Engineering Calculus Notes 383

Engineering Calculus Notes 383 - g ( u ) = 2 u ; we need to...

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3.9. THE PRINCIPAL AXIS THEOREM 371 with determinant 1 > 0. It turns out that we have to look at a minor of the determinant, as well. To understand this, we approach our problem diFerently, taking a clue from the proof of Lemma 3.8.2 : to know that Q is positive de±nite 19 we need to establish that the minimum value of its restriction to the unit sphere S 2 = { −→ u R 3 | b −→ u b = 1 } is positive. This means we need to consider the constrained optimization problem: ±nd the minimum of f ( −→ x ) = Q ( −→ x ) subject to the constraint g ( −→ x ) = −→ x · −→ x = 1 . This can be attacked using Lagrange multipliers: we know that the point −→ u ∈ S 2 where the minimum occurs satis±es the condition −→ f ( −→ u ) = λ −→ g ( −→ u ) for some λ R . We already know that
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Unformatted text preview: g ( u ) = 2 u ; we need to calculate f ( u ). To this end, we write f ( x 1 ,x 2 ,x 3 ) = Q ( x 1 ,x 2 ,x 3 ) in the matrix form f ( x 1 ,x 2 ,x 3 ) = b x 1 x 2 x 3 B a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 x 1 x 2 x 3 = 3 s i =1 3 s j =1 a ij x i x j . To nd f x 1 , we locate all the terms involving x 1 : they are a 11 x 2 1 + a 12 x 1 x 2 + a 13 x 1 x 3 + a 21 x 2 x 1 + a 31 x 3 x 1 ; using the symmetry of A we can combine some of these terms to get a 11 x 2 1 + 2 a 12 x 1 x 2 + 2 a 13 x 1 x 3 . 19 we return to the negative defnite case at the end oF this subsection...
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