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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?

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How are you defining the characteristic polynomial? You can show that it is the product of all the elementary divisors of the associated module, and then the result is obvious. –  wj32 Nov 14 '12 at 21:46
    
@wj32 Characteristic polynomial defined as in here: en.wikipedia.org/wiki/… –  hwhm Nov 14 '12 at 21:53

3 Answers 3

up vote 2 down vote accepted

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.

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For the first question, it is easiest to see through some field theory.

First, it should have already been covered that the minimal polynomial has the same roots as the characteristic polynomial, i.e. every eigenvalue is a root of the minimal polynomial.

Let $q$ denote the minimal polynomial and let $p$ denote the characteristic polynomial.

Theorem: The minimal polynomial has the same irreducible factors as the characteristic polynomial.

Proof: Clearly the characteristic polynomial has all the irreducible factors of the minimal polynomial since $q\mid p$.

On the other hand, the two polynomials have the same roots, so let $\pi$ be an irreducible factor of $p$. Then $\pi$ has a root, possibly in some larger field extension, which is also a root of $q$. It is well known that any polynomial which shares a root with an irreducible polynomial must be divided by that polynomial. Therefore $\pi \mid q$. $\square$

Your second question strikes me as a bit weird. Normally the argument is not presented as wlog but rather it is necessary that the $a_i$'s are not zero. The linear combination summing to $1$ is possible only because the $q_i$s are collectively coprime. If $q_i$ is missing for some $i$, then $p_i$ divides all the remaining terms.

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Thanks! I realized the second question was a bit dumb, but for the first, is there a simple explanation without invoking field extensions? –  hwhm Nov 14 '12 at 23:23
    
I have seen a direct proof through clever tricks and then taking determinants, but that is far from illuminating (to this day I have no idea what that proof really did). Understanding wise, I would say that field extensions are the best way. –  EuYu Nov 14 '12 at 23:42

For the record, here is a very simple argument that the each irreducible factor of the characteristic polynomial also divides the minimal polynomial (the other direction follows from the Cayley-Hamilton theorem). It does not use any field extension of the ground field$~K$, but it uses the invariant factor decomposition from the structure theorem of modules of a P.I.D., in this case $K[X]$. As usual the $K[X]$ module is our vector space$~V$, with multiplication by$~X$ defined by the action of the linear operator$~\phi$ under consideration.

Since $V$ is of finite dimension over$~K$, the invariant factor decomposition of$~V$can only contain torsion summands $V_i=K[X]/(P_i)$, where $P_i$ are the invariant factors, a sequence of monic polynomials in $K[X]$ that satisfy $P_1\mid P_2 \mid\cdots\mid P_k$. Each summand $V_i$ is cyclic, and $P_i$ is both the minimal polynomial and the characteristic polynomial of the restriction of$~\phi$ to$~V_i$. Then the characteristic polynomial$~\chi_\phi$ of$~\phi$ is the product $P_1P_2\ldots P_k$ of the invariant factors, while the minimal polynomial$~\mu_\phi$ of$~\phi$ is their least common multiple, which by the divisibility conditions is in fact the final invariant factor: $\mu_\phi=P_k$. The same divisibility conditions then imply that $\chi_\phi$ divides the power $\mu_\phi^k=P_k^k$ of the minimal polynomial, so that each irreducible factor of$~\chi_\phi$ must divide $\mu_\phi$. [One also sees that $P_k\mid \chi_\phi$, without using Cayley-Hamilton.]

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