Linear Algebra Solutions 72

# Linear Algebra Solutions 72 - 5 1 4 4 1 = 1 4 = 2(b Hence...

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= λ µ λ 2 - 5 λ 4 + 1 4 = λ ( λ - 1) µ λ - 1 4 . (b) Hence the characteristic polynomial has no repeated roots and we can use Theorem 6.2.2 to fnd a non–singular matrix P such that P - 1 AP = diag(1 , 0 , 1 4 ) . We take P = [ X 1 | X 2 | X 3 ], where X 1 , X 2 , X 3 are eigenvectors corresponding to the respective eigenvalues 1 , 0 , 1 4 . Finding X 1 : We have to solve ( A - I 3 ) X = 0. we have A - I 3 = - 1 / 2 1 / 2 0 1 / 4 - 3 / 4 1 / 2 1 / 4 1 / 4 - 1 / 2 1 0 - 1 0 1 - 1 0 0 0 . Hence the eigenspace consists o± vectors X = [ x, y, z ] t satis±ying x = z and y = z , with z arbitrary. Hence X = z z z = z 1 1 1 and we can take X 1 = [1 , 1 , 1] t . Finding X 2 : We solve AX = 0. We have A = 1 / 2 1 / 2 0 1 / 4 1 / 4 1 / 2 1 / 4 1 / 4 1 / 2
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