4 comments on “An Example of Cayley-Hamilton

  1. Abdulkarim says:

    Sir, I would like to be sure if am on the right track in the solving of the xteristic equation below

    -λ^3+15λ^2+18λ=0

    λ^3-15λ^2-18λ=0
    using cayley Hamilton we have

    T^3-15T^2-18T=0
    Multiply by inverse of T gives

    T^2-15T-18=0
    Multiply again by the inverse of T gives
    T-15-18T^-1 = 0

    let I=1
    18T^-1 = T – 15I
    T^-1 = (T-15I)/18
    T^(-1)= 1/I_3 (T^2-I_1T+I_2I)
    Then I_3=18, I_2=-15, I_1=-1

    • oafak says:

      No you are NOT doing it right track. I will respond adequately shortly.

      • oafak says:

        Error number 1. You did not immediately see that the equation actually shows that the determinant, I3= 0. The tensor is therefore singular and has no inverse.
        Error Number 2.
        You are handling a tensor equation as if it were a scalar equation. Cayley Hamilton says that a tensor satisfies its characteristic equation. But what you then have is a tensor equation. It is for that reason that there is an adjustment of the identity tensor in the last term! Apart from not recognizing the singularity of the tensor, your bigger error is in not recognizing tensor character of this equation.
        Error Number 3.
        This arises out of Error Number 2. You CANNOT equate, under any circumstances, that a tensor is equal to a scalar.
        The problem was deliberately tricky to draw your attention to these issues.

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