Activity for 4.1: General Vector Spaces DEFINITION are defined: associating with

Question

Activity for 4.1: General Vector Spaces DEFINITION are defined: associating with each pair of 1 Let V be an arbitrary nonempty set ofobjects on which two operations addition, and multiplication by scalars. By addition we mean a rule for objects u and v in V an object u + v, called the sum of u v; by scalar multiplication we mean a rule for associating with each scalar k and ach object u in V an object ku, called the scalar multiple of u by k. If the following axioms V a vector space and we call the objects in V vectors. 1. If u and v are objects in V, then u+ v is in V are satisfied by all objects u, v, w in V and all scalars k and m, then we call 3, 4. u+(v+w) = (u + v) + w There is an o for all u in V For cach u in V, there is an object -u in V. called a negative of u, such that bject 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u 5. u+(-u) = (-u) + u = 0. 6. If k is any scalar and u is any object in V, then ku is in V 8. 9. 10. (k+m)u=ku + mu k(mu)=(km)(u) 1u=u To Show that a Set with Two Operations is a Vector Space Step I. Identify the set V of objects that will become vectors. Step 2. Identify the addition and scalar multiplication operations on V. Step 3. Verify Axioms 1 and 6; that is, adding two vectors in V produces a vector in V, and multiplying a vector in V by a scalar also produccs a vector in V. Axiom I is called closure under addition, and Axiom 6 is called closure under scalar multiplication Step 4. Confirm that Axioms 2, 3, 4, 5, 7, 8, 9, and 10 hold.

Explanation / Answer

LetX = (x1,y1), and Y = (x2,y2) be 2 arbitrary vectors in S and let k be an arbitrary scalar. Then X+Y = (x1,y1)+ (x2,y2) = (x1+x2,y1+y2) S as x1+x2 and y1+y2 R. Thus S is closed under vector addition. Further, kX = k(x1,y1) = (2kx,2ky) S as 2kx and 2ky R. Thus S is closed under scalar multiplication. We will now check whether the remaining vector space axioms are satisfied ore not:

Axiom 2 is satisfied as addition is commutative.

Axiom 3 is satisfied as addition is associative

Axiom 4 is satisfied as (0,0) = 0 is the additive identity of S.

Axiom 5 is satisfied as –X = (-x,-y) is the additive inverse of X = (x,y).

k(X+Y) = k(x1+x2,y1+y2) = (2k(x1+x2),2k(y1+y2)= (2kx1+2kx2,2ky1+2ky2) = (2kx1,2ky1)+ (2kx2,2ky2) =kX+kY so that the 7th axiom holds.

(k+m)X = (k+m)(x1,y1) = (2(k+m)x1, 2(k+m)y1)= (2kx1+2mx1,2ky1+2my1) =(2kx1,2ky1)+(2mx1,2my1)= kX+mX so that the 8th axiom holds.

k(mX) = k(2mx1,2my1) = (4kmx1,4kmy1). Further, (km)X = (2kmx1,2kmy1) (4kmx1,4kmy1) so that k(mX) (km)X. Thus, the 9th axiom does not hold.

The scalar multiplicative identity is ½ as ½(x1,y1) = (2*(1/2)x1,(2*1/2)y1) = (x1,y1)= X.

Since the 9th axiom does not hold, hence S is not a vector space.

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