# Linear Combinations of Dependent Vectors

1. Feb 14, 2009

### Random Variable

1. The problem statement, all variables and given/known data

If (u,v,w) is a family of linearly dependent vectors in vector space V and vector x is in the span of (u,v,w), then x=αu+βv+γw has infinitely-many choices for α,β, and γ.

2. Relevant equations

If (u,v,w) is linearly dependent, then there exists an α, β, and γ, not all equal to zero, such that αu+βv+γw=0.

3. The attempt at a solution

My first attempt, which didn't go anywhere, was to assume that there were only a finite number of choices and see if that led to a contradiction.

For my second attempt, I started with the fact that (u,v,w) is linearly dependent. Then I multiplied both sides by an arbitrary scalar n. Then I thought I could add x to both sides and manipulate the equation somehow, but that didn't lead anywhere either.

If I was dealing with the vector space R^n, then x=αu+βv+γw would have infinitely-many solutions because you would end up with at least one free variable since at least one of the vectors is just a linear combination of the others. Can you apply that reasoning to all vector spaces?

Last edited: Feb 14, 2009
2. Feb 14, 2009

### HallsofIvy

Staff Emeritus
As you say, since (u,v,w) is linearly independent so there exist $\alpha$, $\beta$, $\gamma$, not all 0, such that $\alpha u+ \beta v+ \gamma w$= 0.

Further, since (u,v,w) spans V there exist, for any x in V, scalars a, b, c such that au+ bv+ cw= x.

Now, what can you say about $au+ bv+ cw+ R(\alpha u+ \beta v+ \gamma w)$ for any real number R?

3. Feb 14, 2009

### Random Variable

It equals x?

(Rα+a)u +(Rβ+b)v +(Rγ+c)w = x

And since Rα or Rβ or Rγ is not zero, the equation has infinitely-many solutions?

4. Feb 14, 2009

### HallsofIvy

Staff Emeritus
Yes, every different choice for R gives a different linear combination but they are all equal to x!

And it is doing exactly what YOU suggested: