The null space of the matrix
$$\begin{bmatrix}
1 & 1 & -1 & -1 & 0 & 0 \\
1 & 1 & 0 & 0 & -1 & -1 \\
0 & 0 & 1 & 1 & -1 & -1 \\
\end{bmatrix}$$
is, by definition, the set of vectors $\mathbf{x}=(x_1,x_2,x_3,x_4,x_5,x_6)$ such that
\begin{align*}
x_1 + x_2 - x_3 - x_4 &= 0 \\
x_1 + x_2 - x_5 - x_6 &= 0 \\
x_3 + x_4 - x_5 - x_6 &= 0
\end{align*}
So, this is the vector space $V$ in question. The matrix has rank $2$, so the dimension of $V$ is $4$ by the Rank Nullity Theorem.
Thus, any $4$ linearly independent vectors in $V$ form a basis. A natural try is $$\{( 1, 1, 1, 1, 1, 1 ),(0, 0, 0, 0, 1, -1),( 0, 0, 1, -1, 0, 0 ),(1, -1, 0, 0, 0, 0)\}.$$ To prove that it is indeed a basis, we check that the matrix $$\begin{bmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 & -1 \\
0 & 0 & 1 & -1 & 0 & 0 \\
1 & -1 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$$ has rank $4$ (e.g. by performing Gaussian elimination).
The column space of the transpose of the above matrix, i.e.
$$\begin{bmatrix}
1 & 0 & 0 & 1 \\
1 & 0 & 0 & -1 \\
1 & 0 & 1 & 0 \\
1 & 0 & -1 & 0 \\
1 & 1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
\end{bmatrix},$$
will be $V$ (or, at least, will be isomorphic to $V$: all the vectors will be transposed); see Wikipedia.
Rouché-Capelli theorem grants us that if $n$ is the number of equations that characterize the subspace $W$, we have
$$\dim(W)=\dim(\mathbb{R^5})-n=4$$
Let v be the vector
$$ v=\ (x_1,x_2,x_3,x_4,x_5)^T $$
If $v \in W$, it must satisfy $x_1-x_3-x_4=0 \implies x_1=x_3+x_4$, so a vector in $W$ must be of the kind
$$(x_3+x_4,x_2,x_3,x_4,x_5)$$
Can you find $4$ linearly independent vectors that satisfy the above constraint?
Best Answer
(Edit: I'm not sure if I read the OP correctly, but to my understanding, except the sums of entries along the main diagonal or main anti-diagonal, all line sums along other diagonals are not part of the definition.)
Denote by $r_i$ the $i$-th row sum, $c_j$ the $j$-th column sum, $d$ the diagonal sum and $a$ the anti-diagonal sum. Every so-called "magic square" in question must satisfy the following 9 constraints: $$r_1=d,\ r_2=d,\ r_3=d,\ r_4=d,\ c_1=d,\ c_2=d,\ c_3=d,\ c_4=d,\ a=d.$$ (I say "so-called" in the above because the definition here deviates from the conventional one --- here, a magic square can have non-integer or even negative entries.)
Clearly the constraint $c_4=d$ is redundant, because the sum of all row sums must be equal to the sum of all column sums. Since a $4\times4$ matrix is specified by 16 entries, you now have 16 unknowns and at most 8 independent constraints. This suggests that the dimension of the vector space in question is at least 16-8=8.
So, to prove that the dimension of the vector space in question is exactly 8, you only need to show that the remaining 8 constraints are indeed linearly independent. This amounts to proving that some $8\times16$ matrix has full row rank. It takes some work but it shouldn't be hard.
Having proven that the dimension is 8, it is not hard to find a basis. All you need is to construct a magic square with nonzero row/column/diagonal/anti-diagonal sums and 7 magic squares with zero row/column/diagonal/anti-diagonal sums. This is easy: \begin{align*} &\pmatrix{1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1}, \ \pmatrix{1&0&0&-1\\ 0&-1&1&0\\ 0&1&-1&0\\ -1&0&0&1},\\ &\pmatrix{1&-1&0&0\\ -1&1&0&0\\ 0&0&-1&1\\ 0&0&1&-1}, \ \pmatrix{1&0&-1&0\\ 0&-1&0&1\\ -1&0&1&0\\ 0&1&0&-1},\\ &\pmatrix{0&0&-1&1\\ 0&0&1&-1\\ 1&-1&0&0\\ -1&1&0&0}, \ \pmatrix{0&-1&0&1\\ 1&0&-1&0\\ 0&1&0&-1\\ -1&0&1&0},\\ &\pmatrix{0&1&-1&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&-1&1&0}, \ \pmatrix{0&0&0&0\\ 1&0&0&-1\\ -1&0&0&1\\ 0&0&0&0}. \end{align*}
By looking at the diagonals and anti-diagonals of their linear combinations, it should be rather obvious that these 8 magic squares are indeed linearly independent.