This is a standard theorem in linear algebra over fields that is still true when working over an integral domain.

For a commutative unital integral domain $R$, let $A$ be an $n\times n$ matrix with entries in $R$. The system of linear equations $A\mathbf{x}=0$ has a non-zero solution if and only if $\det(A) = 0$.

The forward direction of this statement is not true though if we are working over a ring $R$ that is not an integral domain, a counter example being $$ R = \boldsymbol{Z}_6 \qquad A = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \qquad \mathbf{x} = \begin{pmatrix} 2 \\ 2 \end{pmatrix} \,. $$

But what about the other direction? Does $\det(A) = 0$ imply that $A\mathbf{x} = 0$ has a non-zero solution over an arbitrary commutative unital ring? I've played around for a bit, and have yet to find a counterexample.

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    $\begingroup$ How do you define determinant over an arbitrary ring? $\endgroup$ – José Carlos Santos Aug 10 '18 at 15:22
  • $\begingroup$ @JoséCarlosSantos you're right. Let's limit this to commutative unital rings. Looking at the definition of $\det$, the unique alternating $R$-multilinear form that sends $I_n$ to $1_R$ ..., it does require commutative. But now I'm curious how badly things go if you try to define the (left-)determinate in terms of the usual algorithm, where you'd multiply each element of a row on the left by the $\det$ of the corresponding minor, if you don't have a commutative unital ring. $\endgroup$ – Mike Pierce Aug 10 '18 at 15:27
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    $\begingroup$ @MikePierce Determinants cannot be defined over noncommutative rings. There is a similar tool for division rings: en.wikipedia.org/wiki/Dieudonné_determinant $\endgroup$ – egreg Aug 10 '18 at 15:39
  • $\begingroup$ I found this where Sylvia's theorem are trying to prove core.ac.uk/download/pdf/82216288.pdf $\endgroup$ – Upasana Mittal Aug 10 '18 at 15:51
  • $\begingroup$ See this thread: math.stackexchange.com/questions/71740/… $\endgroup$ – Pierre-Yves Gaillard Aug 10 '18 at 23:36

Let $A$ be square of size $n \times n$, and $\det A=0$. Then $A\,\textrm{adj}(A)=0$ where $\textrm{adj}(A)$ is the adjugate of $A$. As long as $\textrm{adj}(A)\neq 0$ then taking some column of $\textrm{adj}(A)$ gives you $x$ with $Ax=0$.

But what if adj$(A)=0$? In this case all minors of $A$ vanish. Find a largest square submatrix of $A$ with nonzero determinant. We can assume that it's size $r$ by $r$ and fills the top left corner of $A$. Let $B$ be the top-left $r+1$ by $r+1$ submatrix of $A$. Then $\det B=0$ but adj$(B)\ne0$. Let $y$ be the $(r+1)$-th column of adj$(B)$ and $z$ be the column vector of height $n$ got by appending zeroes below $y$. Then $z\ne0$ (due to the non-vanishing of the top left $r$ by $r$ determinant) and I claim that $Az=0$.

The top $r+1$ entries of $Az$ are certainly zero. This is the identity $B\,$adj$(B)=0$. If we look at another entry, say the $s$-th then it is zero, essentially by replacing the bottom row of $B$ by the first $r+1$-th entries of the $s$-th row of $A$, and noting that the new matrix has zero determinant, as it is obtained from a submatrix of $A$ of size $r+1$ by $r+1$ by elementary row operations. As all submatrices of $A$ of this size have vanishing determinant, this completes the proof that $Az=0$.

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    $\begingroup$ Which $z$ do you get if you set $$A=\begin{pmatrix}1&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\ ?$$ $\endgroup$ – Pierre-Yves Gaillard Aug 10 '18 at 19:31
  • $\begingroup$ You most certainly mean, "... taking some column of $\mathrm{adj}(A)$ gives you $x$ with $Ax=0$." I've edit your answer. $\endgroup$ – Mike Pierce Aug 10 '18 at 22:35
  • $\begingroup$ @Pierre-YvesGaillard Hope it's right now! $\endgroup$ – Lord Shark the Unknown Aug 11 '18 at 1:52
  • $\begingroup$ Very nice! I'd say that the $s$-th entry is zero because it is the determinant of the submatrix of $A$ of size $r+1$ by $r+1$ obtained by replacing the bottom row of $B$ by the first $r+1$-th entries of the $s$-th row of $A$. See also the references at the end of math.stackexchange.com/a/71755/660. $\endgroup$ – Pierre-Yves Gaillard Aug 11 '18 at 12:22

In a commutative ring (with 1), it turns out that, yes, $Ax=0$ has a non-trivial solution if and only if $\det(A)$ is either zero or a zero divisor. This follows from a theorem of McCoy.

Here is a link to a related question of my own:

MSE -- Do these matrix rings have non-zero elements that are neither units nor zero divisors?

MO -- https://mathoverflow.net/questions/77816/do-these-matrix-rings-have-non-zero-elements-that-are-neither-units-nor-zero-divi

A link to a paper with an account of McCoy's rank theorem: http://math.berkeley.edu/~lam/amspfaff.pdf


Consider $\pmatrix{2&3\cr 2&3}$ in $\mathbb{Z}/6$ its determinant is $0$, and $2x+3y=0$ implies that $2x=3y$. But the multiples of $2$ are $\{0,2,4\}$ and the multiples of $3$ are $\{0,3\}$.

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    $\begingroup$ The solution can use the fact that $2$ and $3$ are zero-divisors though. So $\langle 3,2\rangle$ is a non-zero solution. This is just an example of the statement being true. $\endgroup$ – Mike Pierce Aug 10 '18 at 15:41

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