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There are a handful of longish proofs for Cayley-Hamilton, but I haven't seen one that goes along the following lines. What's wrong in my thinking.

Take a real square matrix $A$, with Jordan decomposition $D=Q^{-1}AQ$. Call the eigenvalues of $A$ (and $D$) $\lambda_1,\lambda_2,\dotsc,\lambda_k$, with respective algebraic multiplicities $n_1,n_2,\dotsc,n_k$. Then $A$ has the characteristic polynomial: $$\Delta(\lambda):=\det(\lambda I-A)=\prod_i(\lambda-\lambda_i)^{n_i}.$$ So, $$\Delta(A)= \prod_i(A-\lambda_i)^{n_i}=Q\left(\prod_i(D-\lambda_i)^{n_i}\right)Q^{-1}.$$

The block(s) associated with $\lambda_i$ for the matrix $(D-\lambda_i)^{n_i}$ are nilpotent (for at least $n_i$ if not smaller) so $\prod_i(D-\lambda_i)^{n_i}=\mathbf{0}$ so $\Delta(A)=\mathbf{0}$.

user1816847
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    It's good! Of course some of $\lambda_i$'s and some entries of $Q$ may be complex, but that's no problem.This also finds the minimal polynomial of $A$. You can also look at Cayley-Hamilton as a bunch of identities with integral coefficients in variables $(a_{ij})$. – orangeskid Oct 13 '14 at 01:34
  • Slight correction: if $D=Q^{-1}AQ$, then $A=QDQ^{-1}$. – Andrew Dudzik Oct 13 '14 at 01:40
  • Slade: Thanks. corrected.

    A thought. Is the proof not presented this way because the most general form of CH is for "a square matrix over a commutative ring" [according to wikipedia] and this proof requires that its over the field of complex numbers?

     – user1816847 Oct 13 '14 at 02:39
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    This is actually one of the most known proofs (and is similar to that in Axler). (Usually accompanied by an argument why studying real/complex matrices is enough.) The only reason why it might be wrong is that you might possibly have already used Cayley-Hamilton when proving the properties of the Jordan normal form that you are using; however, this is unlikely (but check your argument for circularity). – darij grinberg Oct 13 '14 at 02:39
  • Interesting. I just googled for a few proofs, didn't see this and figured I must be thinking about it wrong. Thanks. – user1816847 Oct 13 '14 at 03:13

1 Answers1

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There is nothing wrong with your proof. In fact, this is one of the easiest proofs of this theorem.

However, note that the theorem holds under much weaker assumptions: It is satisfied for every square matrix from $K^{n,n}$, $K$ a field, regardless whether the field $K$ is algebraically closed or not.

If you have Jordan normal form available (or some other normal form), then the proof of Cayley-Hamilton becomes simple. But the way to arrive at the normal form is much longer than an elementary proof of Caley-Hamilton itself, which just uses properties of determinants.

daw
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