only an even dimensional real vector space can admit a complex structure By definition, a complex structure on a real vector space is a linear map $T$ satisfying $T^2 = -1$. Intuitively, this is possible if and only if the vector space is $2n$-dimensional. But how to prove it?
Starting from a nonzero vector $x$, we can get an invariant subspace spanned by $\{x, Tx\}$. Here we can show that $x$ and $Tx$ must be linearly independent. For otherwise, there exist $a,b\in \mathbb{R}$ such that
$$a x + b T x = 0 .  $$
Applying $T$ on the left, we get
$$ - b x + a Tx = 0 . $$
From these two equations we can get $(a^2 + b^2 )x = 0 $, in contradiction with our assumption.
But how to proceed further? Apparently, I now need a complement subspace.
 A: Take any $\ y\not\in\text{span}\{x,Tx\}\ $, and suppose $\ ay+bTy+cx+dTx=0\ $. Then
$$
aTy-by+cTx-dx=0\ ,
$$
from which we get
$$
\big(a^2+b^2)y=(bc-ad)Tx-(ac+bd)x\ .
$$
Because $\ y\not\in\text{span}\{x,Tx\}\ $, it follows that $\ a=b=0\ $, and thence that $\ c=d=0\ $.  So $\ y,Ty,x,Tx\ $ are linearly independent.   Can you see how to use induction to complete the argument?
A: The idea: show that the characteristic polynomial of $T$ has even degree, since the characteristic polynomial's degree equals the dimension of the underlying space.
If $T^2=-1$, then the polynomial $X^2+1$ has $T$ as its root. So the minimal polynomial of $T$ divides $X^2+1$. At the same time, $X^2+1$ is irreducible over $\mathbb R$, so this polynomial already is the minimal polynomial.
Now over $\mathbb C$, it splits into $(X+\mathrm i)(X-\mathrm i)$. But then the characteristic polynomial $\chi_T$ is a product of powers of the two linear factors $X+\mathrm i$ and $X-\mathrm i$. So $\chi_T=(X+\mathrm i)^m(X-\mathrm i)^n$ for positive integers $m,n$.
Now since $T$ acts on a real vector space, the coefficients of its characteristic polynomial are real. And the roots of polynomials with real coefficients come in conjugate pairs, if they're not real. This means $m=n$, implying that the degree of $\chi_T$ is even.
