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.

This is just a small query:

If $A$ is an $n\times n$ square matrix and $p(t)=(t-\lambda_1)(t-\lambda_2)\cdots(t-\lambda_m)$ be a polynomial (with $\lambda_i \in \mathbb{C}$ for all $i=1, \ldots, m$) such that $p(A)=0$, then is it necessary that $\lambda_1,\ldots,\lambda_m$ will be eigenvalues of $A?$

Now I know that if $p(A)=0$ then the minimal polynomial $m_A$ divides $p$; and as $m_A$ is a non-zero polynomial there should be at least one $\lambda_i$ which will be an eigenvector of $A$. But apart from that, I have no idea what to do for this question. The following question is something that I thought I might add with the first one in order to straighten up my understandings in this area.

If not, what extra condition can be imposed on the polynomial and/or on the matrix to make sure that the $\lambda_i's$ are necessarily eigenvalues of $A$?

Thanks and regards.

share|improve this question
    
No, for the simple reason that $p$ may have roots that are not eigenvalues of $A$. –  hardmath Feb 6 '13 at 14:22
    
The first condition to be made is the irreducibility of course. –  awllower Feb 6 '13 at 14:22
    
@awllower Irreducibility of what as a condition to ensure what? –  Did Feb 6 '13 at 14:54
    
Sorry for not specifying. I meant the irreducibility of the polynomial $p(x)$, in order that $p(x)$ contains no other roots than eigenvalues. –  awllower Feb 6 '13 at 15:12
    
@awllower Offtopic, I am afraid. (And, say, not many polynomials on $\mathbb C$ are irreducible...) –  Did Feb 15 '13 at 7:56
show 1 more comment

1 Answer

up vote 8 down vote accepted

No, of course not. $A$ is a root of any polynomial $p$ divisible by $m_A$, and you can always give a polynomial more roots by multiplying it by something. For example, let $A$ be the zero matrix. This has minimal polynomial $t$, and so is a root of $p(t)=t(t-1)(t-2)$ for example, but $1$ and $2$ are not eigenvalues of $A$.

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.