# On degenerate linear systems

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 $XX'=r\in\Bbb N$)?

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

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.

• I am seeking $X$. – Brout Dec 11 '15 at 8:52
• @Turbo In what context? Is this a homework problem where you have a specific matrix A? – Matthew Gunn Dec 11 '15 at 8:58
• no I forgot my linear algebra. I am trying to recollect a good procedure in these situations. – Brout Dec 11 '15 at 8:59
• 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... – Matthew Gunn Dec 11 '15 at 9:03
• that is all I wanted to know and that seems right in retrospective. However what if I replace $\Bbb K$ with a ring $\Bbb R$? – Brout Dec 11 '15 at 9:05