# Thread: exams..help/.. subspace..dimension..

1. ## exams..help/.. subspace..dimension..

Hi I need help with this .
the set X={(x,y,z) which is an element of R^3 such that x=y=2z} is a subspace of R^3
what is the dimension of X?
Thanks.

2. ## Re: exams..help/.. subspace..dimension..

Originally Posted by n22
Hi I need help with this .
the set X={(x,y,z) which is an element of R^3 such that x=y=2z} is a subspace of R^3
what is the dimension of X?
Is this $\{(1,1,0.5)\}$ the basis for $X~?$

If it is then what is dimension of $X~?$

3. ## Re: exams..help/.. subspace..dimension..

x= y= 2z.

So, whatever z is, y= 2z and x= 2z. Such a vector is (2z, 2z, a)= z(2, 2, 1). Now, what do you think a basis for that space is? What is its dimension?

4. ## Re: exams..help/.. subspace..dimension..

Originally Posted by HallsofIvy
x= y= 2z.

So, whatever z is, y= 2z and x= 2z. Such a vector is (2z, 2z, a)= z(2, 2, 1). Now, what do you think a basis for that space is? What is its dimension?
Would the basis just now just be (2,2,1)<--since linearly independent .. and the dimension =1

5. ## Re: exams..help/.. subspace..dimension..

Yes. And I had an obvious typo "Such a vector is (2z, 2z, a)= z(2, 2, 1)" should have been "Such a vector is (2z, 2z, z)= z(2, 2, 1)" but you understood that.

Simlarly, if you have a subspace (x, y, z) satisfying only "y= z", x could be anything so such a vector would be (x, y, y)= (x, 0, 0)+ (0, y, y)= x(1, 0, 0)+ y(0, 1, 1). {(1, 0, 0), (0, 1, 1)} would be a basis and the dimension would be 2.