MAT224 ASSIGNMENT 1 SOLUTIONS DUE BY FRIDAY MAY 22, 2020, 11:59 PM
Each question is worth 5 marks.
Question 1. Let V = R2. For (a1,a2),(b1,b2) ∈ V and c ∈ R define
(0, 0) if c = 0 (a1,a2)+(b1,b2)=(a1+b1,a2+b2)andc(a1,a2)= ca1,a2 ifc̸=0.
c
Is V a vector space with these operations? If yes, prove it. If not, give an example showing one of the axioms fails.
1
Solution. This is not a vector space. Notice that 2(1, 1) = 2, 2 , however, (1, 1) + (1, 1) = (2, 2). That is,
2(1, 1) ̸= (1, 1) + (1, 1), so distributivity fails. There are many other counterexamples.
Question2.LetV beavectorspace.Showthatx=vistheonlysolutiontotheequationx+x=2vinV.Citeall axioms used.
Solution. Ifx=v,thenx+x=v+v=2v. Sox=visasolutiontotheequationx+x=2v. Toshowthatthisis the only solution,
x+x=2v=(1+1)v = 1v+1vbyS3 = v+vbyS5.
Adding−vand−xtobothsidesandusingA2,A4andA5givesx−v=v−x. Sincex−v+(v−x)=0,wehave v−x = −(x−v) (by uniqueness of the negative element). Hence, x−v = −(x−v), that is, (x−v)+(x−v) = 0, and by S3, 2(x − v) = 0, which implies x − v = 0, i.e., x = v (adding v to both sides and using A4 and A5).
Question 3. Suppose n ≥ 1, and let Un = {xg(x) : g(x) ∈ Pn−1}. Is Un a subspace of Pn? If yes, prove it. If not, find a counter example.
Solution. First of all notice that Un ⊆ Pn. Moreover, the zero polynomial f(x) = 0 is in Pn−1, so xf(x) = 0 ∈ Un. If k(x), h(x) ∈ Un, then k(x) = xk1(x) and h(x) = xh1(x) for some k1(x), h1(x) ∈ Pn−1. It follows that k(x) + h(x) = x(k1(x) + h1(x)). Since k1(x) + h1(x) ∈ Pn−1, we have k(x) + h(x) ∈ Un. Moreover, for any real number a ∈ R, ak(x) = axk1(x) = x(ak1(x)). Since ak1(x) ∈ Pn−1, ak(x) ∈ Un. Hence, Un is a subspace of Pn.
Question 4. (a) Suppose U and W are subspaces of a vector space V . Prove that U ∩ W = {v : v ∈ U and v ∈ W } is a subspace of V .
Solution. Since0∈U and0∈W wehave0∈U∩W. Givenu,v∈U∩W anda∈R,
u, v ∈ U ⇒ u + v ∈ U and u, v ∈ W ⇒ u + v ∈ W. Thatis,u+v∈U∩W. Moreover,au∈U andau∈W,hence,au∈U∩W.
(b)GiveanexampleoftwosubspacesU andW andavectorspaceV suchthatU∪W ={v:v∈U orv∈W}isnot a subspace of V .
Let V = R2, U = {(x, 0) : x ∈ R}, and W = {(0, y) : y ∈ R}. Then U and W are subspaces of V . However, U ∪ W is notasubspaceofV. Ifitwere,then(1,1)∈U∪W (since(1,1)=(1,0)+(0,1)and(1,0),(0,1)∈U∪W). Butthen this would imply that (1, 1) ∈ U or (1, 1) ∈ W , however, it is in neither, a contradiction.
1
Question 5. Let U = f ∈ F[0,1] : f is integrable and
answer. Note: You can use without justification your knowledge from first year calculus.
f(x)dx = 0 . Is U a subspace of F[0,1]? Justify your Solution. Let g(x) = 0 for all x ∈ [0, 1] (the zero function on [0, 1]). Then, g(x) is integrable and
1
g(x)dx = 0 ⇒ g(x) ∈ U.
0
1
0
2 MAT224 ASSIGNMENT 1 SOLUTIONS DUE BY FRIDAY MAY 22, 2020, 11:59 PM
Suppose f (x), h(x) ∈ U and a ∈ R. Then, f (x) + g(x) and af (x) are integrable, and 111
(f(x) + h(x))dx = f(x)dx + h(x)dx = 0 + 0 = 0 ⇒ f(x) + h(x) ∈ U 000
1 1
af(x)dx = a f(x)dx=0⇒af(x)∈U.
00
Hence, U is a subspace of F[0, 1].