MATA23H3 Lecture Notes - Lecture 9: Solution Set, Anaplastic Lymphoma Kinase, Row And Column Vectors

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graywildebeest909

19 Jun 2016

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If =(cid:1871)(cid:1868)(cid:4666)(cid:1875)(cid:2869),(cid:1875)(cid:2870), ,(cid:1875)(cid:3038)(cid:4667) is the span of (cid:1863) vectors in , the w is a subspace of . There is at least one (cid:1875) non-empty (cid:3038) (cid:3038) (cid:1838)(cid:1857)(cid:1872) (cid:1873) = (cid:2019)(cid:3036)(cid:1875) (cid:3036) + (cid:2020)(cid:3036)(cid:1875) (cid:3036) (cid:1873) +(cid:1874) = (cid:2019)(cid:3036)(cid:1875) (cid:3036) (cid:3036)=(cid:2869) (cid:3036)=(cid:2869) (cid:3038) (cid:1853)(cid:1866)(cid:1856) (cid:2019) (cid:1838)(cid:1857)(cid:1872) (cid:1873) = (cid:2019)(cid:3036)(cid:1875) (cid:3036) (cid:3036)=(cid:2869) (cid:3038) (cid:3038) (cid:3038) = (cid:2019)(cid:4666)(cid:2019)(cid:3036)(cid:1875) (cid:3036)(cid:4667) (cid:2019)(cid:1873) =(cid:2019) (cid:2019)(cid:3036)(cid:1875) (cid:3036) (cid:3036)=(cid:2869) (cid:3036)=(cid:2869) (cid:3036)=(cid:2869) (cid:2019)(cid:1873) it is closed under scalar multiplication. Since conditions are satisfied, =(cid:1871)(cid:1868)(cid:4666)(cid:1875)(cid:2869),(cid:1875)(cid:2870), ,(cid:1875)(cid:3038)(cid:4667) is a subspace for (cid:1863)>(cid:882) =(cid:1871)(cid:1868)(cid:4666){}(cid:4667) ={(cid:882)} we know that {(cid:882)} is a subspace of . The vectors (cid:1875)(cid:2869),(cid:1875)(cid:2870), ,(cid:1875)(cid:3038), as in the theorem, generate or span the subspace = (cid:1871)(cid:1868)(cid:4666)(cid:1875)(cid:2869),(cid:1875)(cid:2870), ,(cid:1875)(cid:3038)(cid:4667) Let (cid:1827)=[(cid:1853)(cid:3036)(cid:3037)] (cid:1839),(cid:3038)(cid:4666) (cid:4667),there are subspaces that arise naturally from a (cid:1873) +(cid:1874) is a linear combination of the (cid:1875) (cid:3036),=(cid:883),(cid:884), ,(cid:1863) it is closed under vector addition. Row space: the span of row vectors of a.

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Why is the vector [0.25, -0.24]??

2.15

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

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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

MATA23H3 Lecture Notes - Lecture 9: Solution Set, Anaplastic Lymphoma Kinase, Row And Column Vectors

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