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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.

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    $\begingroup$ Perhaps this is what you yourself are noting in the last sentence, but the null space does not fully determine the matrix. $\endgroup$ – hardmath May 26 '15 at 15:44
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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

$$\pmatrix{1 & 1 & -1 \\ -\pi & -\pi & \pi \\ 0 & 0 & 0 \\ 2 & 2 & -2}$$

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  • $\begingroup$ Do you considered that {(1,2,3),(2,5,7)} are columns? Ok, if we considered your solution, when we verify it by finding out the null-space span of the matrix, the result won't be similar to the spans mentioned above, it gives {x1(-1,1,0),x2(1,0,1)} where x1, x2 are scalars. $\endgroup$ – Hozifa Mohammed May 26 '15 at 16:22
  • $\begingroup$ @HozifaMohammed I don't need to know whether they are column or row vectors to take a cross product. And as for that nullspace you found: Consider $(1,2,3) = 2(-1,1,0)+3(1,0,1)$ and $(2,5,7)=5(-1,1,0)+7(1,0,1)$. Does that help you see that $\operatorname{span}[(1,2,3),(2,5,7)]=\operatorname{span}[(-1,1,0),(1,0,1)]$? $\endgroup$ – user137731 May 26 '15 at 16:35
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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.

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  • $\begingroup$ {(1,2,3),(2,5,7)} are columns, not rows, and in that case, if we follows your method, we will gain X3 as a free variable, however the null-space spans in our example are two as we have two vectors {(1,2,3),(2,5,7)} $\endgroup$ – Hozifa Mohammed May 26 '15 at 16:04
  • $\begingroup$ @HozifaMohammed, am i missing something? the matrix $A$ in my answer has the columns $\pmatrix{1\\2\\3}, \pmatrix{2\\5\\7}$ in its null space. $\endgroup$ – abel May 26 '15 at 16:44

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