# 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 $x\neq 0$).