Let $V$ be a finite dimensional vector space over $\mathbb{Q}$ and suppose $T$ is a nonsingular linear transformation of $V$ such that $T^{-1} = T^2 + T$. Prove that the dimension of $V$ is divisible by $3$. If the dimension of $V$ is precisely $3$, prove that all such transformations $T$ are similar.
So applying $T$ to both sides of the given equation gives us $T^3 + T^2 - I = 0$, hence the minimal polynomial $m(x)$ of $T$ divides $x^3 + x^2 + 1$. But this polynomial is irreducible over $\mathbb{Q}$ by the rational root test, hence $m(x) = x^3 + x^2 + 1$. The structure theorem for finitely-generated modules over PIDs tells that $$(V,T) \cong \bigoplus_{i=1}^{t}\frac{\mathbb{Q}[x]}{(a_i(x))},$$ where $t \geq 1$ and $a_i(x) \mid a_{i+1}(x)$ and $a_t(x) = m(x)$. But since $m(x)$ is irreducible, we must have each $a_i(x) = m(x)$, therefore $$(V,T) \cong \left(\frac{\mathbb{Q}[x]}{(f(x))}\right)^t.$$ Now the dimension of $V$ equals $3t$ and is thus divisible by $3$. But doesn't it then also follow that given a dimension $3t$, any two such transformations $T$ make $V$ isomorphic as a $\mathbb{Q}[x]$-module to the above, hence are similar? It seems true in general, not just for $\dim(V) = 3$. Thanks!
