1
$\begingroup$

I have this matrix

$A= \begin{bmatrix} 1 & 2 & 2& 1 \\ 0&1& 0& -1\\ \end{bmatrix} $

which can be reduced to

$A= \begin{bmatrix} 1 & 0 & 2& 3 \\ 0&1& 0& -1\\ \end{bmatrix} $

The basis for the col A would be $(1,0),(2,1)$

the the basis of Row would be the non zero row of a reduced

$(1,0,2,3),(0,1,0,-1)$

But I am not sure if my row space is correct.

$\endgroup$
  • 4
    $\begingroup$ Looks fine to me. Note that you don't need to use the row-reduced rows for your basis, you can of course use the original rows (which correspond to the nonzero rows in your RREF). $\endgroup$ – Cameron Williams Dec 4 '14 at 16:39
  • $\begingroup$ Ok that makes sense thanks. $\endgroup$ – Fernando Martinez Dec 4 '14 at 16:49
  • $\begingroup$ @CameronWilliams Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – Julian Kuelshammer Apr 28 '15 at 20:44
  • $\begingroup$ @JulianKuelshammer I'm aware of the problem with unanswered problems. I just forgot about this post entirely. I actually expected OP to delete it. I'll answer it as Community. $\endgroup$ – Cameron Williams Apr 28 '15 at 21:45
1
$\begingroup$

Turning my comment into an answer: Looks fine to me. Note that you don't need to use the row-reduced rows for your basis, you can of course use the original rows (which correspond to the nonzero rows in your RREF).

$\endgroup$

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.