If all entries of an invertible matrix $A$ are rational, then all the entries of $A^{-1}$ are also rational. Now suppose that all entries of an invertible matrix $A$ are integers. Then it's not necessary that all the entries of $A^{-1}$ are integers. My question is:
What are all the invertible integer matrices such that their inverses are also integer matrices?
On
The inverse of an integer matrix is again an integer matrix iff if the determinant of the matrix is $\pm 1$. Integer matrices of determinant $\pm 1$ form the General Linear Group $GL(n,\mathbb{Z})$
On
Arturo and Sivaram have already given the general condition for integer matrices with integer inverses; here I only note this particular example due to Ericksen that the matrix $\mathbf A$ with entries
$$a_{ij}=\binom{n+j-1}{i-1}$$
where $n$ is an arbitrary nonnegative integer has an integer inverse.
On
Just a comment, a picture of a particularly simple one:
Added:
Iterative solvers have a hard time solving $Ax = b$ with
$b$ a point source $[0 0 0 ... 1 ... 0 0 0]$:
$A^{-1} b$ has to pick out a column of $A^{-1}, \pm \, [1 \, 0 \, -1 \, 0 ...]$.
Also, this $A$ is far from positive-definite -- its eigenvalues are symmetric about $0$.
See also
gmres-for-a-non-diagonalizable-matrix
on scicomp.stack .
On
$$A^{-1} = \frac{1}{|A|}C^T $$ Where $|A|$ denotes the determinant of matrix $A$ and $C$ is the matrix of minors. Each entry in $C^T$, $c_{ji}$, represents the minor removing just the $i$th row and the $j$th column. Each of these determinants is the sum, difference and multiple of integers, so each $c_{ji}$ is integer. This means, of course, that $|A|$ must be $\pm 1$. Since matrices may be of any integer dimension and there are infinite integers, just listing the identity matrices gets us to $\infty$.
Exactly those whose determinant is $1$ or $-1$.
See the previous question about the $2\times 2$ case. The determinant map gives necessity, the adjugate formula for the inverse gives sufficiency.