Instructions: A subset W of some n-space R^n is defined by means of a given cond
ID: 1941024 • Letter: I
Question
Instructions:
A subset W of some n-space R^n is defined by means of a given condition imposed on the typical vector (x1,x2,...xn). Apply Theorem 1 to determine whether or not W is a subspace of R^n.
Theorem 1:
Condition for a Subspace
The nonempty subset W of the vector space V is a subspace of V If and only if it satisfies the following two conditions:
(i) If u and v are vectors in W, then u +v is also in W.
(ii) If u is in W and c is a scalar, then the vector cu is also in W.
Question:
W is the set of all vectors in R^4 such that x1 +2x2 + 3x3 + 4x4 = 0
Explanation / Answer
A subset W of some n-space Rn is defined by means of a given condition imposed on on the typical vector (x1,x2,...,xn). Apply the theorem to determine whether or not W is a subspace of Rn Theorem: The nonempty subset W of the vector space V is a subspace of V iff it satisfies the following two conditions: (i)If u and v are vectors in W, the u + v is also in W. (ii)If u is in W and c is a scalar, the the vector cu is also in W. W is the set of all vectors in R4 such that x1 + x2 = x3 + x4. LET ANY ELEMENT X IN U BE X=[X1,X2,X3,X4] AND ANYELEMENT Y IN V BE Y=[Y1,Y2,Y3,Y4] SO WE HAVE FROM THE GIVEN PROPERTY THAT X1+X2=X3+X4...............................................................1 Y1+Y2=Y3+Y4.......................................................2 NOW LET US APPLY THE 2 TESTS NEEDED FOR THEM TO FORM A SUB SPACE WE GET BEFORE APPLYING THE TESTS WE NOTE THAT W IS NON EMPTY SET SINCE WE CAN FIND SEVERAL VECTORS IN R4 such that x1 + x2 = x3 + x4....VIZ.... [3,2,4,1].....3+2=4+1=5 ETC...SO OK ... (i)If u and v are vectors in W, the u + v is also in W. X+Y=[X1,X2,X3,X4]+[Y1,Y2,Y3,Y4] = [X1+Y1,X2+Y2,X3+Y3,X4+Y4] IN THE RESULTANT VECTOR WE FIND FROM 1 AND 2 THAT (X1+Y1)+(X2+Y2) = (X1+X2)+(Y1+Y2)= (X3+X4)+(Y3+Y4)=(X3+Y3)+(X4+Y4) SO IT SATISFIES THE GIVEN REQUIREMENT AND HENCE IS AN ELEMENT OF W ...OK.. (ii)If u is in W and c is a scalar, the the vector cu is also in W. CX=C[X1,X2,X3,X4]=[CX1,CX2,CX3,CX4] IN THE RESULTANT VECTOR WE FIND FROM 1 THAT CX1+CX2=C[X1+X2]=C[X3+X4]=CX3+CX4... SO IT SATISFIES THE GIVEN REQUIREMENT AND HENCE IS AN ELEMENT OF W ...OK.. HENCE W IS SUB SPACE............PROVED