Vtgyjfy

This is an exercise of Advanced Linear Algebra by Roman:

Let $text{dim}(V)<infty$. If $tau, sigma in mathcal{L}(V)$ prove
that $sigmatau = iota$ implies that $tau$ and $sigma$ are
invertible and that $sigma = p(tau)$ for some polynomial
$p(x) in F[x]$.

The difficulty in this exercise is that it is stated prior to anything related to eigenvalues or the Caley-Hamilton theorem. I want to confirm that the proof below is correct or, by the contrary, if there is some flaw somewhere.

If $tau$ is not invertible then there is some $xin Vsetminus{0}$ such that $tau x=0$, but this would imply that $sigmatau x=0$, what is not possible, so $tau$ is invertible. If $tau$ is invertible then $tau$ is an automorphism, what imply that $sigma$ can take any value, so $sigma$ is invertible by the same reasons than $tau$.

Then $sigma=tau^{-1}$ and we want to show that there is a polynomial $p$ such that $tau^{-1}=p(tau)$. Now suppose that $dim V=n$ and let $v_1,ldots,v_n$ a basis of $V$, then for each $kin{1,ldots,n}$ choose the minimal list that contains $tau^{-1}v_k$ where $tau^{-1}v_k,tau^n v_k,tau^{n+1} v_k,ldots$ is linearly dependent. Then there are $c_jin Fsetminus{0}$, $m_jge 0$ and some $1le n_k< n$ such that
$$
c_{-1}tau^{-1}v_k+sum_{j=1}^{n_k}c_jtau^{n+m_j}v_k=0
$$
Then setting $r_k(x):=c_{-1}x^{-1}+sum_{j=1}^{n_k}c_j x^{n+m_j}$ then the $r_k$ are rational functions and $r:=prod_{k=1}^n r_k$ is a rational function with the form $r(x)=gamma x^{-1}+p(x)$ for some polynomial $pin F[x]$ such that $p(0)=0$ and $gammaneq 0$. Also by construction we find that $r(tau)=0$ what imply that $tau^{-1}=-gamma^{-1} p(tau)$, what finishes the proof.

EDIT: ¡ah! There is a little flaw in the proof above, I would say that $r(x)=gamma x^{-n}+p(x)$ instead, then $tau^{-1}=-gamma^{-1}tau^{n-1}p(x)$ holds.

Also the condition $p(x)=0$ was unnecesary, I dont know what I was thinking to impose this condition.

What about the following argument? (This avoids the use of Caley-Hamilton theorem.) Let us consider $${iota,tau,ldots,tau^{n^2+1}}subset mathcal{L}(V).$$ Since $dim mathcal{L}(V)=n^2<infty$, it follows that $iota,tau,ldots,tau^{n^2+1}$ are linearly dependent and there exists a non-zero polynomial $p(t)inmathbb{F}[t]$ such that $p(tau)=O$. Let $m(t)$ be the monic polynomial of the smallest degree for which $m(tau)=O$ (i.e. $m$ is the minimal polynomial of $tau$.) If $m(0)=0$, then we may write
$$
m(tau)=m^*(tau)cdottau=O
$$ where $m^*(t)=frac{m(t)}{t}neq 0$ is a polynomial of degree $deg m -1$. This impiles $m^*(tau)=O$ and contradicts the minimality of $m$. This shows that $m(0)=lambda neq 0$ and that
$$
m^{**}(tau)cdottau = -lambda iota
$$ for $m^{**}(t)=frac{m(t)-m(0)}{t}$. It follows that $sigma = -frac{1}{lambda}m^{**}(tau)$ as desired.