Problem: Let $V$ be a finite dimensional real vector space. Suppose $T:V \to V$ be a linear transformation such that $T\circ T = -Id$, that is, $T(T(v)) = -v$ for $v \in V$. Then prove that $V$ is even dimensional.
Solution: et the dimension of $V$ be $n$. Choose a basis of $V$ and corresponding to that basis we look at the determinant of the matrix corresponding to the linear transformation $T$. We have
\[
\det (T \circ T) = \det (-Id) \implies \det (T)^2 = (-1)^n.
\]
Since the vector space is over $\mathbb{R} $, the determinant must be real number and hence $\det T^2 \geq 0$. Therefore, $(-1)^n$ must be positive and hence $n$ must be even.