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?

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- Apr 27th 2011, 01:00 AMArronSubsets of R^3
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? - Apr 27th 2011, 01:30 AMDeveno
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. - Apr 27th 2011, 01:31 AMFernandoRevilla
Any vector of S has de form

http://latex.codecogs.com/png.latex?...\in\mathbb{R})

verify

http://latex.codecogs.com/png.latex?z=2x-y

Try the second question.

**Edited**: Sorry, I didn`t see**Deveno**'s post. - Apr 27th 2011, 12:33 PMArron
I need to show that S is a basis for T.

Any ideas anyone? - Apr 27th 2011, 12:38 PMPlato
- May 4th 2011, 08:15 AMArron
I get your point.

How about the following,

First we check that S is linearly independant.

Suppose that

$\displaystyle \alpha $(2,2,2)+$\displaystyle \beta $(2,3,1)=0

that is

(2$\displaystyle \alpha $+2$\displaystyle \beta $,2$\displaystyle \alpha $+3$\displaystyle \beta $,2$\displaystyle \alpha $+$\displaystyle \beta $)=(0,0,0)

Equating corresponding coordinates, we obtain the simultaneous equations

2$\displaystyle \alpha +2\beta =0$

2$\displaystyle \alpha +3\beta =0$

2$\displaystyle \alpha +\beta =0$

Subtracting the third from the first gives $\displaystyle \beta $ =0 and substituting $\displaystyle \beta $ =0 into the third gives $\displaystyle \alpha $ =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 $\displaystyle \alpha ,\beta $ such that

(x,y,2x-y)=$\displaystyle \alpha (2,2,2)+\beta (2,3,1)$

Equating corresponding coordinates, we obtain the simultaneous equations

2$\displaystyle \alpha +2\beta =x$

2$\displaystyle \alpha +3\beta =y$

$\displaystyle 2\alpha +\beta =2x-y$

Subtracting the third equation form the first gives $\displaystyle \beta$ =-x+y, and substituting this into the third gives $\displaystyle \alpha =1/2(3x-2y).$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. - May 9th 2011, 01:56 AMArron
Hi!

Could anyone just confirm that my analysis is correct on the last post.

Thanks - May 9th 2011, 12:38 PMDeveno
your analysis looks fine to me.

- May 10th 2011, 12:13 AMArron
Thanks.

T= {(x,y,2x-y):x,y E R}

I found the dimension of T to be 2.

However I am not sure how to describe it geometrically.

Can anyone help? - May 10th 2011, 01:34 AMDeveno
the plane containing (2,2,2) and (2,3,1)?

- May 13th 2011, 06:13 AMHallsofIvy
And it contains (0, 0, 0). A plane is determined by

**three**points. - May 14th 2011, 01:26 AMDeveno
i was thinking of (2,2,2) and (2,3,1) being vectors, rather than points, and thus (implicitly) including the point (0,0,0). a better choice of words would have been: the plane through the origin...