Linear algebra problem from dummite & foote 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!
Since the polynomial is irreducible, it has distinct roots. Let $a,b,c$ be these distinct roots. Then consider the following two matrices.

\begin{pmatrix}
  a &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; a &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; b &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; b &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; 0 &amp; c &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; c 
\end{pmatrix}

\begin{pmatrix}
  a &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; a &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; a &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; a &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; 0 &amp; b &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; c 
\end{pmatrix}

Obviously, these two matrices have the same minimal polynomial $p = x^3 + x^2 - 1$, however they are not similar.
Since the polynomial is irreducible, it has distinct roots. Let $a,b,c$ be these distinct roots. Then consider the following two matrices.

\begin{pmatrix}
  a &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; a &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; b &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; b &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; 0 &amp; c &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; c 
\end{pmatrix}

\begin{pmatrix}
  a &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; a &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; a &amp; 0 &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; a &amp; 0 &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; 0 &amp; b &amp; 0 \\
  0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; c 
\end{pmatrix}

Obviously, these two matrices have the same minimal polynomial $p = x^3 + x^2 - 1$, however they are not similar.