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Unformatted text preview: Statistics 351 (Fall 2007) Review of Linear Algebra Suppose that A is the symmetric matrix A = 1 1 0 1 2 1 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 6 = 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 nonzero real number α . The nonzero 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 nonzero solution if and only if the matrix A λI is singular (noninvertible). The matrix A λI is invertible if and only if det[ A λI ] 6 = 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 1 2 λ 1 1 3 λ ....
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This note was uploaded on 03/26/2012 for the course STAT 351 taught by Professor Michaelkozdron during the Fall '08 term at University of Regina.
 Fall '08
 MichaelKozdron
 Statistics, Probability

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