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Using Cayley-Hamilton Theorem to Calculate Matrix Powers

  1. Feb 13, 2013 #1
    Given a matrix A (for simplicity assume 2x2) we can use the Cayley-Hamilton theorem to write:

    [tex]A^k = a_0 + a_1 A[/tex]

    for k>= 2.

    So suppose we have a given k and want to find the coefficients [itex]a_0,a_1[/itex]. We can use the fact that the same equation is satisfied by the eigenvalues. That is, for any eigenvalue [itex]\lambda[/itex] we have

    [tex]\text{(1) } \quad \lambda^k=a_0+a_1\lambda.[/tex]

    If A has 2 distinct eigenvalues, we can plug them into this equation and get 2 equations which allow us to solve for [itex] a_0, a_1[/itex]. And we're done.

    If on, the other hand, A has a repeated eigenvalue [itex]\lambda_1[/itex], then we only get 1 equation from the above procedure. To get another equation we can differentiate (1) to get
    [tex]\text{(2) } \quad k \lambda^{k-1} = a_1.[/tex]
    We then plug [itex]\lambda_1[/itex] into (2) to get our second equation. Now we can solve for [itex]a_0, a_1[/itex].

    My question is how we know we can differentiate (1) and still get a valid equation. In order to differentiate, (1) must hold for all [itex]\lambda[/itex] (or at least over some interval). But we only know that it holds for particular discrete values of [itex] \lambda[/itex] (i.e. for the eigenvalues of A).
     
  2. jcsd
  3. Dec 10, 2016 #2
    Ugh... First, huge thanks for your question. I solved my problem via this method.
    I'm not sure that this is still valid question since you posted almost 3yrs ago.
    Think about characteristic polynomial. f(t).
    We differentiate f(t), than substitute eigenvalue after. Than it's a little bit tricky but still no problem .. I guess..
     
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