Cfrgtkky

Since [1 1 1] is a linear combination of [1 0 -1] and [0 1 2], would it be wrong to say that [1 1 1] and [1 0 -1] can serve as the basis for the subspace (assuming that [1 0 -1] and [0 1 2] are the "correct" bases)? If so, why?

I realize that there are an infinite(?) amount of bases for a given subspace (except the subspace of the zero vector), and this situation seems like it should be an example of that, but it also feels wrong somehow. Can anyone confirm or deny?

I think my main confusion stems from the fact that, when you're finding the basis of a column space for example, you eliminate all the column vectors that don't contribute to the vector space because they're dependent on the other vectors. But how do you choose which columns to eliminate and which ones to keep (since all of them can be written as a combination of the others)?

In my example, why do you prioritize keeping [0 1 2] over [1 1 1], even though they can both be written as combinations of each other (with [1 0 -1])?

Basis is not unique.

Yes, you are right, it works. The subspace has dimension $2$. Any two non-zero vectors in the subspace that are not multiples of each other do form a basis.

There are some results that help us to find a basis. For example, if you want to find a basis for a row space, one common strategy is to reduce it to RREF, then we know that the non-zero rows form a basis. If we want to find a basis for the column space, then the columns of the corresponding pivot columns of the RREF is a basis. That gives us a procedure to always being able to find a basis for row space and column space.

Remark:

If we are given two basis, to say that a basis is better than another basis require justification in the sense of what does one mean by "better". Perhaps, by a measure of how other vectors can be expressed as linear combination of them? or by sparseness? If you look at the first set, given a vector in the subspace, we can easily write

$$(x_1, x_2, x_3) = x_1(1,0,-1)+x_2(0,1,2)$$

very quickly.

$$(x_1, x_2, x_3)=(x_1-x_2)(1,0,-1)+x_2(1,1,1)$$

You can see that you still have to compute $x_1-x_2.$

Another possible consideration is perhaps you know that you are collecting data and you know that your data lies inside a subspace. We might purposely want to choose a basis and perhaps even arrange the vector inside the basis as some "directions"
can tell us more information. Your data might tend to lie along a few directions and the other directions are behaving like noise.

If the question is asking us to find a basis, any basis will do. I am just giving you a glimpse of possible applications of linear algebra.

The linear combination of the two vectors will fill different subspaces of $R^3$.
In fact, because the vectors are two, the subspaces will be planes.
Thus, we can say that, for these choice of vectors, the subspaces that are originated will be different.

Let's take another example and analyze the vector space $R^2$.
If we consider the vectors [1,0] and [0,1], their linear combination will fill the whole space $R^2$. Now consider another pair of vectors [1,2] and [2,9]; being the vectors linear indipendent, they will fill the whole space $R^2$.

The vectors both originate the same subspace $R^2$, thus, it does not matter which 'bases' we use to originate the same subspace.

If we repeat this process, we will soon realize that there are infinite set of vectors that fill $R^2$. The only condition is that the two vectors are linear indipendent from each others.