A problem in Werner Greub's, Linear Algebra. The problem states the following: 
Let E be a vector space over a field $\Gamma$ of characteristic 0. Let $j:E \to E$ be a linear transformation such that $j^2=-id_{E}$. Then show that the dimension of E is even.
I can show this for a vector space over the reals or the rationals. However my proof fails for arbitrary fields of characteristic zero. Is the statement still true for arbitrary fields of characteristic zero?
 A: For $\Gamma = \mathbb{C}$ and $E = \mathbb{C}$ the $\mathbb{C}$-linear map $E \to E$, $x \mapsto ix$ a counterexample.
More generally:
The claimed statement holds for any field $\Gamma$ (of arbitrary characteristic) if and only if $-1$ does not have not square root in $\Gamma$:
If such a square root $\omega \in \Gamma$ exists, the one can take $E = \Gamma$ and the $\Gamma$-linear map $E \to E$, $x \mapsto \omega x$ as an counterexample.
If no such square root exists then let $j \colon E \to E$ be a linear transformation with $j^2 = -\operatorname{id}_E$ for some finite-dimensional vector space $E$.
Then
$$
    \det(j)^2
  = \det(j^2)
  = \det(-\operatorname{id}_E)
  = (-1)^{\dim E} \det(\operatorname{id}_E)
  = (-1)^{\dim E},
$$
which shows that $\det(j)$ is a square root of $(-1)^{\dim E}$.
By assumption $-1$ has no square root, so it follows that $(-1)^{\dim E} = 1$.
Note that $\operatorname{char} \Gamma \neq 2$ since otherwise $-1 = 1$ would have a square root.
Hence it follows from $(-1)^{\dim E} = 1$ that $\dim E$ is even.
A: It is false over fields $F$ of characteristic $2$. Just take $\operatorname{Id}\colon F^n\longrightarrow F^n$, for any natural $n$.
