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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.

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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.

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