TITLE: 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?
QUESTION [6 upvotes]: Let $V$ be a real n dimensional vector space & let $T:V→V$ be a linear transformation satisfying $T^2(v)=-v$ for all $v\in V$. Then how can we show that $n$ is even?
I am completely stuck on it. Can anybody help me please?

REPLY [4 votes]: A very simple argument, which basically uses just the existence of the complex numbers as a degree $2$ extension field of the real numbers but nothing else, is as follows.
Make $V$ into an $\mathbf R[X]$-module by having $X$ act as $T$ (and so $(\sum_ka_kX^k)\cdot v=\sum_k a_k T^k(v)$ for arbitrary polynomials). Since $X^2+1$ acts as $T^2+I=0$, the module structure passes to the quotient ring $\mathbf R[X]/(X^2+1)\cong \mathbf C$ and $V$ becomes a complex vector space. But then as a real vector space it is even-dimensional (any basis as complex vector space gives rise to a real basis twice as large, by taking the multiples by $1$ and by $\mathbf i$ of all the basis vectors).
The same reasoning shows that if a $K$-linear operator $T$ satisfies an irreducible polynomial equation over $K$ of degree $d$, the $K$-dimension of the space on which $T$ acts must be divisible by $d$.