Prooving the Cayley-Hamilton Theorem

[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:

<br /> 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?

 

[fresh_42]

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]

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.