# A real vector space with a complex structure is naturally a complex vector space

I am struggling with this exercise from the book 'Tensors and Manifolds: With Application to Physics', by Robert H. Wasserman:

Corresponding to each $a \in \mathbb{K}$ there is a linear operator $\phi$ on $V$ given by $\phi:V\mapsto aV$. In particular, for the space $\mathbb{C}^n$, corresponding to $i=\sqrt{-1}$, we have a linear operator $J$ with the property that $J^2 v = -v$. On the real vector space $\mathbb{R}^{2n}$, the linear operator $J$ defined by

$$J: \left\{\begin{array} ee_k \mapsto e_{k+n} \quad k=1,\dots,n \\ e_k \mapsto e_{k-n} \quad k=n+1,\dots,2n \end{array}\right.$$ where $\{e_1,\dots,e_{2n}\}$ is the natural basis of $\mathbb{R}^{2n}$, has this property. Show that for any real vector space which admits such an opertor, called a complex structure, the same elements can be made into a complex vector space by defining multiplication by a complex scalar by $(a+bi)v = av+bJv$.

I tried to develop some intuition for this problem by considering the special case where $n=1$, i.e., $V=\mathbb{R}^{2}$. The natural basis is simply $\{(1,0),(0,1)\}$. First, by applying the given definition, we have: $$J(1,0)=(0,1)\\ J(0,1)=(1,0)$$ Which implies: $$J^2(1,0)=(1,0)\\ J^2(0,1)=(0,1)$$ This is not the desired property for $J$ in that we should have $J^2(1,0)=-(1,0)$ and so on.

However, if I simply define $J$ as: $$J(v_1,v_2)=(-v_2,v_1)$$ Everything fits together, because $J^2(v_1,v_2)=-(v_1,v_2)$ and we also have that: $$a(v_1,v_2)+bJ(v_1,v_2) = (av_1,av_2)+(-bv_2,bv_1)=(av_1-bv_2,av_2+bv_1)$$ Which is the definition of multiplication by a complex number.

How can I make the definition given in the problem fit in this picture?

The exercise is wrong and you are right. There's a typo, exactly as you predicted. It should define $J(e_k) = -e_{k-n}$ for $k>n$. (As it stands, they defined reflections, rather than rotations.)