# Explicit isomorphism between $(W_\mathbb{R})^\mathbb{C}$ and $W\boxplus\overline{W}$?

Suppose $W$ is a complex vector space. Let $\overline{W}$ denote the complex vector space $W$, but with scalar multiplication replaced by $(z,w)\mapsto \bar{z}\cdot w$.

I want to show that $(W_\mathbb{R})^\mathbb{C}$ is isomorphic (as complex vector spaces) to the external direct product $W\boxplus\overline{W}$ without resorting to dimension arguments. My text isn't clear, but I think $(W_\mathbb{R})^\mathbb{C}$ denotes the complexification of $W$ when viewed as a real vector space.

I tried to define a map $f\colon (W_\mathbb{R})^\Bbb{C}\to W\boxplus\overline{W}$ by $u+vi\mapsto (u,v)$. So $f$ is a bijection, and an additive homomorphism. I tried showing it respects scalar multiplication like this.

I get $$f((a+bi)(u+vi))=f(au-bv+(bu+av)i)=(au-bv,bu+av)$$ and $$(a+bi)f(u+vi)=(a+bi)(u,v)=((a+bi)u,(a-bi)v)$$ so I think my map is wrong. Is there an explicit isomorphism?

By the way, this is problem 2.33(b) of Roman's Advanced Linear Algebra.

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To give a complex linear map $(W_{\mathbb R})^{\mathbb C}$ it is enough to give a real linear map $W_{\mathbb R} \to W\boxplus \overline{W}.$ (The original map is then the unique complex linear extension of this real linear map.) And giving this map is equivalent to giving real linear maps $W_{\mathbb R} \to W$ and $W_{\mathbb R} \to \overline{W}$, and then taking their direct sum.
Since you don't know anything about $W$ (i.e. $W$ is totally arbitrary) it shouldn't be hard to find candidates for these maps. Then working backwards should give you the correct formula for the map you want.
Thanks Matt E. So is it enough to just take the identity embeddings $\iota_1\colon W_\mathbb{R}\to W$ and $\iota_2\colon W_\mathbb{R}\to\overline{W}$? These are both $\mathbb{R}$-linear, so I set $\iota=\iota_1\oplus\iota_2$, and extend it to $\iota^\mathbb{C}$ defined by $\iota^\mathbb{C}(u+vi)=\iota(u)+\iota(v)i$? Please correct me if I understood you wrong. –  Noomi Holloway Dec 18 '12 at 5:07
@NoomiHolloway: Dear Noomi, Yes, that's right. (And note that those "embeddings" are actually the identity map, which is $\mathbb R$-linear whether we regard the target as $W$ or as $\overline{W}$!) Regards, –  Matt E Dec 18 '12 at 5:11