By Hugo. Rossi
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Extra info for Advanced Calculus: Problems and applications to science and engineering
A linear space. A basis of V is be written in the form Let Fbe such that each 47 \ 2cV|Wi=0-WiH Thus, by the independence of (a/, Subspaces ofR" by this computation: 0 k k Linear a . , set 5 of vectors k v in one 2! c% ;=i = and only cl witn one R,\te S e way. Another way of putting this is: a basis for independent and spanning vectors in V. Proposition 12. S is a a linear basis for the linear space V subspace V is a if and only if both set of these conditions hold: S is (i) (ii) an independent set, of S is V.
The system Ax the particular solution (-1, 1, 1). Thus = Taking x3 = -Vi + v2 + v3 =0 20. ,=(2,1,2) (0,3,0) v2 (1,0,4) v3 = = v4 = (0,l,2) independent. Let 1 we have Linear Functions /. 46 Find a linear relation which these vectors must reduce the matrix whose columns A= /l 3 (0 1 \0 If satisfy. we row obtain the matrix we 1/3 -1/ -1/6 1 Now, corresponding 0, and thus of Ax the v's, 1 \ 0 0 are to any value of = xVj x4 Take x4 0. = we obtain 1 Then = . a solution of 1 x4 x3 x2 fr3-3-x4 \ x1= -3x2-x4=-i = = = = Thus Vl 2 ~ iV2 + V3 + V4 = 0 Now, the equivalent form of these two propositions about R", that any members, and any independent set has spanning set of vectors has at least at most members, holds for any linear subspace of R" n Proposition 11.
Let . - point of intersection (x, y) solving The 3y = 0 y = 6 We find x = 2x - 5x+ 9. 18/17, y = Find the line L line L': 8x + 2y = 17. = = must lie on both lines, and thus is the pair 12/17. through the L will be point (7, 3) that is parallel to the given by an equation of the form 26 /. Linear Functions + by point of ax 8x + ax = intersection, = have can so the L and L' must have to L', we must have parallel equations no 17 2y= by + In order to be c. c no common solution. Thus 8_2 a~b Furthermore, since (7, 3) is la + 3b This 6 c 4x + y ?