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MATH136 Lecture Notes - Lecture 5: Empty Set, Linear Independence

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ultramarineelephant806

26 Jul 2018

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University of Waterloo

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Course

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Alejandra Vicente Colmenares

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$MATH136 Full Course Notes$

MATH136 Full Course Notes

Document Summary

Math 136 lecture 5 12 jan, 2018 (cid:1876)(cid:2869)+(cid:1877)(cid:2869)](cid:1488)(cid:2869) (as 1st and 2nd components are same, i. e. to be in (cid:2869), 1st, 2nd, Example: determine which of the following are subspaces of the appropriate . (cid:1853)(cid:4667) (cid:2869)={[(cid:1876)(cid:2869)(cid:1876)(cid:2869)]|(cid:1876)(cid:2869)(cid:1488) } Solution: by definition, (cid:2869) is subset of (cid:2870). By the subspace test, (cid:2869) is subset of (cid:2870) (cid:1854)(cid:4667) (cid:2870)={[(cid:1876)(cid:2869)(cid:1876)(cid:2870)](cid:1488) (cid:2870)|(cid:1876)(cid:2869)(cid:2870) (cid:1876)(cid:2870)(cid:2870)=(cid:882)} (square, not a linear) Not a subspace observe [(cid:883) (cid:883)](cid:1488)(cid:2870) as (cid:883)(cid:2870) (cid:4666) (cid:883)(cid:4667)(cid:2870)=(cid:882), [(cid:883)(cid:883)](cid:1488)(cid:2870) as (cid:883)(cid:2870) (cid:883)(cid:2870)=(cid:882) but, [(cid:883) (cid:883)]+[(cid:883)(cid:883)]=[(cid:884)(cid:882)](cid:1489)(cid:2870) as (cid:884)(cid:2870) (cid:882)(cid:2870)=(cid:882) (cid:1855)(cid:4667) (cid:2871)={[(cid:1876)(cid:2869)(cid:1876)(cid:2870)(cid:1876)(cid:2871)](cid:1488) (cid:2871)|(cid:1876)(cid:2869) (cid:1876)(cid:2870)+(cid:1876)(cid:2871)=(cid:886)} Solution: observe (cid:882) is not in (cid:2871) since (cid:882) (cid:882)+(cid:882) (cid:886) Therefore, not a subspace. must be a subspace] (cid:1856)(cid:4667) (cid:2872)={[(cid:1876)(cid:2869)(cid:1876)(cid:2870)(cid:1876)(cid:2871)](cid:1488) (cid:2871)|(cid:1876)(cid:2869) (cid:1876)(cid:2870)+(cid:1876)(cid:2871)=(cid:882)} [example of a plane passing through origin, therefore. Solution: by definition, (cid:2872) is a subset of (cid:2871), (cid:882) (cid:1488)(cid:2872) since (cid:882) (cid:882)+(cid:882)=(cid:882). Then (cid:1876) and (cid:1877) satisfy (cid:1876)(cid:2869) (cid:1876)(cid:2870)+(cid:1876)(cid:2871)=(cid:882) and (cid:1877)(cid:2869) (cid:1877)(cid:2870)+(cid:1877)(cid:2871)=(cid:882) (cid:1876) +(cid:1877) =[(cid:1876)(cid:2869)+(cid:1877)(cid:2869) (cid:1876)(cid:2870)+(cid:1877)(cid:2870) (cid:1876)(cid:2871)+(cid:1877)(cid:2871)] (cid:4666)(cid:1876)(cid:2869)+(cid:1877)(cid:2869)(cid:4667) (cid:4666)(cid:1876)(cid:2870)+(cid:1877)(cid:2870)(cid:4667)+(cid:4666)(cid:1876)(cid:2871)+(cid:1877)(cid:2871)(cid:4667)=(cid:1876)(cid:2869) (cid:1876)(cid:2870)+(cid:1876)(cid:2871)+(cid:1877)(cid:2869) (cid:1877)(cid:2870)+(cid:1877)(cid:2871)=(cid:882)+(cid:882)=(cid:882) (cid:1876) +(cid:1877) (cid:1488)(cid:2872)

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Related Questions

I have the answer for the following question. But some of theanswers are not correct.

Please do the following question by yourself and check which arethe wrong answers,please post your answers and compare to it.

Let r in and u =(1 1 0), v = (1 4 1), w = (r2 1 r2), b = (3+2r 5+12r 2r). For which values r in is the set {u. v, w} linearly independent? For which values r in is the vector b a linear combination of u. v and w? For which of these values of r can b be written as a linear combination of u. v and w in more than one way? The set 3 of all column vectors of length three, with real entries, is a vector space. Is the subset B = {(x y z) in 3 | xy + yz = 0} a subspace of 3? Justify your answer. Show that the set of all twice differentiable functions f : satisfying the differential equation sin(x) f"(x) + x2f(x) = 0 is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f. Let S be the following subset of the vector space 3 of all real polynomials p of degree at most 3: S = {p in 3 | p(l) = 0, p'(l) = 0}, where p' is the derivative of p. Determine whether S is a subspace of 3. Determine whether the polynomial q(x) = x - 2x2 + x3 is an element of S. u = (1 1 0), v = (1 4 1), w = (r2 1 r2) if {u, v, w} is linearly independent set 1(4r2 - 1)-1(r2 - r2) ne 0 4r2 ne 1 r ne plusmn1/2 this set is linearly independent set for values of e ne plusmn1/2 - - - - - - - (3+2r 5+12r 2r) = a1 (1 1 0) + a2 (1 4 1) + a3 (r2 1 r2) this forms system of equations in a1, a2, a3 if 4r2 - 1 ne 0 then system has unique solution if 4r2 - 1 = 0 and -2 - 4r = 0 rArr r = -1/2, the system has more than one solution if 4r2 - 1 ne 0 b can be written as linear combination of u,v,w as uniquely if r = -1/2 then b can be written as linear combination of u,v,w more than one way - - - - - - - - - (1 1 -1) in B and (1 0 1) in B (1 1 -1) + (1 0 1) = (2 1 0) not in B since 2times1+1times0 ne 0 so addition of vector is failed, this set is not forms subspace consider the following differential equation (sinx)f"(x) + x2f (x) = 0 each solution is continuous function the set of all solutions is subset of space of set of all continuous functions n from R to R let f,g are two solutions (sinx)f"(x) + x2f (x) = 0 (sinx)g"(x)+x2g(x) = 0 ccheck whether af+bg is solution or not (sinx)f"(x) + x2f(x) = 0 rArr (sinx) af"(x) + x2?f (x) = 0 (sinx)g"(x) + x2g (x) = 0 rArr (sinx)bg" (x) + x2bg (x) = 0 (sinx)(af"(x)+bg"(x)) +x2 (af (x) + bg(x)) = 0 af+bg is also solution to differential equation the set of all solutions to differential equation forms subspace - - - - - - - - - - consider the following subset of P3 {p in P3/.p(l)=0, p'(l) = 0} let p, q in S rArr p(l)= 0, p'(l) = 0 and q(l) = 0, q'(l) = 0 (ap + bq) (l) = ap (l) +b q(l) = a (0) + b (0) = 0 (ap + bq)'(1) = ap'(l) +bq'(1) = a (0) + b(0) = 0 p, q in S rArr ap +bq in S then S is subspace q(x) = x - 2x2 + x3 q(1) = 1 - 2 + 1 = 0 q'(x)=1 - 4x + 3x2 q'(l) = 1 - 4 + 3 = 0 q is element of S

I would like to prove the last part (d).