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The title says it all. I'm currently taking an introductory course in linear algebra and this issue has not been adressed spesifically. What I'm wondering is this: Given a transformation matrix $A$ and vector $\vec{x}$ such that $A\vec{x}$ is some transformation of $\vec{x}$ (rotation, scaling, etc), is it true that if matrix $B$ is $A$ after some row operation then $Ax = Bx$? Also why is that? Are there some relevant geometric interpretations?

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  • $\begingroup$ I think I might be missing your question - do $$A=\left[\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right]$$ $$B=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right]$$ qualify? $\endgroup$ – Alfonso Fernandez Feb 25 '13 at 10:43
  • $\begingroup$ Yes that is one example. $\endgroup$ – Andreas Hagen Feb 25 '13 at 10:45
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    $\begingroup$ Then if we take $$x=\left[\begin{array}{c} 0 \\ 1 \end{array}\right]$$ then $$Ax=\left[\begin{array}{c} 1 \\ 0 \end{array}\right]$$ while $Bx=x$ $\endgroup$ – Alfonso Fernandez Feb 25 '13 at 10:49
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It depends on what you mean by $Ax$. Row operations are changes of coordinates, so what is true is that if $A$ and $B$ are related by row operations, then they represent the same operation in different coordinates. However, for some fixed column vector $x$, we will have $Ax\ne Bx$ in general; see the example already given by Ludolila and Alfonso. But this isn't the right thing to check, because in terms of the linear map, we've changed coordinates to get from $A$ to $B$, and the column $x$ represents two different points in the two coordinate systems, so we shouldn't expect equality.

To illustrate with the example already given, let $A=I$ and $B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, obtained from $A$ by swapping the two rows. Thus, while:

$$A\begin{pmatrix}x\\y\end{pmatrix}\ne B\begin{pmatrix}x\\y\end{pmatrix}$$

we do have:

$$A\begin{pmatrix}x\\y\end{pmatrix}=B\begin{pmatrix}y\\x\end{pmatrix}$$

The idea is that the two columns $\begin{pmatrix}x\\y\end{pmatrix}$ and $\begin{pmatrix}y\\x\end{pmatrix}$ represent the same point in different coordinates, as they are related by a row operation.

The key message here is that matrices aren't linear maps, and you can only identify a matrix with a linear map once you have chosen a basis, i.e. a particular set of coordinates.

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This in no true. For example, take $A$ to be a $2\times 2$ identity matrix $I$. Now let $B=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ (it is $A$ after interchanging the rows). Then clearly $Ax \neq Bx$ (for $x\neq 0$).

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