Let $V$ a $\mathbb{Q}$-linear space, $\dim_\mathbb{Q}V<\infty$, $T: V \rightarrow V$ a linear operator such that $T^2 = -Id$. If $V$ has a $T$-invariant proper subspace $W$, $\dim(W) \ge 1$, then the smallest dimension for $V$ is 4.
First of all, the minimal polynomial of $T$ is $m_T(t) = t^2 +1$, right?
I divided in cases:
- If $\dim V = 1$, there is no proper subspace which dimension is $\ge 1$.
- If $\dim V = 2$, then $\dim W = 1$. So, $W = [w]$, $w \in W$. Then $m_T(T)(w) = 0$. It implies that $m_{T,w}|m_T$ (Here $m_{T,w}$ is the minimal monic polynomial that annihilates $T$ at $w$). So, $m_{T,w}(t) = t^2 + 1$, and $\dim W \ge deg(m_{T,w}) = 2$, which is an absurd.
- If $\dim V = 3$, by Cayle-Hamilton Theorem, $m_T | c_T$, and both have the same irreductible factors, which implies $c_T = (m_T)^k, k \in \mathbb{N}$. Then $\deg (c_T) = 2k$ which is even. Since $\dim V = \deg(c_T)$, we have a contradiction.
For the case $\dim V = 4$, let $T:\mathbb{Q}^4 \rightarrow \mathbb{Q}^4$, such that $T(x,y,z,w) = (-y,x,-w,z)$ and let $W = [e_1,e_2]$. Then, $T^2 = -Id$ and $T(W) \subseteq W$.
The characteristic polynomial, by Cayley-Hamilton theorem, has the same prime divisors (because it has the same roots) as $t^2 +1$, which is irreducible. That means that the characteristic polynomial must be a power of $t^2+1$, so it must have even degree. The degree is equal to the dimension of $V$. That excludes the case $\dim V = 3$ which is odd.