Let $V$ be a complex vector space and $J:V\to V$ a map with the following properties:

(i) $J(v+w)=J(v)+J(w)$

(ii) $J(cv)=\overline{c}J(v)$

(iii) $J(J(v))=v$

for all $v,w\in V$ and $c\in\mathbb{C}$. Put $W=\{v\in V\,|\, J(v)=v\}$; this is a real vector space with respect to the operations in $V$ (I have shown this). From here, I am asked to show:

$\bullet$ For each $v\in V$, there exist unique vectors $w,u\in W$ such that $v=w+iu$.

The uniqueness is very straight-forward once you have the decomposition. I've been trying the usual kinds of tricks, like writing $v=\bigl(v-J(v)\bigr)+J(v)$, but this clearly doesn't work. And I understand morally what's happening: we're decomposing $V$ into its formal real part and formal imaginary part. I just don't see how to get the decomposition.


  • $\begingroup$ You need to show $W=\mathbb{R}$ because then you know that $\mathbb{C}=\{a+ib\,\, | \,\, a,b\in\mathbb{R}\}$ $\endgroup$ – Swapnil Tripathi Dec 11 '14 at 2:39
  • $\begingroup$ I don't think that $W=\mathbb{R}$ will be true in general. The space we start with, $V$, is just some arbitrary complex vector space. $\endgroup$ – Bey Dec 11 '14 at 4:31
  • $\begingroup$ I'm so sorry. I mistook $V$ for $\mathbb{C}$ $\endgroup$ – Swapnil Tripathi Dec 11 '14 at 5:43
  • $\begingroup$ Using $J$ to denote the complex conjugation map should be avoided as $J$ is often used to denote an almost complex structure on a vector space (or vector bundle). When I reread my answer below, I thought I had made a huge mistake because I naturally interpreted $J$ as an almost complex structure rather than a complex conjugation. $\endgroup$ – Michael Albanese Apr 3 '15 at 2:54
  • $\begingroup$ Thanks for the follow-up, Michael. For what it's worth, this is an exercise out of Hoffman and Kunze's Linear Algebra text. $\endgroup$ – Bey Apr 3 '15 at 11:36

Given $z \in \mathbb{C}$, we have $\operatorname{Re}(z) = \frac{1}{2}(z + \bar{z})$ and $\operatorname{Im}(z) = \frac{1}{2i}(z - \bar{z})$. Let's try the analogous thing, where in place of $\mathbb{C}$ we have $V$, and instead of conjugation, we have $J$.

Let $w = \frac{1}{2}(v + J(v))$ and $u = \frac{1}{2i}(v - J(v))$; note, we're allowed to multiply by $\frac{1}{2i}$ in the expression for $u$ because $V$ is a complex vector space. Now note that

$$J(w) = J\left(\frac{1}{2}(v + J(v))\right) = \frac{1}{2}(J(v)+J(J(v))) = \frac{1}{2}(J(v) + v) = w$$


$$J(u) = J\left(\frac{1}{2i}(v - J(v))\right) = \frac{-1}{2i}(J(v)-J(J(v))) =\frac{-1}{2i}(J(v) - v) = u.$$

So $w, u \in W$ and

$$w + iu = \frac{1}{2}(v + J(v)) + i\left(\frac{1}{2i}(v - J(v))\right) = \frac{1}{2}(v + J(v)) + \frac{1}{2}(v - J(v)) = v.$$


For completeness, in order to get uniqueness, I would make an argument as follows.

If $z = x + iy = x' + iy'$ where $x, x', y, y' \in W$, then $(x - x') = i(y' - y)$. Since $W$ is a vector space, $x - x' \in W$, so $i(y' - y) \in W$ as well. But $W$ is a vector space over the reals, so if $i(y' - y) \in W$, $y' - y = 0$, which implies $x = x', y = y'$.


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