$GL(n, \mathbb{C})$ is isomorphic to a subgroup of $GL(2n, \mathbb{R})$

Prove that $GL(n, \mathbb{C})$ is isomorphic to a subgroup of $GL(2n, \mathbb{R})$.

My Proof:

For an $A \in GL(2, \mathbb{C})$, $$ A = \begin{bmatrix} a+bi &c+di \\ e+fi & g+hi \end{bmatrix} = \begin{bmatrix} a &c \\ e & g \end{bmatrix} + i \begin{bmatrix} b & d \\ f & h \end{bmatrix} .$$ It follows that $ GL(n, \mathbb{C})$ is a subset of $GL(2, \mathbb{R}) \oplus GL(2, \mathbb{R})$. (it is a subset because the condition that $\det A \neq 0$ puts more restrictions on the elements of $A$ than the condition that the two little matrices are invertible.)

Thus the result holds for $n=2$.

Can this argument be generalized to any $n$?

The hint was to use group actions, of $GL(n, \mathbb{C})$ on $GL(2n, \mathbb{R})$, using the fact that $\mathbb{C}^n \cong \mathbb{R}^{2n}$. I don't understand why a group action would be helpful?

Solution 1:

First your argument for $n=2$ is incorrect: the real and imaginary part of a matrix in $GL_2(\mathbb{C})$ are not in $GL_2(\mathbb{R})$ in general. For instance, if $A\in GL_2(\mathbb{R})$, then $A+i\cdot 0\in GL_2(\mathbb{C})$.

To come back to your hint, for any group $G$, a morphism $G\to GL_m(\mathbb{R})$ is the same thing as a linear action of $G$ on $\mathbb{R}^m$, by definition of a linear action. So since you want an (injective) morphism $GL_n(\mathbb{C})\to GL_{2n}(\mathbb{R})$, it makes sense to make $GL_n(\mathbb{C})$ act on $\mathbb{R}^{2n}$.

Now you can notice that by definition $GL_n(\mathbb{C})$ acts (faithfully and $\mathbb{C}$-linearly) on $\mathbb{C}^n$, and that $\mathbb{C}^n\simeq \mathbb{R}^{2n}$ as $\mathbb{R}$-vector spaces.

Solution 2:

The proof you've given for the $n = 2$ case is incorrect because that map doesn't respect the group operation.

To see why actions are useful, recall that a linear action of $G$ on a vector space $V$ gives a homomorphism $G \to GL(V)$. If $V$ is $d$-dimensional over $\mathbb R$ then picking a basis gives an isomorphism $GL(V) \simeq GL_d(\mathbb R)$. Since $GL_n(\mathbb C)$ acts on $\mathbb C^n \simeq \mathbb R^{2n}$ this would give a homomorphism $GL_n(\mathbb C) \to GL_{2n}(\mathbb R)$. If you prove that this homomorphism is injective then you've given an isomorphism between $GL_n(\mathbb C)$ and the image of that homomorphism.