Set of all square matrices notation

Question

I have a problem where I have to prove or disprove whether $G$ is a group with matrix multiplication. $G$ is defined as: $G=\{A \in \mathbb{R}^{n \times n} \mid \det(A) > 0 \}$ .

I am having problems understanding whether $G$ contains all square matrices with a positive determinant (meaning it would for example contain $3 \times 3$ - matrices as well as $4 \times 4$-matrices etc.) or if it either contains $3 \times 3$-matrices OR $4 \times 4$-matrices OR .... with a positive determinant. (*)

In the first case I would argue it is not a group, because the product would not be defined for certain elements of $G$ (a 3x3 matrix and a 4x4 matrix for example) and there would not exist one $e$ in $G$ which would be the neutral element for all elements of $G$.

Could you tell me how this notation (the definition of $G$) is to be interpreted? And what would be the right way to mathematically define the other of the two variants I mentioned above? (marked with (*))

Answer 1

You're right if you say that all matrices in $G$ have to be of the same size. You are getting a group for every $n \in \mathbb{N}$, this means, you consider a fixed $n$ when looking at $G$. Maybe a possibility to write this down would be to use the notation

$$G_n = \{ A \in \mathbb{R}^{n \times n} : \det(A) > 0 \}.$$

Answer 2

The set $GL^{+}_n(\mathbb{R})=\{A \in M_n(\Bbb R)\mid \det(A)>0\}$ for fixed $n\ge 1$ is a subgroup of $GL_n(\Bbb R)$, because $$ \det(AB)=\det(A)\det(B)>0 $$ for all $A,B$ with $\det(A),\det(B)>0$, and also $\det(A^{-1})=1/\det(A)>0$ for matrices $A$ with $\det(A)>0$.

Your question about different $n$ is already present for $GL_n(\Bbb R)$ itself. We need to multiply two matrices of the same size, and not a matrix of size $3$ with a matrix of size $4$. This has nothing to do with a positive determinant.