The statement of the Cayley-Hamilton Theorem is fairly straight-forward.
I now know how to find characteristic polynomials from a given matrix (or at least a matrix with certain properties that I am unaware of!).
I know that the eigenvalues of the matrix are roots of the polynomial.
But what does having such a polynomial mean? Wikipedia says that the characteristic polynomial "...encodes several important properties of the matrix...", but once we have switched to "matrix form" of the equation, what can we conclude?

In other words, what does the Theorem do for us, besides allowing us to say, "Hey, I know a matrix solution to this polynomial"?? Is there an abstraction of this in abstract algebra (rings, fields, etc.)?

Thanks for your time.

  • 2
    $\begingroup$ I don't understand what you are actually asking. Could you streamline the post a bit and actually ask a question that admits a "right" answer, preferably leaving out irrelevant anecdotes? $\endgroup$
    – Alex B.
    May 3 '11 at 12:09
  • 3
    $\begingroup$ The characteristic polynomial's roots are your matrix's eigenvalues, as a start. $\endgroup$ May 3 '11 at 12:11
  • 1
    $\begingroup$ @Alex: I understand that I am not as refined to/by the MSE culture as others. I deliberately included the tag "soft-question" and the anecdotes to help those who would help me to understand the background. $\endgroup$ May 3 '11 at 12:35
  • 1
    $\begingroup$ @Qiaochu: Applications, in the sense that I'd wondering what we do with the information that $A$ is the root of some polynomial. As far as I can tell, we start with an n x n matrix, calculate the determinant of (lambda*I - A), and find eigenvalues. In the process, we arrive at the characteristic polynomial. While this is a "cool"/"surprising" result, where do we go from here? $\endgroup$ May 3 '11 at 13:39
  • 1
    $\begingroup$ Lots of places. I don't know that there's a short, reasonable answer to this question. An enormous part of mathematics consists of replacing harder problems by linear algebra, and once you've done that you want tools to help you understand the resulting linear algebra, and Cayley-Hamilton is a basic such tool. Applications are everywhere in mathematics. $\endgroup$ May 3 '11 at 14:08

It's simple to see that every matrix $n\times n$ has to be a zero of some polynomial of degree at most $n^2$, simply because the space of $n\times n$ matrices has dimension $n^2$. The Cayley-Hamilton Theorem says that you can find such a polynomial of much smaller degree.

Another way to see this is as follows: For each fixed vector $v$, the vectors $v$, $Av$, ..., $A^n v$ cannot be linearly independent and so there is a polynomial $p$ of degree at most $n$ such that $p(A)v=0$. However, this polynomial depends on $v$. The Cayley-Hamilton Theorem gives you a polynomial that works for all vectors $v$.

Finally, for applications, having a polynomial $p$ such that $p(A)=0$ allows you to compute all powers of $A$ as a linear combination of $I$, $A$, ..., $A^{n-1}$. Indeed, assuming $p$ monic, you can write $p(X)=X^n+q(X)$ with $q$ of degree less than $n$. So $A^n = -q(A)$. Then $A^{n+1}=-A q(A)$. If $A^n$ appears in $A q(A)$, replace it by $-q(A)$. Do the same for $A^{n+2} = A \cdot A^{n+1}$, etc. For a concrete example, see http://en.wikipedia.org/wiki/Cayley-Hamilton_theorem#Illustration_for_specific_dimensions_and_practical_applications.

  • $\begingroup$ @Ihf: Thanks for the answer. Could you elaborate on the last sentence ("...compute all powers of $A$ ...") please? $\endgroup$ May 3 '11 at 13:24
  • $\begingroup$ @The Chaz, done. $\endgroup$
    – lhf
    May 3 '11 at 13:39
  • $\begingroup$ @Ihf: Thanks again! You have been very helpful. $\endgroup$ May 3 '11 at 13:40
  • 1
    $\begingroup$ After you've found a monic polynomial $P$ of degree $d\leq n$ with $P(A)v=0$, also $P(a)A^iv$ for all $i$, and $P(A)$ annihilates a subspace of dimension at least $d$. Restricting further effort to the subspace that is the image of $P(A)$, you find by induction a polynomial of dimension at most $n-d$ that annihilates it, and its product with $P$, which has degree at most$~n$, annihilates the whole space. So you don't really need Cayley-Hamilton, or indeed the characteristic polynomial at all, for this. $\endgroup$ May 21 '13 at 5:43
  • $\begingroup$ What about $e^{At}$? $\endgroup$
    – Marso
    Sep 16 '17 at 5:42

First of all, the Cayley-Hamilton Theorem is a somewhat surprising result on its own. As for uses, it does not immediately give rise to many important statements, but it pops up in the proofs of other results occasionally. It is often involved in proving the Jordan canonical form, and I have also seen it used it to prove the main result regarding Kummer extensions in Galois theory. In particular, if a matrix vanishes on some polynomial, then its eigenvalues are a subset of the roots of the polynomial.

  • 2
    $\begingroup$ putzer user cayley-hamilton theorem to compute the exponential of a matrix without going through the jordan canonical form. it is one of nice application i know. i don't know how to link to one my previous answers on this site. you can or google putzers algorithm. $\endgroup$
    – abel
    May 3 '11 at 14:20
  • 1
    $\begingroup$ I don't think you need the Cayley-Hamilton for proving the existence of Jordan canonical forms: if your minimal (or characteristic) polynomial splits into linear factors, the space is a direct sum of kernels of the occurring powers of those factors (as I explained here), which suffices. Also if $Q(A)=0$ and $\lambda$ is eigenvalue it is obvious that $\lambda$ is root of $Q$; just compute $0=Q(A)v=Q(\lambda)v$ for a corresponding eigenvector. $\endgroup$ May 21 '13 at 5:29

Let $K$ be an algebraically closed field. Let $V$ be a vector space over $K$ of dimension $n$. Let $u\colon V \rightarrow V$ be a $K$-linear map. Let $K[X]$ be the polynomial ring with one variable. $V$ can be regarded as a $K[X]$-module as follows. Let $f(X) \in K[X], x \in V$. We define $f(X)x$ as $f(u)(x)$. It is easy to see that $V$ is a finitely generated torsion $K[X]$-module. Hence $V$ has a composition series: $V = V_0 \supset V_1 \supset\cdots \supset V_{r-1} \supset V_r = 0$. Since each $V_i/V_{i+1}$ is a simple $K[X]$-module, it is isomorphic to $K[X]/(X - \alpha_i)$, where $\alpha_i \in K$. It is easy to see that $f(X) = \prod_i (X - \alpha_i)$ is the characteristic polynomial of $u$. It is also easy to see that $f(X)V = 0$. This is the Cayley-Hamilton theorem.

Now let's consider the following analogy of the above argument. Let $G$ be a finitely generated torsion $\mathbb{Z}$-module, i.e. a finite abelian group. $G$ has a composition series: $G = G_0 \supset G_1 \supset\cdots \supset G_{r-1} \supset G_r = 0$. Each $G_i/G_{i+1}$ is isomorphic to $\mathbb{Z}/p_i\mathbb{Z}$, where $p_i$ is a prime number. It is easy to see that $|G|= \prod_i p_i$. It is also easy to see that $|G|G = 0$.

Therefore, in our analogy, $|G|$ corresponds to a characteristic polynomial and $|G|G = 0$ corresponds to the Cayley-Hamilton theorem.

  • $\begingroup$ This is a good proof. $\endgroup$ Sep 15 '12 at 18:11
  • $\begingroup$ +1 For an nice high level explanation, which explains that the C-H theorem is something natural. I doesn't really provide any answer to the question (what does one use C-H for) though. $\endgroup$ May 21 '13 at 5:34
  • $\begingroup$ $|G|G=0$ is of course Lagrange's theorem. $\endgroup$
    – lhf
    Jul 20 '13 at 15:30

I would interpret the Wikipedia quote as a reference to the trace and determinant of the matrix showing up (up to sign) as the next-to-leading and constant term, respectively, of the polynomial. The other coefficients of the polynomial also have interpretations as matrix invariants, but not as familiar as trace and determinant.

  • 2
    $\begingroup$ There is other important information in the characteristic polynomial beyond its coefficients: It's irreducible factors correspond to irreducible factors of the minimal polynomial/polynomials whose powers occur in the rational canonical form, for example. $\endgroup$ May 3 '11 at 16:06

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.