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# sol63 - Partial Solution Set Leon 6.3 0 1 this case are 1 =...

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Partial Solution Set, Leon § 6.3 6.3.1a We are to ﬁnd a matrix X that diagonalizes A = ± 0 1 1 0 ² . The eigenvalues of A in this case are λ 1 = 1 and λ 2 = - 1. By Theorem 6.3.2, such a matrix X exists, and the columns of X are eigenvectors of A corresponding to the eigenvalues. In the usual fashion, we ﬁnd eigenvectors x 1 = (1 , 1) T and x 2 = (1 , - 1) T . So X = ± 1 1 1 - 1 ² , and X - 1 = 1 2 ± 1 1 1 - 1 ² . It is easy to verify that X - 1 AX = D = ± 1 0 0 - 1 ² , so A = XDX - 1 . 6.3.1b As in (a), but with A = ± 5 6 - 2 - 2 ² . The eigenvalues are λ 1 = 1 and λ 2 = 2. An eigenvector associated with λ 1 is x 1 = ( - 3 , 2) T , and an eigenvector associated with λ 2 is x 2 = ( - 2 , 1) T . So X = ± - 3 - 2 2 1 ² , and X - 1 = ± 1 2 - 2 13 ² . As expected, D = X - 1 AX = ± 1 0 0 2 ² . 6.3.1d As in (a) and (b), but now A = 2 2 1 0 1 2 0 0 - 1 . The eigenvalues are λ 1 = 2 , λ 2 = 1, and and λ 3 = - 1. The associated eigenvectors are (any scalar multiples of) x 1 = (1 , 0 , 0) T , x 2 = ( - 2 , 1 , 0) T , and x 3 = (1 , - 3 , 3) T . So X = 1 - 2 1 0 1 - 3 0 0 3 , and X - 1 = 1 2 5 3 0 1 1 0 0 1 3 . 6.3.4a Given A = ± 2 1 - 2 - 1 ² , ﬁnd a matrix B such that B 2 = A . In other words, ﬁnd a square root for A . This is easily done using diagonalization, i.e., once we factor

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sol63 - Partial Solution Set Leon 6.3 0 1 this case are 1 =...

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