# Prooving the Cayley-Hamilton Theorem

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:

## Answers and Replies

fresh_42
Mentor
Multiply the LHS and then compare for each ##i=0,\ldots,n## the coefficients of ##\lambda^i##. E.g. for ##i=n## we get ##- B_{n-1}=I##, for ##i=n-1## we have ##AB_{n-1}-B_{n-2}=a_{n-1}I##. With ##B_{n-1}=-I## from the previous step, we get ##B_{n-2}=-A-a_{n-1}I## and so on.

• Vuldoraq
Thanks for the reply @fresh_42! I was quite surprised to see one after 11 years, but it's appreciated.

Totally forgot this forum existed, will see if I can brush up on my physics and maths and get more involved.

mathwonk
Science Advisor
Homework Helper
2020 Award
In fact Cayley Hamilton follows immediately from the next to last equation in post # 1, viewed as an equation between polynomials with matrix coefficients, if you know the non commutative root/factor theorem, namely that the fact (A-t) is a (left) factor of P(t) implies that t=A is a (left root, hence also a) root of P(t). This proof occurs in Fundamental Concepts of higher algebra, by A. Adrian Albert, p.84.