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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!

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  • $\begingroup$ It seems to me that you're correct. Not sure why no one has already said so... ? $\endgroup$ – Dustan Levenstein Apr 7 '16 at 20:35
  • $\begingroup$ Yeah I checked with my adviser at school and he agreed with me. Thanks! $\endgroup$ – Ethan Alwaise Apr 8 '16 at 0:38
  • $\begingroup$ I realize that this thread was almost a year ago, but I'm confused as to how you arrived at $m(x) = x^3 + x^2 + 1$ from $T^3 + T^2 - I = 0$. Shouldn't we have arrived at $m(x) = x^3 + x^2 - 1$? $\endgroup$ – Enrico Borba Mar 2 '17 at 19:42
  • $\begingroup$ You're right that it should be $m(x) = x^3 + x^2 -1$, but the rest of the math works out the same anyway. $-1$ and $1$ are still the only possible roots by the rational roots test, and neither of those actually works when you plug them into $m(x)$. $\endgroup$ – Tanner Strunk Apr 18 '17 at 17:49
  • $\begingroup$ @EthanAlwaise: see this $\endgroup$ – Bumblebee Feb 1 '18 at 16:10
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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 & 0 & 0 & 0 & 0 & 0 \\ 0 & a & 0 & 0 & 0 & 0 \\ 0 & 0 & b & 0 & 0 & 0 \\ 0 & 0 & 0 & b & 0 & 0 \\ 0 & 0 & 0 & 0 & c & 0 \\ 0 & 0 & 0 & 0 & 0 & c \end{pmatrix}

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

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

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    $\begingroup$ Ok I see that since their characteristic polynomials are different but I still don't understand where I went wrong. Plus those matrices don't represent operators on $V$ since the entries $a,b,c$ are not in $\mathbb{Q}$. $\endgroup$ – Ethan Alwaise Mar 23 '16 at 23:50

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