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

It makes sense to me that if we have the two 3-vectors $\boldsymbol{v_1}$ and $\boldsymbol{v_2}$ that are linearly independent, they can span any point $[x,y,z]$ in the field $\mathbb{R^3}$. 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.


Two independent vectors do not span the whole space, only a plane. For example $v1=(1,0,0)$ and $v2=(0,1,0)$ span the $z=0$ plane only. In fact the equation that you are given is the equation of a plane going through origin. Suppose for now that $a\ne 0$ and $b\ne 0$. I can rewrite the equation as $ax+by-z=0$. One vector $v1$ can be chosen to have $x=1$, $y=0$, and $z=a$. Similarly, $v2=(0,1,b)$. Any linear combination of $v1$ and $v2$ can be written as $\alpha v1+\beta v2$, with $\alpha,\beta$ real numbers. Such a combination has the form $\alpha (1,0,a)+\beta(0,1,b)=(\alpha, \beta, \alpha a+\beta b)$. 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 $(x,y,z)$ vector obeying your equation can be written in terms of $v1$ and $v2$. So all you need to do is find $\alpha$ and $\beta$. Hint: $\alpha=x$ and $\beta=y$. Make sure that the cases where $a$, or $b$ or both are 0 are still OK.

  • $\begingroup$ To prove that any $[x,y,z]$ vector obeying the equation, $ax+by−z=0$, can be written in terms of v1 and v2. $\alpha[1,0,a]+\beta[0,1,b]=[x,y,z]$ $\alpha*1+\beta*0 = x$ $\alpha = x$ $\alpha*0+\beta*1 = y$ $\beta = y$ $\alpha a + \beta b = z$ substituting x for $\alpha$ and y for $\beta$ we get $ax + by = z$ $\endgroup$ – jeffalltogether Oct 21 '16 at 14:45
  • $\begingroup$ Sorry, that was a mess. I am thinking this is the next step? $\endgroup$ – jeffalltogether Oct 21 '16 at 14:46
  • $\begingroup$ 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 $[x,y,z]$ vector obeying the equation: $ax+by−z=0$, can be written in terms of v1 and v2. We start with this form $\alpha[1,0,a]+\beta[0,1,b]=[x,y,z]$ showing that: $\alpha*1+\beta*0 = x$ or $\alpha = x$; $\alpha*0+\beta*1 = y$ or $\beta = y$; and $\alpha a + \beta b = z$. Finally, substituting x for $\alpha$ and y for $\beta$ we get the original equation $ax + by = z$. $\endgroup$ – jeffalltogether Oct 21 '16 at 14:50
  • $\begingroup$ By the way, you can just edit the comments that you put in. No need to repeat them twice if you made a mistake $\endgroup$ – Andrei Oct 21 '16 at 14:59
  • 1
    $\begingroup$ Because the problem states that $z=ax+by$. I can just move $z$ to the other side, and then $ax+by-z=0$. Now I want to write this as $ax+by+cz+d=0$. I can do that only if for every $(x,y,z)$ combination that obeys the first equation, it obeys the second one as well. So If I choose $x=y=0$ in the first equation, than $z=0$. When I plug in $x=y=z=0$ in the second equation I get $a0+b0+c0+d=0$, therefore $d=0$. I can now choose a random $x\ne0$, and $y$ in such a way that $ax+by=1$. Then $z=1$. Plug in, and you get $1+c1=0$, so $c=-1$ $\endgroup$ – Andrei Feb 17 '17 at 3:16

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