# Prooving the Cayley-Hamilton Theorem

1. Jun 4, 2008

### Vuldoraq

Proving the Cayley-Hamilton Theorem

Hey all,

I'm revising for my linear algebra exam, which is next week, and I got up to the Cayley-Hamilton theorem, but I am stuck on the final leap in the proof. Here is what I understand so far,

Theorem:

Every matrix is a zero of it's characteristic equation$$[P_{A}(\lambda)=det(A-\lambda I)]$$,

For $$\\\\\\\\\\\\\ A \in M_{n}(k)$$ $$P_{A}(A)=0$$

Proof:

$$Let\ P_{A}(\lambda)=det(A-\lambda I)=\lambda^{n}+a_{n-1}\lambda^{n-1}+\ldots+a_{1}\lambda+a_{0}$$

and consider,

$$\phi(\lambda)=adj(A-\lambda I)=B_{n-1}\lambda^{n-1}+ldots+B_{1}\lambda+B_{0}$$

where $$B_{i} \in M_{n}(k)$$

Given that for any $$C \in M_{n}(k); \\ C*adj(C)=det(C)*I.$$

So by letting $$C=A-\lambda I$$ we have,

$$(A- \lambda I)*\phi(\lambda)=det(A-\lambda I)I=P_{A}(\lambda)I$$

Expanding we have,

$$(A- \lambda I)*(B_{n-1}\lambda^{n-1}+ldots+B_{1}\lambda+B_{0})=(\lambda^{n}+a_{n-1}\lambda^{n-1}+ldots+a_{1}\lambda+a_{0})*I$$

The next step has me in tears, my book says compare coefficients and add but I can't see how you would compare these? Please can anyone help me to complete this?

Last edited: Jun 4, 2008