What Happens When Gram-Schmidt Is Applied to Linearly Dependent Vectors?

[mpm]

I have a question about this process.

What will happen if this process is applied to a set of vectors {v1, v2, v3} where v1 and v2 are linearly independent, but v3 belongs to set Span(v1,v2). Will the process fail? If it fails, why does it fail?

 

[mpm]

Doesn't anyone possibly know the answer to this question? Or can anyone even give me some direciton on it?

 

[shmoe]

Have you tried it out on a concrete example? What happens?

 

[BerkMath]

The goal of GM is to take a nonorthogonal set of linearly independent functions and construct an orthogonal basis such that the the span of the original set is contained in the span of the orthonormalized set. This is a key ingridient to proving its most natural corollarly: that every finite dimensional inner product space admits an orthonormal basis. Note that vj is not contained in span(v1,v2,...vj-1) since (v1,...vj) is linearly independent and therefore vj is not in the span(e1,...,ej-1). If vj is in the span(v1,...vj-1) for any j, then carrying out GS merely produces a particular element which lies in the span(e1,...ej-1). It does nothing to further the purpose of GS, however; it does not destroy it, so long as your original set does contain elements which lie outside the span of its companions. But, if j=Dim(V) where V was an arbitrary space, and vk was an element of span(v1,...vk-1) for k<j, then continuing this process would result in an orthonormal set which DOES not form a basis for V, it merely spans/forms a basis or some subset/space of V.

 

[HallsofIvy]

I have a question about this process.
What will happen if this process is applied to a set of vectors {v1, v2, v3} where v1 and v2 are linearly independent, but v3 belongs to set Span(v1,v2). Will the process fail? If it fails, why does it fail?

Eventually, you will wind up having to divide by 0.