0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.