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linalg - Statistics 351(Fall 2007 Review of Linear Algebra...

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Statistics 351 (Fall 2007) Review of Linear Algebra Suppose that A is the symmetric matrix A = 1 - 1 0 - 1 2 1 0 1 3 . Determine the eigenvalues and eigenvectors of A . Recall that a real number λ is an eigenvalue of A if A v = λ v for some vector v = 0. We call v an eigenvector (corresponding to the eigenvalue λ ) of A . Note that if v is an eigenvector of A , then so too is α v for any non-zero real number α . The non-zero vector v is a solution of the equation A v = λ v if and only if v is also a solution of the equation ( A - λI ) v = 0. The equation ( A - λI ) v = 0 has a non-zero solution if and only if the matrix A - λI is singular (non-invertible). The matrix A - λI is invertible if and only if det[ A - λI ] = 0. Therefore, in order to find the eigenvalues of A , we need to find those values of λ such that det[ A - λI ] = 0. (We sometimes call the polynomial equation det[ A - λI ] = 0 the characteristic equation of the matrix A .) Therefore, we consider A - λI = 1 - λ - 1 0 - 1 2 - λ 1 0 1 3 - λ . Since det 1 - λ - 1 0 - 1 2 - λ 1 0 1 3 - λ = (1 - λ )(2 - λ )(3 - λ ) - (1 - λ ) - (3 - λ ) = 2 - 9 λ + 6 λ 2 - λ 3 = (2 - λ )( λ 2 - 4 λ + 1) = (2 - λ )( λ - 2 -
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