Consider the following result and proof about conjugacy of linear transformations:
Here $E$ and $F$ are $n$-dimensional vector spaces, $\varphi:E\to E$ and $\psi:F\to F$. The $\mu$'s are the minimal polynomials. The proposition referenced says that $\varphi$ and $\psi$ are conjugate if and only if (i) their minimal polynomials have the same prime factors $f_1,\ldots,f_r$ and (ii) $r(f_i(\varphi)^j)=r(f_i(\psi)^j)$ for all $1\le i\le r$ and $j\ge 1$, where $r$ denotes the rank.
My question is: why may we assume that $f$ is a power of a single irreducible based on what is presented in the proof? In other words, why "Thus"?
I know the claim is true, but only for reasons that the author has not established at this point in the text, so I feel I am missing something obvious here.
I'm a bit confused by the presentation: in my book the generalised eigenspace of$~\phi$ for a prime factor $f_i$ of its minimal (or characteristic) polynomial is defined as $\ker(f_i[\phi]^{m_i})$, where $m_i$ is the multiplicity of the factor $f_i$ in the minimal polynomial; in other words each $E_i$ would be equal to $\tilde E_i$ by definition. Anyway, the reason one can assume there is just one prime factor, is that $E$ is canonically a direct sum of the generalised eigenspaces; therefore one can find an isomorphism $\psi:E\to F$ intertwining $f$ and $g$ (that is: $\psi\circ f=g\circ\psi$) if and only if one can find individual isomorphisms $\psi_i:E_i\to F_i$ that intertwine the restrictions of $f$ to $E_i$ and of $g$ to $F_i$ (the constructions in both directions are easy). It is then sufficient to focus one the subspaces for one index$~i$.