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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.)

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    $\begingroup$ take determinant of the equation $J^2 = -Id$. $\endgroup$
    – user99914
    Apr 19, 2016 at 18:35
  • $\begingroup$ but what is the connection between determinant and dimension? $\endgroup$
    – DeltaChief
    Apr 19, 2016 at 18:36
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    $\begingroup$ what is $\det (-Id)$? $\endgroup$
    – user99914
    Apr 19, 2016 at 18:37
  • $\begingroup$ "Now $V$ is set to be finite." I believe you mean finite-dimensional. $\endgroup$ Apr 19, 2016 at 18:40
  • $\begingroup$ $\det (-Id)$ is $(-1)^{dimV}$ $\endgroup$
    – DeltaChief
    Apr 19, 2016 at 18:58

2 Answers 2

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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 $\dim V=2p$.

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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.

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