Inverse of an Integer Matrix

I found a problem on the Open Problem Garden which asks about the conditions on a rectangular, full-rank, integer matrix such that its right inverse (given by: $$A^T (AA^T)^{-1}$$ ) is also an integer matrix. The rectangular matrix is constructed in the following way :

Let D be a square diagonal matrix (size $$m \times m$$) with integer elements $$\geq$$ 2 along the main diagonal (in order to ensure full rank and thus existence of a right inverse)

Let X be an integer matrix (size $$m \times n$$) with $$n\geq m$$.

Now, concatenate the matrices to make a new rectangular matrix M = [D X], giving it dimension $$m \times (m+n)$$. I am interested in the right inverse of this matrix M.

I have written code to test some matrices, and I have yet to find even one integral element, let alone an entire matrix. I've done some algebraic analysis on a general 2x4 matrix, and intuitively it looks as though some elements will never be a non-zero integer, but it is difficult to prove. If anyone has any advice on how to proceed or any insight, that would be great.

Edits: Clarified the characterization of the matrices in question. Renamed matrices for consistency.

• Your question is not the same as posted in OPG. There the matrix $A$ has restrictions on its form. And, also, you are posting a particular form for the right inverse. So, are you asking the question as in OPG, or are you asking the more general question you have phrased? Apr 11, 2019 at 17:29
• You are correct, I intended on asking the question which is posed in OPG - I've edited the question to reflect these conditions. Apr 12, 2019 at 0:32
• Thanks. Still, do you want to force the right inverse to be $A^T(AA^T)^{-1}$? Apr 12, 2019 at 0:57
• Also, the condition $n\geq m$ does not appear in the original problem. Apr 12, 2019 at 2:14
• n $>=$ m is in the original problem. The formula for the right inverse is the only way I know to consistently find the right inverse of a given matrix. I suppose I shouldn't force it to be such, but I do not know how else I would go about finding it. Apr 12, 2019 at 3:15

Here's an example for a square 2x2 matrix with each element larger than or equal to 2:

$$\begin{bmatrix}3 & 2\\4 & 3\end{bmatrix}^{-1} = \begin{bmatrix}3 & -2\\-4 & 3\end{bmatrix}$$

• Fortunately the determinant for this matrix is equal $1$ :) Apr 11, 2019 at 17:12
• correct, by construction Apr 11, 2019 at 17:28
• -1: this doesn't answer the question, where the given matrix is not square. Apr 12, 2019 at 1:04

If $$\begin{bmatrix} R\\ S\end{bmatrix}$$ is a right inverse for $$M$$, we have $$I=\begin{bmatrix} D&X\end{bmatrix} \begin{bmatrix} R\\ S\end{bmatrix} =DR+XS.$$ Comparing diagonal entries, we have for all $$k$$ $$\tag1 D_{kk}R_{kk}+\sum_h X_{kh}S_{hk}=1$$ We can see the above as Bézout's Identity, so a necessary condition for all entries of $$R$$ and $$S$$ to be integers is that for all $$k$$ the tuples $$D_{kk}, X_{k1},\ldots,X_{kn}$$ (i.e., the nonzero elements in each row of $$M$$) are coprime.

The remaining equalities are, for $$k\ne j$$, $$\tag2 D_{kk}R_{kj}+\sum_hX_{kh}S_{hj}=0.$$ If each $$S_{hj}$$ is a multiple of $$D_{kk}$$, we can solve for $$R_{kj}$$: so we get a sufficient condition if $$S_{hj}$$ is a multiple of $$\prod_{r\ne j}D_{rr}$$. This requires (by $$(1)$$) that $$D_{11},\ldots,D_{mm}$$ are coprime. Thus

A sufficient condition for $$M$$ to have a right-inverse with integer entries is that

• for each $$k$$, the numbers $$D_{kk}, X_{k1},\ldots,X_{kn}$$ are coprime;

• the numbers $$D_{11},\ldots,D_{mm}$$ are coprime.

To see an example of how to use this, take $$D=\begin{bmatrix} 2&0\\0&3\end{bmatrix},\ \ \ X=\begin{bmatrix}4&5\\6&7 \end{bmatrix} .$$

We want $$S_{hj}=\left(\prod_{r\ne j}D_{rr}\right)\,S_{hj}'$$. So the two equalities in $$(1)$$ become $$\begin{cases}1=2R_{11}+X_{11}\times D_{22}\times S_{11}'+X_{12}\times D_{22}\times S_{21}'\\ 1=3R_{22}+X_{21}\times D_{11}\times S_{12}'+X_{22}\times D_{11}\times S_{22}' \end{cases}$$ That is, we need $$\begin{cases} 1=2R_{11}+4\times 3\times S_{11}'+5\times 3\times S_{21}'\\ 1=3R_{11}+6\times 2\times S_{12}'+ 7\times 2\times S_{22}' \end{cases}$$ For instance (among infinitely many possible choices) $$\begin{cases} 1=2(-13)+4\times 3\times 1+5\times 3\times 1\\ 1=3\times 9+6\times 2\times (-1)+7\times 2\times (-1) \end{cases}$$ This gives us $$R_{11}=-13$$, $$R_{22}=9$$, and $$S_{11}=3,\ S_{21}=3,\ S_{12}=-2,\ S_{22}=-2.$$ Now we can write, from $$(2)$$, $$R_{12}=-(X_{11}S_{12}+X_{12}S_{22})/D_{11}=-(4\times (-2)+5\times (-2))/2=9,$$ $$R_{21}=-(X_{21}S_{11}+X_{22}S_{21})/D_{22}=-(6\times 3+7\times 3)/3=-13.$$ So $$R=\begin{bmatrix} -13&9\\ -13&9\end{bmatrix}$$. Now you can check directly that $$\begin{bmatrix} 2&0&4&5\\ 0&3&6&7\end{bmatrix} \begin{bmatrix} -13&9\\-13&9\\ 3&-2\\3&-2\end{bmatrix} =\begin{bmatrix} 1&0\\0&1\end{bmatrix}.$$

In general we may use that $$(2)$$ can be written as $$\tag3R_{kj}=-\frac{(XS)_{kj}}{D_{kk}},\ \ \ \ \ k\ne j.$$

For another example, if $$D=\begin{bmatrix} 2&0&0\\0&3&0\\0&0&5\end{bmatrix}$$ and $$M=\begin{bmatrix} 5&7\\4&5\\6&7\end{bmatrix}$$ the algorithm, with the choices $$\begin{cases} 1=2(-127) + 5\times 30+ 7\times 15 \\ 1=3(-43)+ 4\times 20 + 5\times 10\\ 1=5\times 23+6\times (-12)+7\times (-6) \end{cases}$$ gives $$R_{11}=-127$$, $$R_{22}=-43$$, $$R_{33}=23$$, $$S=\begin{bmatrix} 30&20&-12\\ 15&10&-6 \end{bmatrix}$$ and then using $$(3)$$ $$\begin{bmatrix} 2&0&0&5&7\\0&3&0&4&5\\0&0&5&6&7\end{bmatrix} \begin{bmatrix} -127&-85&51\\ -65&-43&26\\ -57&-38&23\\ 30&20&-12\\ 15&10&-6\end{bmatrix} =\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$

Finally, note that there are infinitely many choices for the matrix $$S$$ via $$(1)$$, so the algorithm produces infinitely many different right inverses.

You can find a necessary condition for the inverse of a square integer matrix to also be an integer matrix as follows:

If $$A$$ is a square integer matrix then its determinant $$\det (A)$$ is an integer. Similarly if $$A^{-1}$$ exists and is an integer matrix then $$\det (A^{-1})$$ is an integer too. But $$\det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I) = 1$$. So you are looking for two integers whose product is 1 ... there are not many choices ...

• -1: this doesn't answer the question, as the matrix in the question is not square. Apr 12, 2019 at 1:05