# Subsets of R^3

• Apr 27th 2011, 01:00 AM
Arron
Subsets 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 AM
Deveno
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 AM
FernandoRevilla
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 PM
Arron
I need to show that S is a basis for T.

Any ideas anyone?
• Apr 27th 2011, 12:38 PM
Plato
Quote:

Originally Posted by Arron
I need to show that S is a basis for T.

Arron, do you ever plan to show any of your own effort?
I have not seen any in the questions you have posted.
• May 4th 2011, 08:15 AM
Arron

First we check that S is linearly independant.

Suppose that

$\alpha$(2,2,2)+ $\beta$(2,3,1)=0
that is
(2 $\alpha$+2 $\beta$,2 $\alpha$+3 $\beta$,2 $\alpha$+ $\beta$)=(0,0,0)
Equating corresponding coordinates, we obtain the simultaneous equations

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

Subtracting the third from the first gives $\beta$ =0 and substituting $\beta$ =0 into the third gives $\alpha$ =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 $\alpha ,\beta$ such that
(x,y,2x-y)= $\alpha (2,2,2)+\beta (2,3,1)$

Equating corresponding coordinates, we obtain the simultaneous equations

2 $\alpha +2\beta =x$
2 $\alpha +3\beta =y$
$2\alpha +\beta =2x-y$

Subtracting the third equation form the first gives $\beta$ =-x+y, and substituting this into the third gives $\alpha =1/2(3x-2y).$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 AM
Arron
Hi!

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

Thanks
• May 9th 2011, 12:38 PM
Deveno
your analysis looks fine to me.
• May 10th 2011, 12:13 AM
Arron
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 AM
Deveno
the plane containing (2,2,2) and (2,3,1)?
• May 13th 2011, 06:13 AM
HallsofIvy
And it contains (0, 0, 0). A plane is determined by three points.
• May 14th 2011, 01:26 AM
Deveno
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...