Let $V$ be a finite dimensional vector space over $\mathbb{Q}$ and suppose $T$ is a nonsingular linear transformation of $V$ so that $T^{-1} = T^2+T$. Prove that $3 \mid \dim V$. If $\dim V =3$, prove that all such linear $T$ are similar.
What I have/questions:
We know $T^0 = T^3 + T^2$, i.e., $T^3+T^2-1 =0$. How do I calculate the characteristic polynomial $C(x)$ of $T$? I know $m(x)$, the minimal polynomial of $T$, divides the $C(x)$, so if I can show $C(x)$ is irreducible (by some known test for $\mathbb{Q}[x]$), then I get the following by the fundamental theorem of finitely generated modules over a PID:
$V\cong \bigoplus_{i=1}^{k} \frac{\mathbb{Q}[x]}{(a_i(x))}$ where each $a_i(x)=m(x)$. At this point, I'll need to have determined $m(x)$ (via $C(x))$. But still, how do I determine the dimension of $V$ given the degree of $m(x)$?
Finally, for the second part, I want to show that given $\dim V=3$, for another such linear transformation $S$, the $\mathbb{Q}[x]$-module obtained from $V$ via $S$ is isomorphic to that obtained via $T$. How do I show this?
I appreciate the help.
Let's take it for granted that the polynomial $p(x) = x^3 + x^2 - 1$ is irreducible over $\Bbb Q[x]$, as you say, "by some known test" (the rational roots test will suffice).
Usually, you can't determine $c(x)$ completely just by knowing that $m(x) \mid p(x)$. However, since $p$ is irreducible, we actually know that $m(x) = p(x)$ (assuming we take $m$ to be monic). Moreover, for any linear transformation: any irreducible factor of $c(x)$ must also be a factor of $m(x)$. So, we can deduce that $c(x) = [p(x)]^n$ for some integer $n \geq 1$.
Alternatively: if you had determined $m(x)$, then you'd have $V\cong \bigoplus_{i=1}^{k} \frac{\mathbb{Q}[x]}{(a_i(x))}$ where each $a_i(x)=m(x)$. Since $m(x)$ has degree $3$, $\frac{\mathbb{Q}[x]}{(a_i(x))}$ has dimension $3$, and so the dimension of the direct sum is $3k$.
Finally: for $\dim V = 3$, it suffices to consider an isomorphism $\Bbb Q[x]/m(x) \cong V[T]$ with $x \mapsto T$.