Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to understand the similarities and differences between the minimal polynomial and characteristic polynomial of Matrices.

  1. When are the minimal polynomial and characteristic polynomial the same
  2. When are they different
  3. What conditions (eigenvalues/eigenvectors/...) would imply 1 or 2
  4. Please tell me anything else about these two polynomials that is essential in comparing them.
share|improve this question

2 Answers 2

up vote 7 down vote accepted

The minimal polynomial is quite literally the smallest (in the sense of divisibility) nonzero polynomial that the matrix satisfies. That is to say, if $A$ has minimal polynomial $m(t)$ then $m(A)=0$, and if $p(t)$ is a nonzero polynomial with $p(A)=0$ then $m(t)$ divides $p(t)$.

The characteristic polynomial, on the other hand, is defined algebraically. If $A$ is an $n \times n$ matrix then its characteristic polynomial $\chi(t)$ must have degree $n$. This is not true of the minimal polynomial.

It can be proved that if $\lambda$ is an eigenvalue of $A$ then $m(\lambda)=0$. This is reasonably clear: if $\vec v \ne 0$ is a $\lambda$-eigenvector of $A$ then $$m(\lambda) \vec v = m(A) \vec v = 0 \vec v = 0$$ and so $m(\lambda)=0$. The first equality here uses linearity and the fact that $A^n\vec v = \lambda^n \vec v$, which is an easy induction.

It can also be proved that $\chi(A)=0$. In particular that $m(t)\, |\, \chi(t)$.

So one example of when (1) occurs is when $A$ has $n$ distinct eigenvalues. If this is so then $m(t)$ has $n$ roots, so has degree $\ge n$; but it has degree $\le n$ because it divides $\chi(t)$. Thus they must be equal (since they're both monic, have the same roots and the same degree, and one divides the other).

A more complete characterisation of when (1) occurs (and when (2) occurs) can be gained by considering Jordan Normal Form; but I suspect that you've only just learnt about characteristic and minimal polynomials so I don't want to go into JNF.

Let me know if there's anything else you'd like to know; I no doubt missed some things out.

share|improve this answer
    
I am sorry, what is "(1)"? –  blue Sep 4 '13 at 11:28
    
@blue: The numbers refer to the numbers in the question. –  Clive Newstead Sep 4 '13 at 13:34
    
Thank you. (more characters) –  blue Sep 4 '13 at 14:01

The minimal polynomial $m(t)$ is the smallest factor of the characteristic polynomial $f(t)$ such that if $A$ is the matrix, then we still have $m(A) = 0$. The only thing the characteristic polynomial measures is the algebraic multiplicity of an eigenvalue, whereas the minimal polynomial measures the size of the $A$-cycles that form the generalized eigenspaces (a.k.a. the size of the Jordan blocks). These facts can be summarized as follows.

  • If $f(t)$ has a factor $(t - \lambda)^k$, this means that the eigenvalue $\lambda$ has $k$ linearly independent generalized eigenvectors.
  • If $m(t)$ has a factor $(t - \lambda)^p$, this means that the largest $A$-cycle of generalized eigenvectors contains $p$ elements; that is, the largest Jordan block for $\lambda$ is $p \times p$. Notice that this means that $A$ is only diagonalizable if $m(t)$ has only simple roots.
  • Thus $f(t) = m(t)$ if and only if each eigenvalue $\lambda$ corresponds to a single Jordan block, a.k.a each eigenvalue corresponds to a single minimal invariant subspace of generalized eigenvectors.
  • $f(t)$ and $m(t)$ differ if any eigenvalue has more than one Jordan block, a.k.a. if an eigenvalue has more than one generalized eigenspace.
share|improve this answer
    
What definition of generalized eigenspace has room for having more than one of them for a given eigenvalue? I always thought it ways the kernel of $(A-\lambda)^p$ where $p$ is sufficiently large (sufficient meaning that you don't get a larger kernel for higher powers; you can take for $p$ the multiplicity of $t-\lambda$ in $m(t)$ to be definite). –  Marc van Leeuwen Apr 20 '13 at 21:20
    
@MarcvanLeeuwen, sorry, I just saw this message (7 months late!) Yes, I suppose my definition of generalized eigenspace was one that corresponded to a Jordan block, but you're right that that is not the "correct" definition. –  Christopher A. Wong Dec 3 '13 at 19:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.