The theorem of Cayley-Hamilton says that a matrix satisfies it's characteristic polynomial. But can we also make a statement about the eigenvalues if a matrix satisfies a monic polynomial in general?

So let's say that we have a matrix $A \in \mathbb{C}^{m,m}$ and a monic polynomial $p$ of degree $n$ (with $n<m$) for which $p(A)=0$. What can be said about the eigenvalues of $A$ with this information?


By applying the polynomial $p[A]$ in$~A$ term by term to an eigenvector of eigenvalue$~\lambda$, one sees that it acts on it as multiplication by $p[\lambda]$. But on the other hand $p[A]=0$, so one must have $p[\lambda]=0$, and $\lambda$ is a root of$~p$. Every eigenvalue of$~A$ must be a root of$~p$ (but not every root of$~p$ needs to be eigenvalue).

This is elementary, and much easier than Cayley-Hamilton.

  • $\begingroup$ Isn't it true that any polynomial $p$ such that $p(A)=0$ can be factored as $p=rs$, where $r$ is a polynomial and $s$ is the minimal polynomial of $A$? In that case, we would know also something about the minimal algebraic multiplicities of the roots of $p$ which are the eigenvalues of $A$. $\endgroup$ – Algebraic Pavel Jan 28 '14 at 11:19
  • $\begingroup$ So actually (after reading again the OP's question) if the monic polynomial $p(t)=\prod_{i=1}^k(t-\mu_i)^{n_i}$ is given such that $p(A)=0$, then either $\mu_i$ is not an eigenvalue of $A$ or $\mu_i$ is an eigenvalue of $A$ and the dimension of the largest Jordan block associated with $\mu_i$ does not exceed $n_i$. In addition, $\sigma(A)\subset\{\mu_1,\ldots,\mu_k\}$. :-) $\endgroup$ – Algebraic Pavel Jan 28 '14 at 11:25
  • $\begingroup$ Thank you for your comment! $\endgroup$ – user89987 Jan 28 '14 at 11:35
  • $\begingroup$ @AlgebraicPavel: Indeed, the size of the largest Jordan block for$~\lambda$ equals the multiplicity of$~\lambda$ as root of the minimal polynomial, which polynomial divides any annihilating polynomial such as $p$ (so the multiplicity as root of$~p$ can only become bigger). $\endgroup$ – Marc van Leeuwen Jan 28 '14 at 11:59

If $\Bbb F$ is any field, and $A \in \Bbb F^{m, m}$, and $\lambda \in \Bbb F$ is an eigenvalue of $A$ so that

$Av = \lambda v, \tag{1}$

for $0 \ne v \in \Bbb F^m$, then

$A^2 v = A(Av) = A(\lambda v) = \lambda Av = \lambda^2 v, \tag{2}$

and if

$A^kv = \lambda^k v, \tag{3}$


$A(A^k v) = A(\lambda^k v) = \lambda^k (Av) = \lambda^{k + 1} v, \tag{4}$

which, given (1) and (2) basically completes a very simple inductive proof that

$A^n v = \lambda^n v \tag{5}$

for all positive $n \in \Bbb Z$. Taking things one step further, we have

$a_n A^n v = a_n \lambda^n v \tag{6}$

for any $a_n \in \Bbb F$. From (6) we deduce that if $p(x) \in \Bbb F[x]$ with, say,

$p(x) = \sum_0^N a_i x^i, \tag{7}$


$p(A) v = \sum_0^N a_i A^i v = \sum_0^N a_i \lambda^i v = p(\lambda) v, \tag{8}$

so that if $p(A) = 0$, we conclude that $p(\lambda) v = 0$, whence, since $v \ne 0$, we have $p(\lambda) = 0$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!


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