Let S= { (1,1,0),(1,0,1),(0,1,1)} be a subset of the vectorspace F 3 a) Prove th

Question

Let S= { (1,1,0),(1,0,1),(0,1,1)} be a subset of the vectorspace F3 a) Prove that if F=R then S is linearly independent. b) Prove that if F has characteristic 2 then S is linearlydependent. Let S= { (1,1,0),(1,0,1),(0,1,1)} be a subset of the vectorspace F3 a) Prove that if F=R then S is linearly independent. b) Prove that if F has characteristic 2 then S is linearlydependent. a) Prove that if F=R then S is linearly independent. b) Prove that if F has characteristic 2 then S is linearlydependent.

Explanation / Answer

QuestionDetails: Let S= { (1,1,0),(1,0,1),(0,1,1)} be a subset of the vectorspace F3 a) Prove that if F=R then S is linearly independent.
THAT IS THESE ARE 3 VECTORS IN 3D SPACE
LET
X(1,1,0) + Y(1,0,1) +Z(0,1,1)=0
[X+Y , X+Z , Y+Z]=0
X+Y=0...................1
X+Z=0........................2
Y+Z=0.............................3
ADDING ALL
2(X+Y+Z)=0.............X+Y+Z=0........................4
SUBTRACTING EQNS.1,2,3 FROM EQN.4 ,WE GET
X=Y=Z=0
THAT IS IF X(1,1,0) + Y(1,0,1) +Z(0,1,1)=0...THEN IT IMPLIESX=Y=Z=0
HENCE S= { (1,1,0),(1,0,1),(0,1,1)} IS L.I.

b) Prove that if F has characteristic 2 then S is linearlydependent.
S= { (1,1,0),(1,0,1),(0,1,1)}
THAT IS THESE ARE 3 VECTORS IN 2D SPACE .
SO THEY REPRESENT A PLANE IN SPACE
THAT IS IF WE CALL THE 3 VECTORS AS OP,OQ,OR REPRESENTING 3POINTS
P,Q,R RESPECTIVELY WITH O AS ORIGIN , THEN THE PLANEDETERMINED BY
TRIANGLE PQR IS THE 2D SPACE UNDER CONSIDERATION.
IF EQN. OF THAT PLANE IS
AX+BY+CZ=D..............................5
THEN WE GET ON SUBSTITUTING FOR EACH POINT
A+B=D................................6
A+C=D.........................7
B+C=D................................8
ADDING ALL
2(X+Y+Z)=3D.............X+Y+Z=1.5D........................9
SUBTRACTING EQNS.1,2,3 FROM EQN.9 ,WE GET
X=Y=Z=0.5D
PUTTING IN EQN.5 , THE EQN. OF PLANE IS
0.5X+0.5Y+0.5Z=D.............................OR
X+Y+Z=K .........OR ..................Z=K-X-Y........................10
WHERE K IS A CONSTANT
THEN THE 3 GIVEN VECTORS BECOME
S= { (1,1,K-2),(1,0,K-1),(0,1,K-1)}
IF WE TAKE NOW
P(1,1,K-2) + Q(1,0,K-1) + R(0,1,K-1)}=0
P+Q=0...................................................11
P+R=0..............................12
P(K-2)+Q(K-1)+R(K-1)=0..................................13
FROM 11 AND 12 .....Q=R
PUTTING IN 13
P(K-2)+2Q(K-1)=0
P=-2(K-1)Q/(K-2)
HENCE TAKING Q=R=1 AND P = -2(K-1)/(K-2),WE CAN MAKE
P(1,1,K-2) + Q(1,0,K-1) + R(0,1,K-1)}=0
THAT IS
P(1,1,K-2) + Q(1,0,K-1) + R(0,1,K-1)}=0
DOES NOT IMPLY P=Q=R=0
HENCE THEY ARE LINEARLY DEPENDENT
a) Prove that if F=R then S is linearly independent.
THAT IS THESE ARE 3 VECTORS IN 3D SPACE
LET
X(1,1,0) + Y(1,0,1) +Z(0,1,1)=0
[X+Y , X+Z , Y+Z]=0
X+Y=0...................1
X+Z=0........................2
Y+Z=0.............................3
ADDING ALL
2(X+Y+Z)=0.............X+Y+Z=0........................4
SUBTRACTING EQNS.1,2,3 FROM EQN.4 ,WE GET
X=Y=Z=0
THAT IS IF X(1,1,0) + Y(1,0,1) +Z(0,1,1)=0...THEN IT IMPLIESX=Y=Z=0
HENCE S= { (1,1,0),(1,0,1),(0,1,1)} IS L.I.

b) Prove that if F has characteristic 2 then S is linearlydependent.
S= { (1,1,0),(1,0,1),(0,1,1)}
THAT IS THESE ARE 3 VECTORS IN 2D SPACE .
SO THEY REPRESENT A PLANE IN SPACE
THAT IS IF WE CALL THE 3 VECTORS AS OP,OQ,OR REPRESENTING 3POINTS
P,Q,R RESPECTIVELY WITH O AS ORIGIN , THEN THE PLANEDETERMINED BY
TRIANGLE PQR IS THE 2D SPACE UNDER CONSIDERATION.
IF EQN. OF THAT PLANE IS
AX+BY+CZ=D..............................5
THEN WE GET ON SUBSTITUTING FOR EACH POINT
A+B=D................................6
A+C=D.........................7
B+C=D................................8
ADDING ALL
2(X+Y+Z)=3D.............X+Y+Z=1.5D........................9
SUBTRACTING EQNS.1,2,3 FROM EQN.9 ,WE GET
X=Y=Z=0.5D
PUTTING IN EQN.5 , THE EQN. OF PLANE IS
0.5X+0.5Y+0.5Z=D.............................OR
X+Y+Z=K .........OR ..................Z=K-X-Y........................10
WHERE K IS A CONSTANT
THEN THE 3 GIVEN VECTORS BECOME
S= { (1,1,K-2),(1,0,K-1),(0,1,K-1)}
IF WE TAKE NOW
P(1,1,K-2) + Q(1,0,K-1) + R(0,1,K-1)}=0
P+Q=0...................................................11
P+R=0..............................12
P(K-2)+Q(K-1)+R(K-1)=0..................................13
FROM 11 AND 12 .....Q=R
PUTTING IN 13
P(K-2)+2Q(K-1)=0
P=-2(K-1)Q/(K-2)
HENCE TAKING Q=R=1 AND P = -2(K-1)/(K-2),WE CAN MAKE
P(1,1,K-2) + Q(1,0,K-1) + R(0,1,K-1)}=0
THAT IS
P(1,1,K-2) + Q(1,0,K-1) + R(0,1,K-1)}=0
DOES NOT IMPLY P=Q=R=0
HENCE THEY ARE LINEARLY DEPENDENT

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