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    Suppose A is diagonalizable with distinct eigenvalues...

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    © BrainMass Inc. brainmass.com September 28, 2022, 8:52 pm ad1c9bdddf


    Solution Preview

    I will use the notation a1...ak for the eigenvalues (rather than using lambda, which is difficult to write).
    I will also denote just by m the minimal polynomial of A. We know that because A is diagonalisable:
    So P_j(t)=(t-a1)*...*(t-ak)/c_j where in the product the factor (t-aj) is missing, as it ...

    Solution Summary

    This is a proof regarding a diagonalizable matrix and orthogonal projections.