# Thread: Are linearly independent subsets unique?

1. ## Are linearly independent subsets unique?

Problem:

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

Question:

For these problems, you set up a homogeneous system of equations and find a dependency relationship between the vectors. This seems to imply that the answer is not unique, i.e., depending on what vector you solve for, you can remove that particular one from the original set. However, in the following example, my text says there is only one solution, while I think there are different ones depending on what equation you solve for.

Example:

V=P1, {2-5x,3+7x,4-x}

Now, the dependency relationship turns out to be:

−3p1(x) + p2(x) + 2p3(x) = 0

The text solves for p2(x) and removes it from the original set. I don't see why I can't solve for any one of the equations and remove it from the original set?

Any help would be much appreciated!

2. Originally Posted by divinelogos
V=P1, {2-5x,3+7x,4-x}. Now, the dependency relationship turns out to be: −3p1(x) + p2(x) + 2p3(x) = 0

How do you get that relationship?

The text solves for p2(x) and removes it from the original set. I don't see why I can't solve for any one of the equations and remove it from the original set?

Any pair of vectors of the given family are linearly independent and span $V$ .

3. Originally Posted by divinelogos
Problem:

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

Question:

For these problems, you set up a homogeneous system of equations and find a dependency relationship between the vectors. This seems to imply that the answer is not unique, i.e., depending on what vector you solve for, you can remove that particular one from the original set. However, in the following example, my text says there is only one solution, while I think there are different ones depending on what equation you solve for.

Example:

V=P1, {2-5x,3+7x,4-x}

Now, the dependency relationship turns out to be:

−3p1(x) + p2(x) + 2p3(x) = 0
No, it isn't. Assuming you mean p1= 2- 5x, p2= 3+ 7x, p3= 4- x, then 3p1+ p2+ 2p3= 3(2- 5x)+ (3+ 7X)+ 2(4- x)= 6- 15x+ 3+ 7x+ 8- 2x= (6+ 3+ 8)+ (-16+ 7- 2)x= 17-11x, not 0.

The text solves for p2(x) and removes it from the original set. I don't see why I can't solve for any one of the equations and remove it from the original set?

Any help would be much appreciated!
Yes, you could. In fact, because the space of linear functions is two dimensional, with doing any "solving" (perhaps your text wanted to prove that the space was two dimensional), we know this set of 3 vectors must be dependent. But since no two of these are multiples of one another, any two will be independent and form a basis.