1
$\begingroup$

Given that I have an augmented matrix in Row Echelon Form or Reduced Row Echelon form, and the bottom row(s) contain only zeroes.

Generally speaking (disregarding fringe cases if any?):

$\endgroup$
  • 1
    $\begingroup$ They don't, just means you have a lineraly dependant vector. Yes, if you have more variables than equations. Yes, because its free, it means it can be anything, hence infinite number of soltuons $\endgroup$ – XRBtoTheMOON Feb 11 '18 at 23:45
  • $\begingroup$ So it's incorrect to say that : if there is a row of zeroes, then there are infinitely many solutions $\endgroup$ – mathguy Feb 11 '18 at 23:46
4
$\begingroup$

Why do rows of only zero imply the presence of a free variable?

This is not true, consider matrices with more rows than columns like $$\left(\begin{matrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{matrix}\right).$$ Free variables are implied by columns without pivot element.

Can you have a free variable without the bottom row(s) containing only zeroes?

Yes, consider $$\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0\end{matrix}\right).$$

Does the presence of a free variable always mean you have an infinite number of solutions?

This depends on the size of the base field. If it is infinite (like $\mathbb{Q}$ or $\mathbb{R}$) then yes. Otherwise no.

$\endgroup$
  • $\begingroup$ What about this thread math.stackexchange.com/questions/309724/… is the accepted answer saying that a row of zeroes implies the presence of free variables? $\endgroup$ – mathguy Feb 12 '18 at 0:08
  • $\begingroup$ I don't see an accepted answer there. But as I wrote: Free variables are implied by columns without pivot element. This is the correct characterization. $\endgroup$ – azimut Feb 12 '18 at 0:11
  • $\begingroup$ Sorry, i meant the top answer, but point taken ty $\endgroup$ – mathguy Feb 12 '18 at 0:12

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.