Prove that $GL_n(ℝ)$ is a subgroup of $GL_n(ℂ)$. I know that since $\mathbb{R} ⊆ \mathbb{C}$ is a subfield, $GL_n(R)$ is a subgroup of $GL_n(C)$. So any real matrix representation of G is also a complex matrix representation of $G$.
Can someone give a hint on how to prove $GL_n(R)$ is a subgroup of $GL_n(C)$
 A: Why do you want to prove something? You know that $GL_n(ℝ)$ and $GL_n(ℂ)$ are groups with respect to matrix multiplication and that $GL_n(ℝ)$ is a subset of $GL_n(ℂ)$. Since the matrix product $A B$ of two matrices $A,B  \in GL_n(ℝ)$  agrees with the matrix product of these matrices regarded as elements of $GL_n(ℂ)$ , nothing remains to be shown.
A: Well, you have the inclusion $GL(\mathbb{R}) \subset GL(\mathbb{C})$. In order to prove it is a subgroup you only have to check that given $A,B \in GL(\mathbb{R})$ then $AB^{-1}\in GL(\mathbb{R})$. But this is indeed the case, since every nonsingular real matrix has an inverse with real entries, and since the product of two nonsingular matrices with real entries is a nonsingular matrix with real entries.
A: For any subring $A$ of a ring $B$, the group $A^\times$ of units of $A$ is a subgroup of $B^\times$, since if $x\in A$ has an inverse $y$ in $A$, then $x,y$ are also mutual inverses in $B$ (because $xy=yx=1_A=1_B$).
Apply this to $A=M_n(\mathbb R)$ and $B=M_n(\mathbb C).$
A: Let's check the subgroup conditions listed here for your problem: Basic Subgroup Conditions
$1)$ $e=I\in GL_n(\Bbb{R})$. True.
$2)$ Let $A,B\in GL_n(\Bbb{R})$. Because of matrix maltuplication, the multplication of two real matrices is real matrix, that is, it has real entries. Also, If $A$ and $B$ are invertible then their determinants are non-zero. Then since $\det(AB)=\det(A)\det(B)$, the determinant of $AB$ is also non-zero. So. $AB$ is invertible. That is, $AB\in GL_n(\Bbb{R})$.
$3)$ Let $A\in GL_n(\Bbb{R})$. Since, $det(A)\neq 0$, $A^{-1}$ exists and it is a real matrix. It can be found by using its co-factors which are real.
See: https://en.wikipedia.org/wiki/Minor_(linear_algebra)#Inverse_of_a_matrix
Therefore, $A^{-1}\in GL_n(\Bbb{R})$.
