# 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?

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?

-
Yeah, it's called "the epsilon relation", but it has a better symbol than $\epsilon$. You can use \in. – Asaf Karagila Dec 3 '12 at 13:09

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 ↯.

-
Wow! Brilliant! – Inquest Dec 3 '12 at 15:09

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 $\det (T) \in \mathbb{R}$.

-

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 $d$.

-
This is really the most natural proof, 1+. – Martin Brandenburg Jan 26 '13 at 12:01

The following argument does not make use of the order completeness of ${\mathbb R}$ or of the fundamental theorem of algebra. Therefore it works also with ${\mathbb R}$ replaced by ${\mathbb Q}$.

Assume $U$ is an arbitrary invariant subspace of $T$. Given any $y\in V\setminus U$ (in particular $y\ne0$) we can form the span $W:=\langle y, Ty\rangle$, which is obviously an invariant subspace, too. Suppose that for some real $\alpha$, $\beta$ we have $$z:=\alpha y+\beta Ty \in U\ .$$ Then $\alpha z -\beta Tz=(\alpha^2+\beta^2) y$ lies in $U$ as well, which implies $\alpha=\beta=0$. Applying this to the case $U=0$ we infer that ${\rm dim}(W)=2$. For arbitrary invariant $U$ we conclude that $U\cap W=0$, so that $U':=U\oplus W$ is an invariant subspace of $T$ of dimension ${\rm dim}(U')={\rm dim}(U)+2$.

This allows the construction of an ascending chain of even dimensional invariant subspaces $U_{2k}$ $\ (k\geq 0)$ of dimension $2k$, which eventually has to end. At this point $U_{2n}=V$.

-