Linear Algebra Solutions 62

# Linear Algebra Solutions 62 - Section 6.3 4 3 Then A has...

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Section 6.3 1. Let A = 4 - 3 1 0 . Then A has characteristic equation λ 2 - 4 λ + 3 = 0 or ( λ - 3)( λ - 1) = 0. Hence the eigenvalues of A are λ 1 = 3 and λ 2 = 1. λ 1 = 3. The corresponding eigenvectors satisfy ( A - λ 1 I 2 ) X = 0, or 1 - 3 1 - 3 = 0 0 , or equivalently x - 3 y = 0. Hence x y = 3 y y = y 3 1 and we take X 1 = 3 1 . Similarly for λ 2 = 1 we find the eigenvector X 2 = 1 1 . Hence if P = [ X 1 | X 2 ] = 3 1 1 1 , then P is non–singular and
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