# Proving that vector space is a linearly independent subset of bigger vector space

So I have this problem:

Problem

$\left\{x,y,z\right\}$ is a linearly independent subset of another vector V, find the constants a and b such that $\left\{x-ay,ay-z,z-by\right\}$ is also a linearly independent subset of V.

I know that if $\left\{x,y,z\right\}$ is linearly independent then:

$\alpha(x) + \delta(y) + \beta(z) = \vec{0} \rightarrow \alpha = \delta = \beta = 0$

I would assume the same holds true for this:

$\alpha(x-ay) + \delta(ay-z) + \beta(z-by) = \vec{0}$

and I also know $\alpha,\delta,\beta = 0$ since it is linearly independent. However, how do I relate those two combinations, I'm not so sure about how to get a and b, at first I thought I'd just do a system of equations to find a and b but I'm not sure. Can someone help me out with this?

• You omitted something: If $\{x,y,z\}$ is linearly independent then $\alpha x + \delta y + \beta b =\vec 0$ implies that $\alpha = \delta = \beta = 0$. – aes May 1 '15 at 19:50

Hint: Write $$\alpha(x-ay)+\delta(ay-z)+\beta(z-by)=\alpha x+(-a\alpha+a\delta-b\beta)y+(-\delta+\beta)z$$ Now use the fact that $\alpha x+\beta y+\gamma z=0$ implies $\alpha=\beta=\gamma=0$.