Thread: Linearly independent set that spans same subspace as original set of vectors

Problem:

Determine a linearly independent set of vectors that spans the same subspace of V as that spanned by the original set of vectors.

V=R3, {(1,1,1),(1,-1,1),(1,-3,1),(3,1,2)}

Question:

There is a theorem that says, "If one of the vectors in the set is a linear combination of the other vectors in the set, then that vector can be deleted from the given set of vectors".

I found the following dependency relationship among these vectors to be:

v1-2v2+v3=0

Now, I can solve for any one of the vectors in terms of the other two. Does this mean I can remove one, two, or three vectors from the original set, or am I limited to just one "vector removal"?

Thanks for any help!

Originally Posted by divinelogos

Problem:

Determine a linearly independent set of vectors that spans the same subspace of V as that spanned by the original set of vectors.

V=R3, {(1,1,1),(1,-1,1),(1,-3,1),(3,1,2)}

Question:

There is a theorem that says, "If one of the vectors in the set is a linear combination of the other vectors in the set, then that vector can be deleted from the given set of vectors".

I found the following dependency relationship among these vectors to be:

v1-2v2+v3=0

Now, I can solve for any one of the vectors in terms of the other two. Does this mean I can remove one, two, or three vectors from the original set, or am I limited to just one "vector removal"?

Thanks for any help!

So these vectors are linearly independent so they span all of

Since they only have to span the same space why not just pick the standard basis vectors.

Originally Posted by TheEmptySet

So these vectors are linearly independent so they span all of

Since they only have to span the same space why not just pick the standard basis vectors.

Can you elaborate a little? By standard basis vectors do you mean multiples of (1,0,0) (0,1,0) and (0,0,1)?

Originally Posted by divinelogos

Can you elaborate a little? By standard basis vectors do you mean multiples of (1,0,0) (0,1,0) and (0,0,1)?

That's correct.