Let $T:V→V$ be a linear transformation satisfying $T^2(v)=-v$ for all $v\in V$. How can we show that $n$ is even?
Since $T^2(v)=-v \ \ \forall v \in V \Rightarrow T$ does not have any real eigenvalues (if $T(v)=\lambda v, \ v\neq0\Rightarrow T^2(v)=\lambda^2 v = -v \Rightarrow \lambda^2=-1$ ).
If $n$ is odd the characteristic polynomial of $T$ has odd degree therefore at least one real eigenvalue ↯.
Since $T^2(v) = -v$ for every $v \in V$, we have $T^2 = -I_n$, where $I_n$ is the $n\times n$ identity matrix. Therefore, $$\det(T)^2 = \det(T^2) = \det(-I^n) = (-1)^n.$$ If $n$ is odd then $(-1)^n = -1$ and thus $\det(T)^2 = -1$. But this is impossible since $\det (T) \in \mathbb{R}$.