We write $0$ or $1$ in every cell of a $m \times n$ table. Show that exists a "cross" such that the sum of integers written in the cross is odd. 
There is a $m \times n$ table and you write $0$ or $1$ in every cell of the table. Show that there exists a "cross"(:= union of a row and a column) such that the sum of integers written in the cross is odd. There is at least one cell that has $1$ written on it. Integers $n$ and $m$ are even.

We consider working in the field $\mathbb{Z}_2$. We can imagine this table as $m\times n$ matrix $A =[a_{i,j}]\neq 0$, where $a_{i,j}\in\{0,1\}$. Now let $s_1,...s_m$ be a columns and $v_1,...v_n$ rows of this matrix and make new matrix $M$ with $m+n+1$ columns and $mn$ rows: $$M =\pmatrix{v_1,\;\;s_1^{\top},\;\;a_{1,1}\\v_1,\;\;s_2^{\top},\;\;a_{2,1}\\ \vdots \\v_n,\;\;s_m^{\top},\;\;a_{m,n}\\}$$
Now we need to show that if $M$ acts on vector $\vec{1}^{\top}$ where $\vec{1} = \underbrace{(1,1,\ldots 1)}_{m+n+1} $ we don't get null vector $\vec{0}$. Clearly as $A\ne 0$ we have $M\ne 0$ and if we go with a contradiction we see $\ker (M)$ is not trivial. Also we see, by rank nullity theorem, that rank $(M)$ and def $(M)$ are of oposite parity since $m+n+1$ is odd.
Any idea how to end?
 A: Here's a proof using a slightly different approach. You could use this to construct a $\ 1\times mn\ $ row vector $\ v\ $, with entries in $\ \mathbb{Z}_2\ $, such that
$$
vM\vec{1}^T\ne0\ ,
$$
so your approach can certainly be made to work (although this would be an unnecessarily round about way of doing it).
Let
\begin{align}
T&=\sum_{i=1}^m\sum_{j=1}^n a_{ij}\ ,\\
a_{xy}&=1\ \text{, and}\\
c_{st}&=\sum_{j=1}^na_{sj}+\sum_{i=1}^ma_{it}-a_{st}\ .\\
\end{align}
Then $\ c_{st}\ $ is the sum of the entries in the cross formed by the union of the $\ s^\text{th}\ $ row and $\ t^\text{th}\ $ column of $\ A\ $. Now we have
$$
c_{xy}=\sum_{j=1}^na_{xj}+\sum_{i=1}^ma_{iy}-1\ ,
$$
and
\begin{align}
\sum_{s=1}^mc_{sy}+\sum_{t=1}^nc_{xt}-c_{xy}&=\sum_{s=1}^m\Big(\sum_{j=1}^n a_{sj}+\sum_{i=1}^ma_{iy}-a_{sy}\Big)\\
&\hspace {1em}+\sum_{t=1}^n\Big(\sum_{j=1}^n a_{xj}+\sum_{i=1}^ma_{it}-a_{xt}\Big)-c_{xy}\\
&=2T+(m-1)\sum_{i=1}^ma_{iy}\\
&\hspace{1em}+(n-1)\sum_{j=1}^na_{xj}-c_{xy}\\
&=2T+m\sum_{i=1}^ma_{iy}+n\sum_{j=1}^na_{xj}\\
&\hspace{1em}-\sum_{i=1}^ma_{iy}-\sum_{j=1}^na_{xj}-c_{xy}\\
&=2T+m\sum_{i=1}^ma_{iy}+n\sum_{j=1}^na_{xj}-2c_{xy}+1\ .
\end{align}
The final expression on the right of these identities is odd, and the first expression on their left is the sum of the $\ c_{st}\ $ over
$$(s,t)\in\big\{(1,y),(2,y),\dots,(m,y)\big\}\cup\big\{(x,1),(x, 2),\dots,(x,n)\big\}\ .
$$
It follows that $\ c_{st}\ $ must be odd for at least one $\ (s,t)\ $ in that set.
A: Let $A=[a_{ij}]$ be the original table. This is an element of $\mathbb{Z}_2^{m\times n}$.
Let $B=[b_{ij}]$ be a table containing cross sums of $A$ modulo $2$, i.e.
$$b_{ij}=\sum_{r=i\ \vee\ s=j} a_{rs} \pmod 2$$
This too is an element of $\mathbb{Z}_2^{m\times n}$.
This defines a linear mapping $f:\mathbb{Z}_2^{m\times n} \rightarrow \mathbb{Z}_2^{m\times n} $ that takes each $A$ to its corresponding $B$.
Let's take a closer look at the image of $f$. It is straightforward to check that if $A$ contains $1$s on row $i$ and in column $j$, and $0$s everywhere else, then $B=f(A)$ has only a single $1$, namely $b_{ij}=1$, and $0$ elsewhere. So the image of $f$ contains every possible table with just a single $1$. By linearity, the image therefore contains every possible table, i.e. $f$ is onto.
Since the domain and image of $f$ have the same finite cardinality and $f$ is onto, $f$ must also be one-to-one. Obviously the zero table maps to the zero table, and therefore all non-zero tables will map to non-zero tables.
