This page on theorem 8.2 states that, Neither of the operations of the gaussian elimination changes the row space of an $m \times n$ matrix after applying the operation. It says later that this is only true about the row space and not the column space.

I can clearly see how multiplying and adding two vectors does not change the row space. Let's assume any pair of two dimensional non parallel non zero vectors. These vectors span $R^2$. Thus it does not matter how we combine them linearly, they will still span $R^2$.

The column space for any two non zero non parallel vectors can be thought of as being two dimensional vectors, spanning another two dimensional space again. Lets call this one $R^2_c$.

Now here is my question, doing gaussian elimination on the columns of a matrix, will firstly, do nothing to the span of the column space, $R^2_c$, because that space still can be spanned with the two new column vectors, and secondly, it will result in two new row vectors in $R^2$ that can still span $R^2$. So it seems to me by doing linear combinations on the column space, neither the row nor the column space change.

What am I doing wrong?


Consider the matrix $$\begin{pmatrix}1&5\\2&10\end{pmatrix}.$$ Its column space is one-dimensional, spanned by $\binom12$ (or equivalently by $\binom5{10}$ since these vectors are proportional). After a row operation, subtracting twice the first row from the second, the matrix becomes $$\begin{pmatrix}1&5\\0&0\end{pmatrix}.$$ The column space of this is still one-dimensional, but it's a quite different one-dimensional space from before; the new one is spanned by $\binom10$. (Edit to correct an error in a comment: The column space of the original matrix was a line of slope $2$ (not $5$); the column space of the new matrix has slope $0$.)

  • $\begingroup$ Yes it is spanned by different vectors, but both still span $R$, right? $\endgroup$ – plumSemPy Jul 7 '16 at 22:22
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    $\begingroup$ No, neither spans $R$. They span two different, one-dimensional subspaces of $R^2$. One is a line with slope 5, and the other is the $x$-axis. $\endgroup$ – Andreas Blass Jul 8 '16 at 0:39
  • $\begingroup$ I might have fundamental misunderstandings then. How is the x-axis different from $R$? and how are the x-axis and a line in $R^2$ different subspaces? how are they not the same set? $\endgroup$ – plumSemPy Jul 8 '16 at 1:50
  • $\begingroup$ The $x$ axis contains the vector $\binom10$ as well as all vectors of the form $\binom t0$ for any real number $t$. None of those vectors except $\binom00$ are elements of the other subspace, the one spanned by $\binom15$. So these two sets are not the same because they have different elements. (To show that two sets are different, all you need is a single element that's in one but not in the other; here you have infinitely many such elements.) And $R$ is yet another set entirely; its elements are real numbers, not two-component vectors. $\endgroup$ – Andreas Blass Jul 8 '16 at 1:57
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    $\begingroup$ @rb612 You're right; it seems I was looking at the matrix sideways. Since I can't edit a comment, I'll put a correction into the answer. $\endgroup$ – Andreas Blass Feb 11 '18 at 11:41

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