Lin Algebra - Find a basis for the given subspaces

[jinksys]

Find a basis for the given subspaces of R3 and R4.

a) All vectors of the form (a, b, c) where a =0.

My attempt:

I know that I need to find vectors that are linearly independent and satisfy the given restrictions, so...

(0, 1, 1) and (0, 0, 1)

The vectors aren't scalar multiples of each other and therefore not linearly dependent. What I am unsure of is how many vectors I'm supposed to list. From a previous question I posted here which talked about "Find a basis for the subspace W = Span(s)..." it was shown that since W was a subspace of R3, that its dimension was smaller than R3 so I would need less than three vectors.

 

[vela]

You need to find enough vectors so that they span the subspace but not so many that the vectors are linearly dependent. So far, you know you don't have too many because the two vectors are linearly independent. You just need to convince yourself that they span the subspace. If they do, you have a basis.

 

[jinksys]

So you're saying that I need to find a set of vectors whose linear combination can represent any other vector in the set?

So, (0,0,1) and (0,1,0) satisfy the requirements of the problem (a=0, b and c are reals) and there exists a linear combination for any other (0,b,c) vector that we'd like to represent. Am I on the right track?

 

[KIDRoach]

I'm pretty sure both of your answers are correct:

{(0,0,1), (0,1,0)}
{(0,1,1), (0,0,1)}

Both of these sets can be the basis of the space that you are referring to.

 

[jinksys]

Another thing, if I may. Is the number of vectors, (dimension?), of a basis going to depend on the restrictions placed upon it?

For example, W is a subspace of R3 with vectors in the form (a,b,c), where a=b=c.

Only one vector can be a part of that basis since every other vector would be a multiple of that vector. Therefore the dim is 1 and the basis only has one vector.

Would this be correct?

 

[murmillo]

Yes, that's one way to think about it. For the case of W, if you "take care" of the first component a, you automatically take care of the other components. Whereas in your initial question, you have the first component fixed, and you have to take care of the second and third components separately. That is why your basis has 2 vectors.

 

[vela]

Another thing, if I may. Is the number of vectors, (dimension?), of a basis going to depend on the restrictions placed upon it?

For example, W is a subspace of R3 with vectors in the form (a,b,c), where a=b=c.

Only one vector can be a part of that basis since every other vector would be a multiple of that vector. Therefore the dim is 1 and the basis only has one vector.

Would this be correct?

Yes. Generally, each independent constraint will reduce the dimension by 1. In the first problem, you have just one constraint, a=0, so the dimension of the subspace is 3-1=2. In the second problem, you have two constraints, a=b and a=c (or any other two equivalent constraints, e.g., a=b and b=c), so the dimension of the resulting subspace is 3-2=1.

As KIDRoach mentioned, in your first problem, the first two vectors you chose are a basis for the subspace because you can express (0,b,c) as a linear combination of (0,1,1) and (0,0,1):

(0,b,c) = b (0,1,1) + (c-b) (0,0,1)