Prooving the Cayley-Hamilton Theorem

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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[tex][P_{A}(\lambda)=det(A-\lambda I)][/tex],

For [tex]\\\\\\\\\\\\\ A \in M_{n}(k)[/tex] [tex] P_{A}(A)=0[/tex]


Proof:

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

and consider,

[tex]\phi(\lambda)=adj(A-\lambda I)=B_{n-1}\lambda^{n-1}+ldots+B_{1}\lambda+B_{0}[/tex]

where [tex] B_{i} \in M_{n}(k)[/tex]

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

So by letting [tex] C=A-\lambda I[/tex] we have,

[tex](A- \lambda I)*\phi(\lambda)=det(A-\lambda I)I=P_{A}(\lambda)I[/tex]

Expanding we have,

[tex](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[/tex]

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:

fresh_42

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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.
 
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

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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.
 

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