# Linear independence

1. Oct 18, 2010

### theRukus

1. The problem statement, all variables and given/known data
Let {X, Y, Z} be linearly independent in Rn. If {X, Y, Z, W} is linearly dependent, show that W $$\epsilon$$ span{X, Y, Z}. NB: You must SHOW this.

2. Relevant equations

3. The attempt at a solution
For W to belong to the span of {X,Y,Z}, W = aX + bY + cZ where a, b, c are some real numbers. Since {X,Y,Z,W} is linearly dependent, one of the four vectors can be shown as a linear combination of the other three, ie. W = aX + bY + cZ, where a, b, c are some real numbers.

On my assignment, I have the previous statement, and I've also represented X, Y, Z, and W with {x1, x2, x..., xn}, etc. I've taken these equations and multiplied the real numbers in.

I know what I'm trying to say here, but I don't exactly know how to say it, if you know what I mean. Is my explanation enough? Does it SHOW that W belongs to the span of {X, Y, Z}?

Thanks,

Connor Bode

2. Oct 18, 2010

### jbunniii

Yes, at least one of the vectors can be expressed as a linear combination of the other three, but you haven't demonstrated that W is one of the vectors that can be so expressed.

What you can say is that there exist real numbers $a,b,c,d$ (at least one of them nonzero) such that

$$aW + bX + cY + dZ = 0$$

To proceed from here, examine whether $a = 0$ is a possibility, given that $X$, $Y$, and $Z$ are linearly independent. If you can exclude that possibility, then rearrange the equation to the desired form.

3. Oct 18, 2010

### vela

Staff Emeritus
This is one of those proofs where you're trying to show something "obvious," so it's kind of hard to identify exactly what you need to show. So what you have is {X, Y, Z} are linearly independent. This means aX+bY+cZ=0 has only the trivial solution a=b=c=0. The fact that {X, Y, Z, W} are linearly dependent means that eX+fY+gZ+hW=0 has a non-trivial solution. You need to get from that statement to W=qX+rY+sZ. Think about the coefficients and when they can or can not be 0.

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