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Prooving the Cayley-Hamilton Theorem

  1. Jun 4, 2008 #1
    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,


    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]


    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: Jun 4, 2008
  2. jcsd
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