Are transformation matrices invariant over row operations?

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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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? – Alfonso Fernandez Feb 25 '13 at 10:43
Yes that is one example. – Andreas Hagen Feb 25 '13 at 10:45
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 $Bx=x$ – Alfonso Fernandez Feb 25 '13 at 10:49

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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Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Are transformation matrices invariant over row operations?

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?

-
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? – Alfonso Fernandez Feb 25 '13 at 10:43
Yes that is one example. – Andreas Hagen Feb 25 '13 at 10:45
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 $Bx=x$ – Alfonso Fernandez Feb 25 '13 at 10:49

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.

-

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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user
The text passage is a discussion about the invariance of transformation matrices under row operations in the context of linear algebra. It includes a question and several responses providing examples and explanations.

TYPE: NO_QUESTION

However, this classification is not entirely accurate because the passage does contain a question. The more appropriate classification should be:

TYPE: SINGLE_QUESTION_WITH_ANSWER

The initial post poses a question about the invariance of transformation matrices under row operations, and the subsequent comments and responses provide answers and examples to address this question.