^{n-1} C_j(\Phi(A)), $$ So $A$ is uniquely determined by $\Phi(A)$, showing that $\Phi$ in injective.

A variant on Jendrik Stelzner's excellent answer is to note that if a matrix $a$ has row and column sum zero, then for each $i$ $$ a_{i,n} = - \sum_{s=1}^{n-1} a_{i,s}, $$ for each $j$ $$ a_{m,j} = - \sum_{t=1}^{m-1} a_{t,j}, $$ and finally $$ a_{m,n} = \sum_{s=1}^{n-1} \sum_{t=1}^{m-1} a_{t, s}. $$ Conversely, these three conditions imply that $a$ has row and column sum zero.

Now it is immediate that a basis of the space of this matrices is given by the $(m-1)(n-1)$ matrices $$ e_{i,j} - e_{i,n} - e_{m, j} + e_{m, n}, $$ for $0 \le i < m$, $0 \le j < n$, where $e_{s, t}$ is the usual matrix which has zero everywhere except for a $1$ in position $s, t$. The above show that they span the space, and it suffices to look at the first $n-1$ and $m-1$ rows to see that they are independent.

For instance, when $m = n = 3$ you get $$ \begin{bmatrix} 1 & 0 & -1\\ 0 & 0 & 0\\ -1 & 0 & 1\\ \end{bmatrix}, \quad \begin{bmatrix} 0 & 0 & 0\\ 1 & 0 & -1\\ -1 & 0 & 1\\ \end{bmatrix}, \quad \begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & -1\\ 0 & -1& 1\\ \end{bmatrix}, \quad \begin{bmatrix} 0 & 1 & -1\\ 0 & 0 & 0\\ 0 & -1& 1\\ \end{bmatrix}. $$ Note that if you look at the first two rows and columns, you get the usual base of the space of $2 \times 2$ matrices.