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We're given $V$ a finite dimensional vector space over $\mathbb{Q}$, $T$ a non-singular linear transformation of $V$ such that $T^{-1} = T^{2} + T$. The question has two parts. If I understand part (a), I should be able to get part (b), so right now I'm looking for a hint on part (a):

(a) Prove that the dimension of $V$ is divisible by 3. (b) Prove that if the dimension is exactly 3, then all such transformations $T$ are similar.

I'm trying to think of $T$ as a matrix. I can get $v = A^{3}v + A^{2}v$ for any $v$, but I don't know where to go from there.

Xam
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BMac
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    Can you prove that the polynomial $x^3+x^2-1$ is irreducible over $\Bbb{Q}$? – Jyrki Lahtonen Aug 20 '17 at 21:50
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    Anyway, an attack I could spot is that given $v\in V, v\neq0$, the three vectors ${v,Tv,T^2v}$ are necessarily linearly independent. Furthermore, the 3-d subspace $W_v$ they span is stable under $T$. This, in turn, implies that if you pick another vector $u\notin W_v$, the same recipe gives another 3-d subspace $W_u$ intersecting trivially with $W_v$. Eventually you get all of $V$ as a direct sum of such 3-d subspaces. – Jyrki Lahtonen Aug 20 '17 at 21:57
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    If you didn't see all the details: a key point is that a "new" 3-d subspace $W_u$ cannot intersect non-trivially with the direct sum of the earlier ones, because then that intersection would be stable under $T$, but a vector $w$ from it still gives rise to a 3-d subspace $W_w$. – Jyrki Lahtonen Aug 20 '17 at 22:06
  • Going back to your polynomial suggestion, is the idea to think in field extension terms by showing that this is the minimal polynomial for our linear transformation (in matrix terms, $A^{3} + A^{2} - I$ is 0, and no lower power of $A$ will suffice), so the dimension of the vector space is the degree of this polynomial? – BMac Aug 20 '17 at 22:52
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    $V$ is a vector space over the field $K = \mathbb{Q}(X) / (X^3 + X^2 - 1)$ and $\mathrm{dim}_{\mathbb{Q}}(V) = 3 \cdot \mathrm{dim}_K(V)$, since $K/\mathbb{Q}$ has dimension $3$. Here $X \in K$ acts on $V$ by $T$. – user399601 Aug 20 '17 at 22:54
  • This looks like what I was thinking, except I must have gotten confused, because I thought we should think of our field extension $K$ as being our $V$. – BMac Aug 20 '17 at 22:56
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    @user399601's way is surely the simplest (if you have covered field extensions). – Jyrki Lahtonen Aug 21 '17 at 05:13

1 Answers1

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(a) We can rewrite the given condition as $T^3+T^2-I=0$ and $x^3+x^2-1$ is irreducible over $\Bbb{Q}.$ Therefore this is the minimal polynomial $m_T(x)$ satisfies by $T.$

Since characteristic polynomial $C_T(x)$ and minimal polynomial has same zeros, here the only possibility is $C_T(x)=(m_T(x))^r.$ Also $\dim_{\Bbb{Q}}V=\deg(C_T(x))=3r.$

OR: we can note that $m_T(x)$ is the only invariant factor of $T.$ Hence $$V=\Bbb{Q(x)}/(m_T(x))\oplus\Bbb{Q(x)}/(m_T(x))\oplus\cdots \oplus\Bbb{Q(x)}/(m_T(x))$$ (finite direct sum) and further $\Bbb{Q(x)}/(m_T(x))=\langle 1, x, x^2\rangle_{\Bbb{Q}}$ implies $3|\dim_{\Bbb{Q}}V.$

(b) To answer this part, we can easily see that $C_T(x)=m_T(x).$ Therefore this define a unique Smith normal form $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & x^3+x^2-1 \\ \end{pmatrix} $$ and unique Frobenius normal form $$ \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & -1 \\ \end{pmatrix} .$$ Hence the matrix representation of all such transformations has to be similar to the above second matrix.

Bumblebee
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