If $V$ is finite-dimensional with $J : V \to V$ such that $J^2 = -id$, then $V$ has even dimension

Question

Let $V$ be a $\Bbb R$-vector space, with $J$ being an endomorphism $J: V \to V$ with $J^2=-id$ (identity).

I already had to show that $V$ became a $\Bbb C$-vector space with the scalar multiplication: $$(a+bi)\cdot v=av+bJ(v).$$

Now $V$ is set to be finite-dimensional. The problem is the following:

Show that $\operatorname{dim}_{\Bbb R} V$ is even.

Any ideas on how to show this? (I've seen the post If $V$ is a vector space, then, proving that... which has the same problem but I don't really get the solution that is offered there, they write $J$ as a matrix, so far so good, but from there I don't get it.)

Answer 1

As $V$ is a finite-dimensional $\mathbb{C}$-vector space, $\operatorname{dim}_{\mathbb{C}}V$ is an integer, so $\operatorname{dim}_{\mathbb{R}}V = 2\operatorname{dim}_{\mathbb{C}}V$ is even.

By the way, $J$ is sometimes called a linear complex structure.

Answer 2

Use this

$$\det(-id)=(-1)^{\dim V}=\det(J^2)=(\det(J))^2\ge0$$

Alternative Other way to prove the result is to consider the polynomial $P=x^2+1$ which annihilate $J$ so the only complex eigenalues of $J$ are $\pm i$ and since $J$ is real then the characteristic polynomial has real coefficients and $P$ divides it so

$$\chi_J(x)=(x^2+1)^p$$ and then $\dim V=2p$.