I have a question about the following 2 subset of R^3
S = ((2,2,2),(2,3,1)),
T = ((x,y,2x-y):x,y which are elements of real numbers).
Show S is contained within T and write down a vector in R^3 that does not belong to T.
Can anyone help?
if (2,2,2) is contained in T, then we have:
2 = x
2 = y
2 = 2x - y = 2(2) - 2 = 4 - 2 = 2, which is true. see if you can find an x and y so that (x,y,2x-y) = (2,3,1).
to find a vector NOT in T, pick some x, and some y, and calculate 2x-y. pick anything BUT that (value for 2x-y) for the 3rd coordinate.
I get your point.
How about the following,
First we check that S is linearly independant.
Suppose that
(2,2,2)+ (2,3,1)=0
that is
(2 +2 ,2 +3 ,2 + )=(0,0,0)
Equating corresponding coordinates, we obtain the simultaneous equations
2
2
2
Subtracting the third from the first gives =0 and substituting =0 into the third gives =0, so the set is linearly independent.
Alternatively the set {(2,2,2),(2,3,1)} is linearly indepentent, as the second vector is not a multiple of the first.
We then check whether each vector in T can be written as a linear combination of the vectors in the set. Choosing (x,y,2x-y) as the vector in T, we seek numbers such that
(x,y,2x-y)=
Equating corresponding coordinates, we obtain the simultaneous equations
2
2
Subtracting the third equation form the first gives =-x+y, and substituting this into the third gives So
(x,y,2x-y)=1/2(3x-2y)(2,2,2)+(-x+y)(2,3,1)
and the set of vecotrs S spans T.
Since the two conditions hold, the set S is a basis for T.