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Let $V$ be a finite dimensional vector space over $\Bbb Q$ and let $T$ be non-singular linear transform on $V$ such that $T^{-1}=T^2+T$

Prove that the dimension of $V$ is divisible by $3$.

If $\dim V=3$ then show that all such linear transformations are similar.

This question is exactly a duplicate of this question.

But the problem is I have encountered this problem only after having a course in Linear Algebra and so I don't know what is meant by finitely genearted modules over a PID and so on.

Can't it be solved using Linear Algebra techniques only.

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  • Mark has deleted his answer, but it's basically correct. It's not really important that $m(x)$ be the minimal polynomial of $T$, only that $\phi_T(x)$ has the same roots as an irreducible polynomial. Since $m(x)$ is irreducible, whenever one is a root of a given polynomial, the other two are as well. So you can keep dividing out copies of $m(x)$ until $\phi_T(x)/m(x)^r$ has no more roots (has degree zero). This is entirely analogous to the idea that complex roots of a real polynomial always appear in conjugate pairs. – Erick Wong Dec 17 '16 at 05:31
  • see this – Bumblebee Feb 01 '18 at 16:07

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