Alg1231W2T2 - Subsets of a vector space

Tuesday, 24 July 2018

10:30 PM

Machine generated alternative text: Using the GeoGebra applet above, consider the movable vector v along with its multiples For example if v then: Note: the Maple notation for the vector w = 2v, u 1,2

 

Machine generated alternative text: By moving v around, we see that [select all that apply] the vectors w, u, z always lie on a line through the origin (i.e though the zero vector O) Cl the vectors w, u, z always point in different directions the vectors w, u, z always belong to the set {Av : e R}. We also notice that no matter what v is, we always have the relations [select all that apply] —9 Cl W = 3u 2 3 — 2z

 

Machine generated alternative text: Let S be the set Let's show that S is a subspace of R by using the Subspace Theorem. y + 2z A subset S of a vector space V is a subspace precisely when S i) contains the zero vector from V, ii) is closed under vector addition and iii) is closed under multiplication by scalars Please help us fill out the argument to show that the S above is indeed a subspace Step 1. We firstly verify that S contains the vector 0 e R because Step 2. Now we show that S is closed under vector addition Suppose that belong to s VI ¯ Then and Yl + Y2 We want to show that VI + v2 belongs to S. We do this by checking that the components of VI + v2 satisfy the condition for being in S _ The left hand side of this condition is: 301 + x2) ¯ (Yl +Y2) +2(z1 + Splitting this into the components of VI and v2 gives (3c1 ¯YI + El) + (3x2 ¯Y2 + 2z2)• But 3x1 — Yl + 2z1 = 0 because O , and 3:c2 — + 2z2 = O because V2ES 3(X1 + X2) — (Yl +Y2) + 2(z1 + z2) ¯ and the condition for inclusion in S is satisfied. So we have shown S is closed under vector addition. VIES O Hence

 

Machine generated alternative text: Step 3. Now we show that S is closed under multiplication by scalars Suppose that y belongs to S 0 and that is a scalar, meaning is a number O Then We want to show that Av belongs to S _ We do this by checking that the components of Av satisfy the condition for being in S _ The left hand side of this condition is: Factoring out gives Since v e S then 3m — y + 2z 3(Ax) - (Ay) + O Hence 3(Ax) - (Ay) + 2(Az) - O, and the condition for inclusion in S is satisfied. So S is closed under multiplication by scalars. Finally by the subspace theorem 3 , we can conclude that S is indeed a subspace of R o

 

Machine generated alternative text: Step 1. We firstly verify that T contains the vector 0 e IR because O can be written as r Step 2. Now we show that T is closed under vector addition. Suppose that VI and v2 belong to T this, firstly note that we know these vectors must be of the form Let T be the set In other words; to test that a vector v is in T, we need to show it can be written in the form r Let's show that T is a subspace of R by using the Subspace Theorem: VI ¯ 3 —1 2 3 —1 2 :reR for some r e R (compare this to the requirement in the previous questions). A subset T of a vector space V is a subspace precisely when T i) contains the zero vector from V, ii) is closed under vector addition and iii) is closed under multiplication by scalars Please help us fill out the argument to show that the T above is indeed a subspace. 3 —1 2 for r 3 —1 2 O e IR. _ We want to show that VI + v2 also belongs to T To show for some s, t e R. So we have where 3 —1 2 3 2 and = t 3 —1 2 3 —1 2 O e IR. Thus VI + v2 is of the form required; and so we have shown T is closed under vector addition

 

Machine generated alternative text: Step 3. Now we show that T is closed under multiplication by scalars. Suppose that v belongs to T O and that is a scalar, meaning is a number for some r e R. So O We now want to show that Av must also belong to T. Again we start by noting that v must be for the form 3 —1 2 3 —1 2 3 —1 2 where Finally, hr O e IR. by the subspace theorem O Thus is of the form required, and so we have shown T is closed under multiplication by scalars. 3 we can conclude that T is indeed a subspace of R

 

Machine generated alternative text: The subspace that we considered in question 2 passes through three distinct points Notes: enter your three points using Maple's set notation; i.e. something of the form { [O , 1 , 2] Which of the following diagrams is a picture of S? c c 0.31 ,1.-11 1-2.031 13 Y+ 2z=o 0] If ,-2.01 [.1.0.11 Y

 

Machine generated alternative text: Consider the plane It meets the x-axis at the point : 3m — y + 2z O It meets the y-axis at the point O and it meets the z-axis at point O Does it also pass though the origin? No Note: enter points using the Maple notation, e.g. [1 , 2 , 3] This subset P ofR3 is [select all that apply]: also a subspace of R not a subspace of R not a subspace of R not a subspace of R not a subspace of R by the subspace theorem because it is empty because it is not closed under scalar multiplication because it is not closed under vector addition 3 because it doesn't contain the zero vector of IR

 

 

Created with Microsoft OneNote 2016.