# Prooving the Cayley-Hamilton Theorem

## Main Question or Discussion Point

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:

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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