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Is it true that if in $$AX=0$$ where $A$ is $n\times n$ matrix over field $\Bbb K$ and $X$ is a length $n$ vector of variables if $rank(A)=n-1$ we will have an unique solution up to constant factors?

How do you solve for such a solution (assume $XX'=r\in\Bbb N$)?

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    $\begingroup$ Yes, in this case it's true, because $\text{rank}(A) = n-1 \Leftrightarrow \text{dim}(\text{Ker} (A) ) = 1$ $\endgroup$ – Tryss Dec 11 '15 at 8:34
  • $\begingroup$ how do you solve? Assume $XX'=r$. $\endgroup$ – Brout Dec 11 '15 at 8:35
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You're basically asking for the null space of $A$.

A standard way to find a basis for the null space is to take the singular value decomposition:

$$ U \Sigma V^* = A $$

where $\Sigma$ is a diagonal matrix containing what is known as the singular values.

  • The number of non-zero singular values give you the rank of matrix A.
  • The number of zero singular values gives you the dimension of the null space.
  • Columns of $V$ associated with zero singular values form a basis for the null space of A.

Note: this is how many numerical linear algebra systems (eg. Matlab, numpy) calculate rank and null space.

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  • $\begingroup$ I am seeking $X$. $\endgroup$ – Brout Dec 11 '15 at 8:52
  • $\begingroup$ @Turbo In what context? Is this a homework problem where you have a specific matrix A? $\endgroup$ – Matthew Gunn Dec 11 '15 at 8:58
  • $\begingroup$ no I forgot my linear algebra. I am trying to recollect a good procedure in these situations. $\endgroup$ – Brout Dec 11 '15 at 8:59
  • $\begingroup$ If you're doing something by hand for a small matrix, you'd just find the solutions to the system $Ax=0$ by hand. If you're doing this for a large matrix on computer, you'd take the singular variable decomposition. Typically, svd routines put zero singular values last in the $\Sigma$ matrix and what you're calling $x$ would be the final column of my matrix $V$. If the null space were two dimensional, it would be the final 2 columns etc... $\endgroup$ – Matthew Gunn Dec 11 '15 at 9:03
  • $\begingroup$ that is all I wanted to know and that seems right in retrospective. However what if I replace $\Bbb K$ with a ring $\Bbb R$? $\endgroup$ – Brout Dec 11 '15 at 9:05

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