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Engineering Calculus Notes 396

# Engineering Calculus Notes 396 - 384 CHAPTER 3 REAL-VALUED...

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384 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION sides need not be parallel to the coordinate planes), and (changing the numbering of the eigenvectors if necessary) we can assume that it is positively oriented, so the signed volume of this cube is −→ V ( E ) = 1. It follows that −→ V ( AE ) = det A . But the sides of AE are the vectors λ i −→ u i , so this is again a cube, whose signed volume is the product of the eigenvalues : −→ V ( AE ) = λ 1 λ 2 λ 3 . Thus det A = λ 1 λ 2 λ 3 and we have the following observation: If Q is positive-definite, then Δ 3 = det A> 0. Of course, the product of the eigenvalues ( i.e. , det A ) can also be positive if we have one positive and two negative eigenvalues, so we need to know more to determine whether or not Q is positive-definite. If Q is positive-definite, we know that its restriction to any plane in R 3 is also positive-definite. In particular, we can consider its restriction to the xy -plane, that is, to all vectors of the form −→ x = ( x,y, 0). It is easy to
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