homework - Find a basis for the space of all vectors in R6 with x1 + x2 = x3+ x4 = x5+ x6 a) Find a basis for the space of all vectors in $\mathbb{R}^6 $ with $x_1 + x_2 = x_3 + x_4 = x_5 + x_6$.
b) Find a matrix with that subspace as null space.
c) Find a matrix with that subspace as column space.
I'm having a really hard time for parts a and b. I've found a set of vectors of dimension 4 that suit $\mathbb{R}^6$ by treating $x_1 \ldots x_2$ as the nullspace of a vector and $x_1 + x_2 = x_3 + x_4 \ldots$ but I'm having trouble showing why it works. I'm also having trouble with part b in that I don't know how to find a matrix (of rank 2) with that subspace as nullspace.
I think for part c all I need to do is treat the basis vectors I found in part a as column vectors and form a matrix out of that, but again I'm not sure how to explain it or if it's even correct.
 A: 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.
