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

Question

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?

Answer 1

I'd like to expand on a piece of Jim's answer:

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 $GL_n(\mathbb C) \to GL_{2n}(\mathbb R)$.

Let's consider in detail the case of $n=2$. We begin by selecting a basis for the real-vector space $\Bbb C^{2}$: $$ \mathcal B = \{v_1,v_2,v_3,v_4\} = \{(1,0),(i,0),(0,1),(0,i)\}. $$ Now, suppose that we are given a matrix $$ A = \pmatrix{b_{11} & b_{12}\\b_{21} & b_{22}} + i\pmatrix{c_{11} & c_{12}\\ c_{21} & c_{22}}. $$ The referenced isomorphism from $GL_2(\Bbb C)$ to $GL_{4}(\Bbb R)$ that comes from picking a basis is the map that produces the matrix of the transformation $x \mapsto Ax$ relative to $\mathcal B$.

We can see what this matrix looks like by seeing what $x \mapsto Ax$ does to each column-vector. For instance, we have $$ Av_1 = \left(\pmatrix{b_{11} & b_{12}\\b_{21} & b_{22}} + i\pmatrix{c_{11} & c_{12}\\ c_{21} & c_{22}}\right) \pmatrix{1\\0} \\ = \pmatrix{b_{11} + c_{11}i\\ b_{21} + c_{21}i} = b_{11}v_1 + c_{11}v_2 + b_{21}v_3 + c_{21} v_4 $$ and can therefore see that the first column of the matrix of $x \mapsto Ax$ should be $(b_{11}, c_{11},b_{21},c_{21})^T$. Proceeding in a like fashion, we can see that the full matrix for this map will be $$ \left[\begin{array}{cc|cc}b_{11} & -c_{11} & b_{12} & -c_{12}\\ c_{11} & b_{11} & c_{12} & b_{12}\\ \hline b_{21} & -c_{21} & b_{22} & -c_{22}\\ c_{21} & b_{21} & c_{22} & b_{22}\\ \end{array}\right] $$ In other words, one version of the isomorphism that you're looking for is $$ \pmatrix{b_{11} & b_{12}\\b_{21} & b_{22}} + i\pmatrix{c_{11} & c_{12}\\ c_{21} & c_{22}} \mapsto \pmatrix{b_{11} & -c_{11} & b_{12} & -c_{12}\\ c_{11} & b_{11} & c_{12} & b_{12}\\ b_{21} & -c_{21} & b_{22} & -c_{22}\\ c_{21} & b_{21} & c_{22} & b_{22}\\}. $$