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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?

maininformer
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2 Answers2

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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 $0$.)

Andreas Blass
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  • Yes it is spanned by different vectors, but both still span $R$, right? – maininformer Jul 07 '16 at 22:22
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    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 08 '16 at 00:39
  • 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? – maininformer Jul 08 '16 at 01:50
  • 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 08 '16 at 01:57
  • Thank you very much, this was very helpful. – maininformer Jul 08 '16 at 01:59
  • Well another problem actually, why can one not make the same argument about the row space? $\left[ {\begin{array}{cc} 1 & 2 \ 3 & 4 \ \end{array} } \right]$ has a row space spanned by ${t \choose 2t}$ and ${3t \choose 4t}$. If we subtract double the first row form the second, then it is spanned by ${t \choose 2t}$ and ${t \choose 0}$, Which is the x-axis. But the theorem says the row space does not change. – maininformer Jul 08 '16 at 23:40
  • No, the space spanned by $\binom t{2t}$ and $\binom t0$ is the whole two-dimensional $R^2$. So is the space spanned by $\binom t{2t}$ and $\binom{3t}{4t}$. What gave you the idea that one of these spaces is just the $x$-axis? For example, both of the spaces in question contain hte vector $\binom12$, which is not on the $x$-axis. – Andreas Blass Jul 09 '16 at 00:12
  • Ah I see the confusion now, correct me if I'm wrong. The theorem is that performing row operation will not change the row space but will change the column space. My impression was that performing row operation on the row space will not change it, but performing "column"wise operations on the columns will change the column space. Which is what I was not happy about. Thus doing column wise operation will not change the column space, doing row-wise will not change the row space but will change the column space. – maininformer Jul 09 '16 at 00:30
  • @AndreasBlass I know this was a while ago, but in your comment, you said one is a line with slope 5. Isn't it slope 2? – rb612 Feb 11 '18 at 07:27
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    @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
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First of all, you should clarify two defination: Vector space $R^n$ and Subspace .

Vector space $R^n$ : The space $R^n$ consists of all column vectors v with n components.

Subspace : A subspace of a vector space is a set of vectors (including 0 (i.e. the Null vector)) that satisfies two requirements:

If v and w are vectors in the subspace and c is any scalar, then

(1) v + w is in the subspace

(2) c * v is in the subspace

Consider a matrix by 5*2. The combination of first column and second column forms the column space. The column space is a subspace of $R^5$, not $R^n$ itself！

When doing Gaussian elimination, you do change the inner structure of each column, then you do change the column space. the column space is a subspace of $R^n$, not $R^n$.

tarmas99
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Dian He
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