only an even dimensional real vector space can admit a complex structure

Solution 1:

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?

Solution 2:

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.