I was wondering what the relationship between the minimal polynomial and the Jordan Canonical Form is. Given a matrix, all one needs to do is to compute the characteristic polynomial to determine the Jordan Canonical Form, and using a dot diagram (a la Friedberg), it is unique. Thus, I am wondering what additional information the minimal polynomial gives.

For example, we have the following question.

Find all possible $7 \times 7$ Jordan Canonical Forms for a matrix with characteristic polynomial $$\chi(t) = t^2(2-t)^3(3-t)(4-t)$$ whose minimal polynomial is the same as the characteristic polynomial.


See Chapter 7.3, Problem 13 from Friedberg.

If $T$ is in $\mathcal L(V)$ and $\chi_T(t)$ splits, let $λ_1, λ_2, \dots, λ_k$ be the distinct eigenvalues of $T.$ For each integer $1 \leq i \leq k,$ let $p_i$ be the order of the largest Jordan block corresponding to $λ_i$ in a Jordan Canonical Form of $T$. Then, the minimal polynomial of $T$ is $$(t-λ_1)^{p_1}(t-λ_2)^{p_2} \cdots (t-λ_k)^{p_k}.$$

  • $\begingroup$ I see Jordan Canonical Form is a matrix of blocks of complex number $\lambda$ on the diagonal and 1 'above' it, (mathworld.wolfram.com/JordanCanonicalForm.html), would you tell what it's mainly for? $\endgroup$ – Charlie Chang Jul 31 at 14:47
  • $\begingroup$ @CharlieChang, the Jordan Canonical Form of a matrix whose characteristic polynomial splits (or a matrix over an algebraically closed field) exists and is unique, and similar matrices have the same Jordan Canonical Form. Further, the Jordan Canonical Form is the sum of a diagonal matrix and a nilpotent matrix, so computations with the Jordan Canonical Form are usually much easier than with the original matrix itself. $\endgroup$ – Carlo Aug 1 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.