To each representation on $\mathbb{C}^n $ corresponds a representation on $\mathbb{R}^{2n} $

So, I am supposed to show that to each representation $\rho$ of a group $G$ on $\mathbb{C}^n $ corresponds a representation $\tilde{\rho}$, of the same group $G$, on $\mathbb{R}^{2n} $.

Additionally: Show that, if the representation $(\tilde{\rho},\mathbb{R}^{2n}) $ is irreducible, then the representation $(\rho,\mathbb{C}^{n} )$ is also irreducible.

My thinking:

For the first part:

Suppose we have the following map: $\phi(z):\mathbb{C}^{n}\rightarrow \mathbb{R}^{2n}, \space \phi(z)= \begin{bmatrix} Re(z) \\ Im(z) \\ \end{bmatrix}$. This is obviously a bijection, additionally it follows that $\phi(z_1+z_2)=\phi(z_1)+\phi(z_2)$. My idea then was to find a bijection $\eta$ between the sets $\rho(g)$ and $\tilde{\rho}(g)$, for $g \in G$ .

And this is what I came up with (I'll denote $\rho(g)$ as $\rho_g$):

$$\eta:\mathbb{R}^{2n}\rightarrow \mathbb{C}^{n}, \space \eta:\space \tilde{\rho}_g(.)\rightarrow \phi^{-1}(\tilde{\rho}_g(\phi(.)))=\rho_g(.)$$

The map $\rho_g (.)=\phi^{-1}(\tilde{\rho}_g(\phi(.)))$ is clearly $\mathbb{C}^{n} \rightarrow \mathbb{C}^{n}$, it is also a bijection and it follows that $\rho_g(z_1z_2)=\rho_g(z_1)\rho_g(z_2)$, all together meaning that $\rho_g $ is a representation of $G$ on $\mathbb{C}^{n}$. I have therefore shown that to each representation $(\tilde{\rho},\mathbb{R}^{2n})$ corresponds a representation $(\rho,\mathbb{C}^{n})$, via $\eta$.

For the second part: (proof by contradiction)

Irreducible means that for $\tilde{\rho}$ there is no (non-trivial) invariant subspace in $\mathbb{R}^{2n}$. Lets take a (non-trivial) subset $R\subset \mathbb{R}^{2n}$, so that for a subset $W \subset \mathbb{C}^{n}: \space$ $\phi(W)=R$ and $\phi^{-1}(R)=W$. We suppose that the opposite holds ($\rho$ is reducible): that $W$ is some non-trivial invariant subset for $(\rho,\mathbb{C}^{n} )$. This means that $\rho_g(W) \subset W$ for each $g \in G$. But since $\phi^{-1}(\tilde{\rho}_g(\phi(W)))=\rho_g(W) \subset W$, it follows that $\tilde{\rho}(\phi(W))\subset R $ and using $\phi(W)=R$, we get that $\tilde{\rho}(R)\subset R$. So we found a (non-trivial) invariant subspace for $\tilde{\rho}$, but this contradicts the above statements that $\tilde{\rho}$ is irreducible, therefore $\rho$ must be irreducible.

My question:

Is my thinking correct? Could you please suggest some improvements or maybe show a different, more compact way to prove the above statements.

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