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**Additional info for Linear Algebra Examples c-4**

**Example text**

2 2 √ √ It follows from (−1, 1, 4) = 1 + 1 + 16 = 3 2 that the orthonormed eigenvectors are v2 − 1 v1 2 (v2 · v1 )v1 = (0, 1, 2) − Please click the advert 1 q1 = √ (1, 1, 0) 2 and 1 q2 = √ (−1, 1, 4). com 45 Linear Algebra Examples c-4 If λ2 = 9, then A − λ2 I 5. Conical surfaces ⎛ ⎞ ⎛ ⎞ −5 −4 2 1 −1 −4 = ⎝ −4 −5 −2 ⎠ ∼ ⎝ 0 −9 −18 ⎠ 2 −2 −8 0 −9 −18 ⎛ ⎞ ⎛ ⎞ 1 −1 −4 1 0 −2 1 2 ⎠∼⎝ 0 1 2 ⎠. g. v3 = (2, −2, 1) of length 3, hence q3 = 1 (2, −2, 1). 3 Applying the transformation ⎛ ⎞ ⎛ √1 1 − 3√ x 2 2 1 1 √ ⎝ y ⎠=⎜ ⎝ √2 3 2 4 √ z 0 3 2 2 3 − 23 1 3 ⎞⎛ ⎞ x1 ⎟⎝ y1 ⎠ ⎠ z1 we get −36x + 18y = 36 36 18 18 √ + √ −√ + √ x1 + 2 2 3 2 3 2 √ √ = −9 2 x1 + 9 2 y1 − 36 z1 , y1 + − 72 36 − 3 3 z1 and the equation is transferred into √ 0 = 0 · x21 + 0 · y12 + 9z12 − 9 2(x1 − y1 ) − 36z1 + 90 √ = 9{z12 − 4z1 + 4 − 4} − 9 2(x1 − y1 ) + 90, which we reduce to √ (z1 − 2)2 − 2(x1 − y1 ) + 6 = 0, hence to 1 1 √ (x1 − y1 ) − 3 = (z1 − 2)2 .

1. The corresponding ⎛ 0 2 A=⎝ 2 0 0 0 matrix ⎞ 0 0 ⎠ a has the characteristic polynomial (λ − 2)(λ + 2)(λ − a), thus the eigenvalues are λ1 = 2, λ2 = −2 and λ3 = a. Hence, one can in some other coordinates reduce to 2x21 − 2y12 + az12 = 1. 2 It is quite ⎛ √1 − √12 2 1 ⎝ √ √1 Q= 2 2 0 0 easy to prove that one may choose’ the orthogonal substitution as ⎞ 0 0 ⎠. ♦ 1 2. The form describes a rotational surface, if a = ±2. (a) If a = 2, then 2(x21 + z12 ) − 2y12 = 1, which is the equation of a rotational hyperboloid of 1 sheet.

If a = −b2 < 0, the surface is an hyperboloid of 1 sheet. In the plane x = 1 we ﬁnd the two straight lines z = ±by, hence √ √ : (1, y, −a y) and m : (1, y, − −a y), y ∈ R, Please click the advert are two lines on the surface. We get all such straight lines by rotating these around the Y -axis. com 31 Linear Algebra Examples c-4 4. 3 Given in an ordinary rectangular coordinate system XY Z of positive orientation in space a conical surface of the equation 50x2 + 25y 2 + az 2 − 100x − 200y − 2az = 0, where a ∈ R.