The 1 Loo wciastic matrix er 4 Exercise Set Chapter Vector Spaces , he the set o
ID: 3114196 • Letter: T
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The 1 Loo wciastic matrix er 4 Exercise Set Chapter Vector Spaces , he the set of all ordered pairs of real numbers, and spa the der the following addition and scalar multiplication op- he foll erations on u = (ul , 112) and v = (ul , U2): u + v = (ul + ul, u2 + U2), ku = (0, ku2) (a) Compute u + v and ku for u = (2,4), v = (1,-3), and k=5. (b) In words, explain why V is closed under addition and scalar multiplication. (c) Since addition on V is the standard addition operation on R2, certain vector space axioms hold for V because they are known to hold for R2. Which axioms are they? (d) Show that Axioms 7, 8, and 9 hold. (e) Show that Axiom 10 fails and hence that V is not a vector space under the given operations. 2. Let V be the set of all ordered pairs of real numbers, and consider the following addition and scalar multiplication op- erations on u (111, 112) and v = (ui, U2): ku = (kui, kuz) 1( 1 -t u-l, u2 + U2-1),Explanation / Answer
1 (a). We have u+v = (2,4)+(1,-3) = (2+1,4-3) = (3,1).Also, 5(2,4) = (0,5*4) = (0,20).
(b). V is the set of all ordered pairs of real numbers. Given the way addition and scalar multiplication are defined on V, we have, for u = (u1,u2) and v = (v1,v2), u =v = (u1+v1,u2+v2)and ku = (0,ku2) where k is a scalar. Since (u1+v1,u2+v2) is an ordered pair of real numbers and so is (0,ku2), hence V is closed under addition and scalar multiplication.
(c ) & (d). The conditions that a vector space S must satisfy are as under:
1. For all X, Y , X+Y is in S.
2. For all X, Y , X+Y = Y+X ( commutativity of vector addition).
3. For all X, Y, Z , (X+Y)+Z=X+(Y+Z) (Associativity of vector addition).
4. For all x, 0+X = X+0 = X ( Existence of Additive identity)
5. For any X, there exists a -X such that X+(-X)= 0 (Existence of additive inverse)
6. For any scalar k, kX is in S.
7. For all scalars r and vectors X,Y, r(X+Y)=rX+rY (Distributivity of vector addition).
8. For all scalars r,s and vectors X , (r+s)X=rX+sX (Distributivity of scalaraddition).
9. For all scalars r,s and vectors X, r(sX)=(rs)X( Associativity of scalar multiplication).
10. For all vectors X, 1X=X ( Existence of Scalar multiplication identity).
We may observe that u+v, being an ordered pair of real numbers is in V. Further, u+v = (u1+v1,u2+v2) = (v1+u1,v2+u2) =v+u. Also, if w = (w1,w2), then (u+v)+w = u+(v+w) ( as addition is associative), 0+u = (0,0) +(u1,u2)= (0+u1,0+u2) = (u1,u2) = u. Thus, the first 4 axioms of a vector space are satisfied. Since –u has not been defined in V, we cannot say anything about the 5th axiom. Further, ru = (0,ru2), being an ordered pair of real numbers is in V so that the 6th axiom is also satisfied. As regards the 7th axiom, we have r(u+v) = r(u1+v1,u2+v2) = (0,r(u2+v2) = (0,ru2+rv2) =(0,ru2)+ (0,rv2)= ru+rv so that the 7th axiom is satisfied. Further, (r+s)u = (0, (r+s)u2) = (0,ru2)+(0,su2) = ru+su so that the 8th axiom is satisfied. Also, r (su) = r(0,su2) = (0,rsu2) = (rs)u so that the 9th axiom is satisfied. However, 1*u = 1*(u1,u2) = (0,1*u2) = (0,u2)u, the 10th axiom is not satisfied.
(e ). As may be observed from the above, the 10th axiom fails so that V is not a vector space.
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