Here are two questions that I currently have about the minimum polynomial and characteristic polynomial of a linear endomorphism on a finite dimensional space:
- If the characteristic polynomial $c = p_1^{d_1} p_2^{d_2} \dotsm p_k^{d_k}$, where each $p_i$ is irreducible and $d_i \neq 0$, we know the minimum polynomial $m = p_1^{e_1} p_2^{e_2} \dotsm p_k^{e_k}$. Why can it not be the case that some $e_i = 0$?
- A proof of the primary decomposition theorem starts like this: For $k > 1$, let $q_i = m/p_i^{e_i}$. Then the $q_i$'s are coprime, so there are polynomials $a_i$ such that $a_1 q_1 + \dotsb + a_k q_k = 1$. Why it is wlog that all $a_i \neq 0$?
Thank you for your help!
Edit: Is there a simple explanation without extending the ground field?
Best Answer
For your first question, assume that the characteristic polynomial had an irreducible factor $p_i$ that does not divide the minimal polynomial. Then extending the ground field (by writing down a matrix for the endomorphism and interpreting it as one of a linear transformation $\phi$ over the larger field) we can obtain a root of $p_i$, which would be an eigenvalue of $\phi$ (a root of its characteristic polynomial, which hasn't changed by the field extension) without being a root of its minimal polynomial (which hasn't changed either). But every eigenvalue of $\phi$ must be a root of its minimal polynomial, since any polynomial $P(\phi)$ of $\phi$ acts on an eigenvector $v$ with eigenvalue $\lambda$ by the scalar $P(\lambda)$, and when $P$ is the minimal polynomial one has $P(\phi)=0$, whence $P(\lambda)=0$.
For your second question, $p_i$ divides every $q_j$ with $j\neq i$, so if $a_i=0$ then one would have that $p_i$ divides $a_1 q_1 + \dotsb + a_k q_k = 1$, which is absurd. (You don't need to apply wlog to get $a_i\neq0$.)
In reply to the added question, here is a proof for the first question entirely working over the original field. It will be a hardly inspiring induction on the dimension, where in addition it is hard to see where exactly is the crux of the proof, but that is because of a real difficulty: what exactly can we say about the characteristic polynomial $\chi_\phi$ of $\phi\in\operatorname{End}(V)$ without mentioning its roots, the eigenvalues (which may not exist over the original field)? I will use the following fact, which follows from the computation of the characteristic polynomial of a block triangular matrix: if $W$ is a $\phi$-stable subspace of $V$, then $\chi_\phi$ is the product of the characteristic polynomials of the endomorphisms $\phi|_W$ of $W$ and $\phi_{V/W}$ of $V/W$ defined by $\phi$. In this situation it is also clear that if some polynomial of $\phi$ vanishes (on $V$) then the same polynomial of $\phi|_W$ and of $\phi_{V/W}$ also vanish (on $W$ and $V/W$ respectively; the converse may fail), so the minimal polynomials of $\phi|_W$ and of $\phi_{V/W}$ divide that of $\phi$.
Now I will prove "every irreducible factor $p_i$ of $\chi_\phi$ divides the minimal polynomial of $\phi$" by induction on $\dim V$. If $\dim V=0$ then both polynomials are unity and there is nothing to prove. Otherwise choose a nonzero vector $w$ and let $W$ be the span of all vectors $\phi^i(w)$ for $i\geq0$ (the smallest $\phi$-stable subspace containing $w$). If $d$ is minimal such that $\phi^d(w)$ is linearly dependent on $w,\phi(w),\ldots,\phi^{d-1}(w)$ then $d=\dim W$ is nonzero since $w\neq0$, and the minimal polynomial $P$ of $\phi|_W$ cannot be of degree less than $d$; therefore it is of degree $d$ and equal to the characteristic polynomial of $\phi|_W$. Since $P$ divides the minimal polynomial of $\phi$, we are done if $p_i$ divides $P$, so assume this is not the case. But then $p_i$ divides the characterisitic polynomial of $\phi_{V/W}$; by the induction hypothesis it divides the minimal polynomial of $\phi_{V/W}$, and therefore the one of $\phi$.
Although I appeared to use the Cayley-Hamilton theorem in concluding equality of minimal and characteristic polynomials of $\phi|_W$, this can in fact be proved by a direct calculation (the matrix in the basis $w,\phi(w),\ldots,\phi^{d-1}(w)$ is a companion matrix), and a proof of the Cayley-Hamilton theorem along these lines is given in this answer.