# Thread: Finding Basis and Dimension

1. ## Finding Basis and Dimension

Hello!

I have two questions that are giving me some trouble,

1) I am trying to find a basis for a subspace W = {(s+4t, t, s, 2s-t)} for reals s, t.
Then the dimension will be the number of vectors in the basis. It appears to me there should be two vectors in the basis?

2) Similarily, and perhaps easier, is finding a basis (& dimension) for W = {(5t, t, -t}}.
Also, (seemingly unrelated?) is describing the geometric representation of W. Isn't it a plane?

1) Would a basis be {(1,0,1,2),(4,1,0,-1)} because of the parameters? (and thus dimension 2)

2) Similarily, {(5,1,-1)}? (dimension one?)
However I still am not sure of it's graphical represenation!

Thanks!!

2. Originally Posted by matt.qmar
Hello!

I have two questions that are giving me some trouble,

1) I am trying to find a basis for a subspace W = {(s+4t, t, s, 2s-t)} for reals s, t.
Then the dimension will be the number of vectors in the basis. It appears to me there should be two vectors in the basis?

2) Similarily, and perhaps easier, is finding a basis (& dimension) for W = {(5t, t, -t}}.
(5t, t, -t)= t(5, 1, -1).

Also, (seemingly unrelated?) is describing the geometric representation of W. Isn't it a plane?

1) Would a basis be {(1,0,1,2),(4,1,0,-1)} because of the parameters? (and thus dimension 2)
Yes, just write (s+ 4t, t, s, 2s- t)= (s, 0, s, 2s)+ (4t, t, 0, -t)= s(1, 0, 1, 2)+ t(4, 1, 0, -1) and the basis should be obvious.

2) Similarily, {(5,1,-1)}? (dimension one?)
However I still am not sure of it's graphical represenation!
(5t, t, -t)= t(5, 1, -1). However, if this is the "W" referred to, it is NOT a plane. It only has dimension 1, just as you say.

[/quote]Thanks!![/QUOTE]