vector spaces: how to prove the linear combination of V1V_1 and V2V_2 solve z=ax+byz = ax+by

I am working through a linear algebra book [1] that has a practice question for which I am a little lost. The question is:

“Let a,ba,b be real numbers. Consider the equation z=ax+byz = ax+by. Prove that there are two 3-vectors v1\boldsymbol{v_1}, v2\boldsymbol{v_2} such that the set of points [x,y,z][x, y, z] satisfying the equation is exactly the set of linear combinations of v1\boldsymbol{v_1} and v2\boldsymbol{v_2}. (Hint: Specify the vectors using formulas involving a, b.).”

It makes sense to me that if we have the two 3-vectors v1\boldsymbol{v_1} and v2\boldsymbol{v_2} that are linearly independent, they can span any point [x,y,z][x,y,z] in the field R3\mathbb{R^3}. However, I am missing something here to answer this proof.

[1] Klein, Philip. Coding the Matrix: Linear Algebra through Computer Science Applications (Page 204). Newtonian Press.

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1 Answer
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Two independent vectors do not span the whole space, only a plane. For example v1=(1,0,0)v1=(1,0,0) and v2=(0,1,0)v2=(0,1,0) span the z=0z=0 plane only. In fact the equation that you are given is the equation of a plane going through origin.
Suppose for now that a≠0a\ne 0 and b≠0b\ne 0. I can rewrite the equation as ax+by−z=0ax+by-z=0. One vector v1v1 can be chosen to have x=1x=1, y=0y=0, and z=az=a. Similarly, v2=(0,1,b)v2=(0,1,b). Any linear combination of v1v1 and v2v2 can be written as αv1+βv2\alpha v1+\beta v2, with α,β\alpha,\beta real numbers. Such a combination has the form α(1,0,a)+β(0,1,b)=(α,β,αa+βb)\alpha (1,0,a)+\beta(0,1,b)=(\alpha, \beta, \alpha a+\beta b). You can just plug this into your equation, to check that the linear combination is still part of the plane. Next step is to prove that any (x,y,z)(x,y,z) vector obeying your equation can be written in terms of v1v1 and v2v2. So all you need to do is find α\alpha and β\beta. Hint: α=x\alpha=x and β=y\beta=y. Make sure that the cases where aa, or bb or both are 0 are still OK.

  

 

To prove that any [x,y,z][x,y,z] vector obeying the equation, ax+byâˆ′z=0ax+by−z=0, can be written in terms of v1 and v2. \alpha[1,0,a]+\beta[0,1,b]=[x,y,z]\alpha[1,0,a]+\beta[0,1,b]=[x,y,z] \alpha*1+\beta*0 = x\alpha*1+\beta*0 = x \alpha = x\alpha = x \alpha*0+\beta*1 = y\alpha*0+\beta*1 = y \beta = y\beta = y \alpha a + \beta b = z\alpha a + \beta b = z substituting x for \alpha\alpha and y for \beta\beta we get ax + by = zax + by = z
– jeffalltogether
2 days ago

  

 

Sorry, that was a mess. I am thinking this is the next step?
– jeffalltogether
2 days ago

  

 

I’m new to this game, apparently enter sends the comment immediately. Anyway, here is the first comment in a cleaner form: To prove that any [x,y,z][x,y,z] vector obeying the equation: ax+by−z=0ax+by−z=0, can be written in terms of v1 and v2. We start with this form \alpha[1,0,a]+\beta[0,1,b]=[x,y,z]\alpha[1,0,a]+\beta[0,1,b]=[x,y,z] showing that: \alpha*1+\beta*0 = x\alpha*1+\beta*0 = x or \alpha = x\alpha = x; \alpha*0+\beta*1 = y\alpha*0+\beta*1 = y or \beta = y\beta = y; and \alpha a + \beta b = z\alpha a + \beta b = z. Finally, substituting x for \alpha\alpha and y for \beta\beta we get the original equation ax + by = zax + by = z.
– jeffalltogether
2 days ago

  

 

By the way, you can just edit the comments that you put in. No need to repeat them twice if you made a mistake
– Andrei
2 days ago