How do we know if a matrix is in a general linear group?

Question

I have that $$A = \begin{bmatrix}3&1\\2&2\end{bmatrix}$$ $$B = \begin{bmatrix}3&-2\\2&-1\end{bmatrix}$$$$ C = \begin{bmatrix}3&2&2\\2&1&2\\-1&3&1\end{bmatrix}$$$$D = \begin{bmatrix}-3&-1&2\\1&-3&-1\\3&0&-2\end{bmatrix}$$ My question is how do we know if a matrix is in G$L_n$($\mathbb{Z})$? My understanding of how general linear groups is that G$L_n$($\mathbb{Z})$ is the set of matrices such that {$n\times n$ matrices with real coefficients and $\det(A) \neq 0$}. Therefore, since all of them have real coefficients and all of their determinants are nonzero, are all of them in G$L_n$($\mathbb{Z})$?

Answer 1

One of the requirements for a set to be a group is that every element have an inverse element in the group.

$GL_n(\mathbb Z)$ is the set of matrices with integer entries that also has an inverse with integer entries.

$GL_n(\mathbb R)$ is the set of $n\times n$ matrices with real entries and non-zero determinants. The non-zero determinant means that every matrix in the group has an inverse matrix that is also in the group.

Looking at $GL_n(\mathbb Z)$ we can't take all matrices with non-zero determinants. It is not a sufficiently restrictive condition. This should illustrate the problem.

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

$GL_n(\mathbb Z)$ is the set of matrices with integer entries that have determinants equal to $\pm 1$ will do the trick.

Answer 2

It will be in $GL_n(\mathbb{Z})$ if and only if its determinant is $\pm1$. This is part of a general result that a matrix with coefficients in $R$ is invertible over $R$ (that is, has an inverse with coefficients in $R$) if and only if its determinant is a unit in $R$. The proof is the same as the one you know for a field. The formula for the inverse of a matrix requires only that the determinant be invertible as an element of the ring.