Dimension of a subspace

Matrix form.
$$F = \ker\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \end{pmatrix}.$$

Rank. The two rows are not proportional, so the rank is $2$.

Rank-nullity. $\dim F = 4 - 2 = 2$.

Explicit basis. Solve the system:
$$x+y+z+t=0, \qquad x-y+z-t=0.$$
Adding: $2x+2z=0$, so $z=-x$. Subtracting: $2y+2t=0$, so $t=-y$. The solutions are $(x,y,-x,-y)=x(1,0,-1,0)+y(0,1,0,-1)$. A basis is $((1,0,-1,0),(0,1,0,-1))$.

Conclusion. $\dim F=2$.