# Why Gaussian elimination on the columns changes the column space?

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?

## 1 Answer

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 $0$.)

• Yes it is spanned by different vectors, but both still span $R$, right? – plumSemPy Jul 7 '16 at 22:22
• 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. – Andreas Blass Jul 8 '16 at 0:39
• 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? – plumSemPy Jul 8 '16 at 1:50
• 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. – Andreas Blass Jul 8 '16 at 1:57
• @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. – Andreas Blass Feb 11 '18 at 11:41