6. Let F be the integers mod 7, and V (our vector space) is the set of 2-tuples
ID: 2941322 • Letter: 6
Question
6. Let F be the integers mod 7, and V (our vector space) is the set of 2-tuples - i.e.,(a,b) with a and b elements of F.
a) How many vectors ar there in V?
b) Given B1=(2,3) and B2=(3,4) as basis vectors, write the element (1,1) as a linear combination of these vectors (the scalars in this linear combination are elements of the field F). You can do this by solving a (fairly simple) system of equations, but calculations must be done mod 7.
c) Would B1=(2,3) and B2=(3,1) work as a basis for this vector space? Give a reason for your answer.
Explanation / Answer
6. Let F be the integers mod 7, and V (our vector space) is the set of 2-tuples - i.e.,(a,b) with a and b elements of F.
F=[0,1,2,3,4,5,6]
V=[(0,0),(0,1),(1,0),(0,2).......(6,6)]
a) How many vectors ar there in V?...
EACH OF 7 INTEGERS (0 TO 7) CAN BE PAIRED WITH EACH OF THE 7 INTEGERS (0 TO 7)
HENCE WE HAVE 7*7=49 VECTORS IN V
b) Given B1=(2,3) and B2=(3,4) as basis vectors, write the element (1,1) as a linear combination of these vectors (the scalars in this linear combination are elements of the field F). You can do this by solving a (fairly simple) system of equations, but calculations must be done mod 7.
LET [1,1]=X[2,3]+Y[3,4]=[2X+3Y,3X+4Y]
2X+3Y=1 OR 7P+1......................................1
3X+4Y=1 OR 7Q+1.....................................2
EQN.1 - EQN. 2 GIVES
X+Y=0....OR .....7R
SOLVING THIS WITH EQN. 1 GIVES...X=6 AND Y=1
HENCE....
[1,1]=6[2,3]+1[3,4]=[15,22]=[1,1]
c) Would B1=(2,3) and B2=(3,1) work as a basis for this vector space? Give a reason for your answer.
V IS VECTOR PACE WITH DIMENSION =2
TO FORM A BASIS OF V WE NEED 2 VECTORS WHICH SHALL BE L.I.
WE FIND THAT.... 5B1=5[2,3]=[10,15]=[3,1]=B2.....SO THESE 2 VECTORS ARE L.D.
SO THEY CAN NOT FORM A BASIS FOR THE VECTOR SPACE V
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