# If $A$ is a matrix and $p$ is a polynomial such that $p(A)=0$ then must the roots of $p$ necessarily be eigenvalues of $A$?

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.

• 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

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 $A$.