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.

Let $M$ be a $n$ by $n$ matrix over a field $F$.

  1. $M$ is diagonizable, i.e. $M=P DP^{-1}$ for some invertible matrix $P$ and some diagonal matrix $D$, if and only if there exists an eigenbasis.

    I wonder if $M$ can be similar to some special matrix if and only if there exists a generalized eigenbasis. Note a generalized eigenvector $v$ for an eigenvalue $\lambda$ with algebraic multiplicity $c$ is defined as a vector which satisfies $ (M - \lambda I)^c v = 0$.

  2. $M$ admits a Jordan decomposition $M=P JP^{-1}$ for some invertible matrix $P$ and some Jordan canonical form $J$, if and only if the characteristic polynomial of $M$ can split into linear factors over $F$.

    Since columns of $P$ form a generalized eigenbasis, "the characteristic polynomial of $M$ can split into linear factors over $F$" implies "there exists a generalized eigenbasis", but I wonder if the reverse is false, i.e. "there exists a generalized eigenbasis" doesn't necessarily imply "the characteristic polynomial of $M$ can split into linear factors over $F$"?

Thanks!

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

Having a generalized eigenbasis is equivalent to the existence of Jordan decomposition.

One way to see this is by noting that $\prod_{\lambda}(M-\lambda I)^{c(\lambda)}=0$ because the image of each vector in the generalized eigenbasis is 0 under this function.

Now, the minimal polynomial of $M$ divides this, so it splits in linear factors, but the characteristic polynomial shares all irreducible factors with the minimal polynomial (albeit with possibly different multiplicities), so the characteristic polynomial also splits into linear factors.

share|improve this answer
add comment

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.