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V 1 = (1, 0, −1) and v2 = (1, −1, 1) where v1 = 2, and v1 · v2 = 0, and v2 = 3, thus the subspace of all eigenvectors corresponding to λ1 = 1 is spanned by the orthonormal eigenvectors 1 q1 = √ (1, 0, −1) 2 2. It follows from Now, since ⎛ 2 2 ⎝ 2 5 1 2 and 1 q2 = √ (1, −1, 1). 3 the construction above that (1, 2, 1) must be perpendicular to both q 1 and q2 ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 1 7 1 2 ⎠ ⎝ 2 ⎠ = ⎝ 14 ⎠ = 7 ⎝ 2 ⎠ , 2 1 7 1 1 we see that q3 = √ (1, 2, 1) is an eigenvector corresponding to the eigenvalue λ3 = 7.
The corresponding matrix ⎛ ⎞ a 3 4 A=⎝ 3 a 0 ⎠ 4 0 a has the characteristic polynomial det(A − λI) = = 4 a−λ 3 4 3 a−λ 0 3 4 a−λ 0 4 0 a−λ + (a − λ) a−λ 3 3 a−λ = −16(a − λ) + (a − λ){(λ − a)2 − 9} = −(λ){(λ − a)2 − 25} = −(λ − a)(λ − a − 5)(λ − a + 5). com 64 Linear Algebra Examples c-4 6. Quardratic forms We have three different eigenvalues, λ1 = a − 5, λ2 = a, λ3 = a + 5. In order to find the type of the surface we analyze the signs pf the λ, λ1 x21 + λ2 y12 + λ3 z12 = 1 a < −5 a = −5 −5 < a < 0 a=0 0
1. 2) to a form without product terms by applying Q as the matrix of change of variables. 2. Prove that (4) in an ordinary rectangular coordinate system XY Z in space of positive orientation describes a cylinder surface, and describe the type and the direction of the generator of this surface. 1. It follows from 22 + 32 + 62 = 4 + 9 + 36 = 49 = 72 , that all columns in Q are unit vectors. Then we calculate the three possible inner products [without the factor 17 ], to get (2, 3, 6) · (3, −6, 2) (2, 3, 6) · (6, 2, −3) (3, −6, 2) · (6, 2, −3) = 6 − 18 + 12 = 0, = 12 + 6 − 18 = 0, = 18 − 12 − 6 = 0.