methods of constructing a matrix from its null space span I have a matrix of size $4\times3$ and its null-space span is $\{(1,2,3), (2,5,7)\}$.
How can I find the original matrix? It is not obvious from the span which vectors are free.
 A: you can construct the rows of the matrix $A$ whose null space is panned by $\{(1,2,3)^\top, (2,5, 7)^\top\}$ by finding rows orthogonal to these basis vectors. that is finding the null space of $$\pmatrix{1&2&3\\2&5&7} \to \pmatrix{1&0&1\\0&1&1} $$ you find that null space of the latter is $$(1, 1, -1)^\top. $$
therefore, one $4 \times 3$ matrix is $$A = \pmatrix{1&1&-1\\0&0&0\\0&0&0\\0&0&0\\} $$ so are any matrix of the form $BA$ where $B$ is any $4 \times 4$ invertible matrix.
A: The nullspace of a matrix is the orthogonal complement of its rowspace.  So you just need a set of vectors that are orthogonal to $(1,2,3)$ and $(2,5,7)$.  Those are two linearly independent vectors in $\Bbb R^3$, so the orthogonal complement of them will just be a line.  I.e. you just need to find $1$ vector orthogonal to both of them.  So why not just use the cross product to do that?
$$(1,2,3) \times (2,5,7) = (14-15, 6-7, 5-4) = (-1,-1,1)$$
So just fill up the rows of a $4 \times 3$ matrix with scalar multiples of this vector.  One such example is 
$$\pmatrix{1 & 1 & -1 \\ -\pi & -\pi & \pi \\ 0 & 0 & 0 \\ 2 & 2 & -2}$$
