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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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up vote 3 down vote accepted

The map you wrote down is not complex linear, so that is a problem.

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.

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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

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