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GR0877 Problem 86

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) The eigenvalues of the matrix A are the solutions \lambda to the equation \det(A - \lambda I) = 0, where I is identity matrix. Thus, A -\lambda I = \begin{pmatrix} 2 - \lambda & i\\ -i & 2- \lambda \end{pmatrix} and \det(A - \lambda I) = (2-\lambda)^2 + i^2 \equiv 0. From this, one has \lambda_1 = 1, \lambda_2 = 3.

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Categories: 0877 Tags: , , , ,
  1. Brent
    09.03.2012 at 07:26

    The eigenvalues of a Hermitian matrix are always real, so we can immediately rule out D and E.
    The trace of any matrix is invariant, so Tr(A) = \Sigma \lambda_i. This rules out A (and also D and E if you forgot the first property).
    \det(A) is simply the product of the eigenvalues. This leaves us with B

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