# Determining consistency of a general overdetermined linear system

For $m > 2$, consider the $m \times 2$ (overdetermined) linear system $$A \mathbf{x} = \mathbf{b}$$ with (general) coefficients in a field $\mathbb{F}$; in components we write the system as $$\left(\begin{array}{cc}a_{11} & a_{12} \\ \vdots & \vdots \\ a_{m1} & a_{m2} \end{array}\right) \left(\begin{array}{c}x_1 \\ x_2\end{array}\right) = \left(\begin{array}{c}b_1 \\ \vdots \\ b_n\end{array}\right),$$ where $m > 2$, so that the system is overdetermined.

If $m = 3$ and the system is consistent, (equivalently, $\mathbf{b}$ is in the column space of $A$), then the columns of the augmented matrix $\pmatrix{ A \mid {\bf b}}$ are linearly dependent, and so $$\det \pmatrix{ A \mid {\bf b}} = 0.$$ In particular, we have produced a polynomial in the components $(a_{ij}, b_j)$ of the linear system for which vanishing is a necessary condition for system's consistency. I'll call such polynomials polynomial obstructions for the system.

If $m > 3$, then we can produce ${m}\choose{3}$ such polynomials by considering the determinants of the $3 \times 3$ minors of $\pmatrix{ A \mid {\bf b}}$.

Are essentially all polynomials obstructions to the system essentially given by these, or are there others? Put more precisely: By definition the polynomial obstructions comprise an ideal in the polynomial ring $\mathbb{F}[a_{11}, \ldots, a_{m2}, b_1, \ldots b_m]$---do the determinants of the $3 \times 3$ minors generate this ideal? If not, how does one produce a complete set of generators?

More generally, for an $m \times n$ overdetermined linear system (so that $m > n$) $$A \mathbf{x} = \mathbf{b},$$ we can produce polynomial obstructions by taking the determinants of the ${m}\choose{n+1}$ minors (of size $(n + 1) \times (n + 1)$). What are the answers to the obvious analogues to the above questions in the $n = 2$ case?

## 2 Answers

When $m=3$ and $\mathbb F$ is infinite, there are no other obstructions besides the determinant. When $\mathbb F$ is finite, there are many others : for example if we put $\chi_{\mathbb F}(X)=\prod_{t\in {\mathbb F}^*} (X-t)$, $\chi(t)$ is zero iff $t$ is nonzero, so that the following $n$ polynomials are all obstructions :

$$w_i=b_i\prod_{j=1}^n\chi_{\mathbb F}(a_{ij})$$

For the infinite case, one can use the following lemma :

Generalized Euclidean division. Let $A$ and $B$ be two polynomials in ${\mathbb F}[X_1,X_2,\ldots,X_n,Y]$. Let $a={\sf deg}_Y(A)$, $b={\sf deg}_Y(B)$, and let $L$ be the leading coefficient of $B$ with respect to $Y$ (so that $L\in{\mathbb F}[X_1,X_2,\ldots,X_n]$ and $B-LY^b$ has degree $<b$ in $Y$). Then if $a \geq b$, there are two polynomials $Q,R\in {\mathbb F}[X_1,X_2,\ldots,X_n,Y]$ such that $L^{a-b+1}A=QB+R$ and ${\sf deg}_Y(R)<b$.

Proof. Let ${\mathbb K}={\mathbb F}(X_1,X_2,\ldots,X_n)$. We can view $A$ and $B$ as members of ${\mathbb K}[Y]$, and perform ordinary euclidian division ; this yields $Q^{\sharp},R^{\sharp}\in {\mathbb K}[Y]$ such that $A=Q^{\sharp}B+R^{\sharp}$. Since the division process involves $a-b+1$ divisions by $L$, we see that $Q^{\sharp}$ and $R^{\sharp}$ are of the form $\frac{Q}{L^{a-b+1}}$ and $\frac{R}{L^{a-b+1}}$ with $Q,R\in {\mathbb F}[X_1,X_2,\ldots,X_n,Y]$. This concludes the proof of the lemma.

Let us now explain how this can be used when $m=3$. Let $I$ be the ideal (in the ring ${\mathfrak R}={\mathbb F}(A_{11},A_{12},A_{13},A_{21},A_{22},A_{23},B_1,B_2,B_3)$ of all obstructions. In particular, the determinant

$$\Delta=(A_{12}A_{23}-A_{13}A_{22})B_1+ (A_{13}A_{21}-A_{11}A_{23})B_2+ (A_{11}A_{22}-A_{12}A_{21})B_3 \tag{1}$$

is a member of $I$. Let $P\in I$, and let $p={\sf deg}_{B_3}(P)$. By the generalized Euclidean division property above, there are polynomials in $Q,R$ in $\mathfrak R$ such that $(A_{11}A_{22}-A_{12}A_{21})^p P=\Delta Q+R$, such that $R$ does not contain the variable $B_3$ (note that we need $p\geq 1$ in order to apply the lemma ; but if $p=0$, we can simply take $Q=0,R=P$). Then $R\in I$. Consider the set

$$W=\bigg\lbrace (a_{11},a_{12},a_{13},a_{21},a_{22},a_{23},b_1,b_2) \in {\mathbb F}^{8} \ \bigg| \ a_{11}a_{22}-a_{12}a_{21} \neq 0\bigg\rbrace \tag{2}$$

Since $\mathbb F$ is infinite, $W$ is a Zariski-dense open subset of ${\mathbb F}^{8}$. We have a natural map $\phi : W \to V(I)$, defined by

$$\phi(a_{11},a_{12},a_{13},a_{21},a_{22},a_{23},b_1,b_2)= \bigg(a_{11},a_{12},a_{13},a_{21},a_{22},a_{23},b_1,b_2, -\frac{(a_{12}a_{23}-a_{13}a_{22})b_1+ (a_{13}a_{21}-a_{11}a_{23})b_2}{a_{11}a_{22}-a_{12}a_{21}}\bigg) \tag{3}$$

For any $w\in W$, we have $R(\phi(w))=0$ since $R\in I$. We deduce $R(w)=0$ for all $w\in W$. Since $W$ is Zariski-dense, $R$ is zero eveywhere. So $R$ must be the zero polynomial, $(A_{11}A_{22}-A_{12}A_{21})^p P=\Delta Q$. Since $A_{11}A_{22}-A_{12}A_{21}$ and $\Delta$ have no common factors, we see that $\Delta$ divides $P$.

• Thanks for this, Ewan. In order to apply the lemma, we must assume that $p \geq 1$, but if it is not, by symmetry we can replace $B_3$ with any generator that does occur in $P$. Why does $R \in I$ imply that $R(\phi(w)) = 0$? Also, it seems like this argument must be modified significantly when $m > n + 1$ (rather than equal as in this case), as in that case there are multiple determinant obstructions---or is there an obvious and easy modification I'm not seeing? – Travis Willse Sep 29 '14 at 3:22
• @Travis "we must assume that $p\geq 1$, but if it is not, by symmetry we can replace $B_3$ with any generator that does occur in $P$" Indeed, or you can use another argument I’ve just inserted in my answer, see above. – Ewan Delanoy Sep 29 '14 at 5:30
• @Travis "Why does $R\in I$ imply that $R(\phi(w))=0$" ? Because $\phi(w)\in V(I)$ by construction of $\phi$ ; what $\phi$ does is choose the unique value of $b_3$ so that $b$ will indeed be a linear combination of the columns of $A$. Then, by definition of $V(I)$, all the polynomials of $I$ must be zero on $V(I)$. – Ewan Delanoy Sep 29 '14 at 5:35
• @Travis "This argument must be modified significantly when $m>n+1$ as in that case there are multiple determinant obstructions" Indeed, I’m not even certain that there are no other obstructions in this case. What is needed, probably, is showing that the set of determinental obstructions is a Groebner basis with respect to your favorite monomial ordering. – Ewan Delanoy Sep 29 '14 at 5:44
• Thanks for these comments, Ewan, the outstanding points are very clear to me now. As for finding a Groebner basis this seems out of reach without another insight, as I don't have a characterization of the ideal of obstructions (for general $m$) besides the definition. I'll attempt some variations of your argument for $m = 3$ (which works just as well for $m = n + 1$ I think), but perhaps this is harder than it seems. – Travis Willse Sep 29 '14 at 6:51

We assume that $b\not=0$ and that $\mathbb{F}$ is an infinite field. The required condition is $rank(A)=rank(A|b)$ or equivalently $rank(A|b)\leq rank(A)$. If $A$ is a generic matrix, then $rank(A)=n$ and your conditions are sufficient because they are equivalent to $rank(A|b)\leq n=rank(A)$. Yet, if $A$ is not generic and $rank(A)=r<n$, then it is not sufficient -for instance assume that the first column of $A$ is zero-. You must extract from $A$ a $r\times r$ invertible submatrix $A'$; we can assume that $A'$ is formed using the first $r$ columns and rows of $A$. After you construct, for $r<i\leq m$, the $(r+1)\times (r+1)$ matrices $U_i=\begin{pmatrix}A'&[b_1,\cdots,b_r]^T\\L_i&b_i\end{pmatrix}$ where $L_i=[a_{i,1},\cdots,a_{i,r}]$ and you say that their determinants are zero. Of course, if you know $r$ and not $A'$, then you must consider all the $(r+1)\times (r+1)$ matrices, extracted from $(A|b)$, containing in the last column a part of the vector $b$. If you do not know $r$, that is if $A$ is a formal matrix, then $r=0$ or $r=1$ or ...$r=n$ and you apply that is said above.

• Thanks, loup blanc. To be clear, I am interested in conditions on the entries of $A, b$ that are necessary (and not necessarily sufficient) for consistency for all $A, b$. – Travis Willse Sep 29 '14 at 3:27