5
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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

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

then

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

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,

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

then

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

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!!!

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.