Math 215.01, Spring Term 1, 2021 Writing Assignment #4
Turn your le in on Pweb in pdf format under \Writing Assign- ments”. There is no need to put your name on the document since I will grade anonymously, and Pweb will keep track of the authorship of the document.
You will be graded on your writing (use of quantiers, state- ment and use of denitions, and other mathematical language) as well as the validity and completeness of your mathematical arguments.
1. Let V be a vector space. For this problem, be very explicit and mention which property and/or result you are using in each step of your argument.
Supposethatc2Rand~v2V aresuchthatc~v=~0. Show that either c = 0 or ~v = ~0.
Hint: In mathematics, one of the standard ways to prove a statement of the form \A or B” is to assume that A is false, and use this to prove that B must be true (or alternatively to assume that B is false, and use this to prove that A must be true). This allows you to use an additional assumption, which is extremely useful. In this case, I suggest that you also assume that c 6= 0 (in addition to c ~v = ~0), and then prove that ~v = ~0.
Denition 4.1.1: A vector space is a nonempty set V of ob- jects, called vectors, equipped with operations of addition and scalar multiplication, along with an element ~0, such
that the following properties hold:
1. Forall~v,w~ 2V,wehave~v +w~ 2V (closureunder addition).
2.Forall~v2V andallc2R,wehavec~v2V (closure under scalar multiplication).
3. Forall~v,w~ 2V,wehave~v +w~ =w~ +~v (commuta- tivity of addition).
4. Forallu~,~v,w~ 2V,wehave(u~+~v)+w~ =u~+(~v + w~) (associativity of addition).
5. Forall~v 2V ,wehave~v +~0=~v (~0isanadditive identity).
6. Forall~v 2V,thereexistsw~ 2V suchthat~v +w~ =~0 (existence of additive inverses).
7. Forall~v,w~ 2V andallc2R,wehavec(~v +w~)= c ~v + c w~ .
8. For all ~v 2 V and all c ; d 2 R, we have (c + d ) ~v = c ~v + d ~v .
9. Forall~v2V andallc;d2R,wehavec(d~v)= ( c d ) ~v .
10.Forall~v 2V,wehave1~v =~v.
Proposition 4.1.11. Let V be a vector space. 1. 0 ~v = ~0 for all ~v 2 V .
2. c ~0 = ~0 for all c 2 R.
3. (-1) ~v = ~v for all ~v 2 V .
Since we need to proof a statement of the form A or B, we can assume that A is false, and use this to prove that B must be true.
Proof: Let V be a vector space. Suppose that c 2 R and ~v 2 V are such that c ~v = ~0. If c 6= 0, then ~v 2 V = ~0.
Let V be a vector space. Let c 2 R be arbitrary such that c 6= 0, and let ~v 2 V be arbitrary such that c ~v = ~0. Since c 6= 0, we can x n 2 R to be multiplicative inverse of c such that nc = 1. By multiplying n on both side of theequation~0=c~v,wehaven~0=n(c~v). Since n 2 R and according to proposition 4.1.11, item 2, we know n ~0 = ~0 which gives us ~0 = n (c ~v ).
Now notice that
~0 = n ( c ~v )
= (n c ) ~v (By property 9: associativity of multiplication) = 1 ~v (Since n is multiplicative inverse of c )
= ~v (By property 10: scalar identity):
Since c 2 R is arbitrary nonzero scalar and v 2 V is arbi- trary, we have shown that: Suppose c 2 R and ~v 2 V are suchthatc~v =~0. Ifc 6=0,then~v 2V =~0. Baseonthe proof above, we can conclude: Suppose c 2 R and ~v 2 V are such that c ~v =~0. Then either c = 0 or ~v =~0.
2. SupposeV isavectorspaceandthatc;d2Randv2V have the property
c ~v = d ~v :
Show that if ~v 6= ~0, then c = d. Just as in Problem 1, be very explicit and mention which property and/or result you are using in each step of your argument.
Denition 4.1.1: A vector space is a nonempty set V of ob- jects, called vectors, equipped with operations of addition and scalar multiplication, along with an element ~0, such that the following properties hold:
1. Forall~v,w~ 2V,wehave~v +w~ 2V (closureunder addition).
2.Forall~v2V andallc2R,wehavec~v2V (closure under scalar multiplication).
3. Forall~v,w~ 2V,wehave~v +w~ =w~ +~v (commuta- tivity of addition).
4. Forallu~,~v,w~ 2V,wehave(u~+~v)+w~ =u~+(~v + w~) (associativity of addition).
5. Forall~v 2V ,wehave~v +~0=~v (~0isanadditive identity).
6. Forall~v 2V,thereexistsw~ 2V suchthat~v +w~ =~0 (existence of additive inverses).
7. Forall~v,w~ 2V andallc2R,wehavec(~v +w~)= c ~v + c w~ .
8. For all ~v 2 V and all c ; d 2 R, we have (c + d ) ~v = c ~v + d ~v .
9. Forall~v2V andallc;d2R,wehavec(d~v)= ( c d ) ~v .
10.Forall~v 2V,wehave1~v =~v.
Proposition 4.1.11. Let V be a vector space.
1. 0 ~v = ~0 for all ~v 2 V .
2. c ~0 = ~0 for all c 2 R.
3. (-1) ~v = ~v for all ~v 2 V .
Let V be a vector space and let ~v 2 V be arbitrary such that~v 6=~0. Letc;d 2Rbearbitrarysuchthatc~v =d~v. Since d 2 R, we know d is the additive inverse of d. By adding (d)~v on both side of the equation c ~v = d ~v, we have c ~v + (d ) ~v = d ~v + (d ) ~v .
Now notice that
c ~v + ( d ) ~v = d ~v + ( d ) ~v
(c + (d )) ~v = (d + (d )) ~v (By property 8: distributivity of multiplication)
= 0 ~v (Since (d ) is the additive inverse of d ) = ~0 (By proposition 4.1.11: item 1)
In problem 1, we have proved that: Let W be a vector space. Suppose a 2 R and w~ 2 W are such that aw~ =~0. Theneithera=0orw~ =~0. Since(c+(d))2Rand ~v 2 V is a nonzero vector, then (c + (d )) must be zero in order to satisfy (c + (d)) ~v = ~0.
Then we have
c + (d) = 0
c+(d)+d=0+d (Byaddingdonbothsideoftheequation)
c + ((d) + d) = 0 + d (Associativity of addition)
c +0 = 0+d (Since (d) is the additive inverse of d)
c = d: (Additive identity)
Since ~v 2 V is arbitrary nonzero vector and c;d 2 R are arbitrary, we have shown that: Suppose c ~v = d ~v . If ~v 6= ~0, then c = d.
3. Let V be a vector space. Suppose that U and W are both subspaces of V . Let U \ W be the intersection of U and W, i.e.
U\W =f~v 2V :~v 2U and~v 2Wg:
Is U \ W necessarily a subspace of V ? Give a proof, or construct a specic counterexample with two subspaces U and W in a specic vector space V .
Denition 4.1.12: Let V be a vector space. A subspace of V is a subset W V with the following properties:
1.~0 2 W .
2. For all w~1; w~2 2 W , we have w~1 + w~2 2 W .
3. For all w~ 2 W and all m 2 R, we have m w~ 2 W .
Let V be a vector space and let U and W be subspaces of V. Let U \W be the intersection of U and W. Since U and W are subspaces and according to denition 4.1.12, property1,wehave~02Uand~02W. Since~02U and ~0 2 W and according to the denition of \, we have ~02U\W. Since~02U\W,U\W isnotanempty intersection. Let ~x ; y~ 2 U \ W be arbitrary. Since ~x ; y~ 2 U \ W , we know ~x ; y~ 2 U and ~x ; y~ 2 W . Since U and W are subspaces and according to denition 4.1.12, property 2, we have ~x + y~ 2 U and ~x + y~ 2 W . Since ~x + y~ 2 U and
~x + y~ 2 V and according to the denition of \, we can say that~x+y~2U\W. Since~x;y~2U\W arearbitrary,we have shown that U and W is closed under vector addition. Let c 2 R be arbitrary and let n~ 2 U \ W be arbitrary. Sincen~2U\W,weknown~2Uandn~2W. SinceU and W are subspaces and according to denition 4.1.12, property 3, we have cn~ 2 U and cn~ 2 W. Since cn~ 2 U and c n~ 2 W and according to the denition of \, we have cn~2U\W. Sincec2Risarbitraryandn~2U\W is arbitrary, we have shown U \ W is closed under scalar multiplication. Since U and W are subspaces of V and all three properties listed in denition 4.1.12 are satised by U \W, we can conclude U \W is a subspace of V.
4. Let V be a vector space. Suppose that U and W are sub- spacesofV. LetU[W betheunionofU andW,i.e.
U [ W = f ~v 2 V : ~v 2 U o r ~v 2 W g :
Is U [ W necessarily a subspace of V ? Give a proof, or construct a specic counterexample with two subspaces U and W in a specic vector space V .
Denition 4.1.12: Let V be a vector space. A subspace of V is a subset W V with the following properties:
1.~0 2 W .
2. For all w~1; w~2 2 W , we have w~1 + w~2 2 W .
3. For all w~ 2 W and all m 2 R, we have m w~ 2 W .
LetV beavectorspacesuchthatV =R2 =f(x;y): x;y 2Rg. LetU andW besubspacesofV suchthatU = f(n;0): n2RgandW =f(0;m): m2Rg. LetU[W be the union of U and W.
To show U is a subspace of R2: Let n = 0, then we have U = (0; 0) which indicates ~0 2 U. Let n1; n2 2 R be arbitrary with (n1; 0); (n2; 0) 2 U. By using linear combination, we have (n1; 0)+(n2; 0) = (n1 +n2; 0+0) = (n1 +n2; 0). Since n1;n2 2R,wehaven1+n2 2R. Sincen1+n2 2R,we know (n1 + n2; 0) 2 U. Since n1; n2 2 R are arbitrary, we have shown U is closed under vector addition. Let c 2 R
be arbitrary and let n3 2 R be arbitrary with (n3;0) 2 U.
Now notice c (n3;0) = (c n3;c 0) = (c n3;0). Since
c2Randn3 2R,weknowcn3 2R. Sincecn3 2R,
we know that (c n3;0) 2 U. Since c 2 R is arbitrary and
n3 2 R is arbitrary, we have shown U is closed under scalar
multiplication. Since U satises the three properties listed
in denition 4.1.12, we can conclude U is a subspace of V =R2.
ToshowW isasubspaceofR2: Letm=0,thenwehave W = (0;0) which indicates ~0 2 W. Let m1;m2 2 R be arbitrary with (0; m1 ); (0; m2 ) 2 W . By using linear com- bination, we have (0; m1) + (0; m2) = (0 + 0; m1 + m2) = (0;m1 +m2). Since m1;m2 2 R, we have m1 +m2 2 R. Since m1 + m2 2 R and m1;m2 2 R are arbitrary, we know (0;m1 + m2) 2 W. Thus we have shown W is closed under vector addition. Let b 2 R be arbitrary and let m3 2 R be arbitrary with (0;m3) 2 W. Now notice b(0;m3) = (b0;bm3) = (0;bm3). Since b 2 R andm3 2R,weknowbm3 2R. Sincebm3 2R,we know that (0;b m3) 2 W. Since b 2 R is arbitrary and
m3 2 R is arbitrary, we have shown W is closed under scalar
multiplication. Since W satises the three properties listed
in denition 4.1.12, we can conclude W is a subspace of V =R2.
Since 12 R and U = f(n;0) : n 2 Rg and W = f(0;m) : m2Rg,weknow(1,0)2Uand(0,1)2W. Since (1, 0) 2 U and (0, 1) 2 W and according to the denition of [, we have (1, 0), (0, 1) 2 U [W. By applying linear combination, we have (1, 0) + (0, 1) = (1, 1). If (1; 1) 2 U, we will have (1;1) = (n;0) which indicates 1 = 0. Since 1 = 0 is a contradiction, we know that (1;1) 62 U. If (1;1) 2 W, we will have (1;1) = (0;m) which indicates 1 = 0. Since 1 = 0 is a contradiction, we know that (1;1) 62 W: Since (1, 1) 62 U and (1, 1) 62 W and according to the denition of [, we have shown (1, 1) 62 U[W which indicates U [ W is not closed under vector addition in this example. Since U [ W doesn’t satisfy the second property listed in denition 4.1.12 in this specic example, we know U [ W is not a subspace of V = R2 when U = f(n; 0) :
n2RgandW =f(0;m): m2Rg. Sincewehavefound onespeciccasewhereU[W isnotasubspaceofV,we can say U [ W is not necessarily a subspace of V .
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5. Comment on working with partner(s): We completed the problem set individually and checked our work together. Although we took dierent approaches to prove these prob- lems, we had the same results in the end. We conclude our works are mostly valid.