# Why is the matrix used to describe a linear transformation different if we use row vectors?

Lets say some linear transformation $$T$$ is defined by: $$T(x, y , z) = (x+2y, 2x+3y-4z, 4x-4y)$$

Now this can be expressed as $$\begin{pmatrix}1&2&0\\2&3&-4\\4&-4&0\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix}$$.

A friend of mine asked me why cant we represent this transformation as $$(x, y, ,z) M$$ where $$M$$ is the matrix given above.

Doing that computation gives the output to be $$(x+2y+4z, 2x+3y-4z,-4y)$$ (commas used to seperate elements).

Is it just the case the matrix representing linear transformations are different if the vectors are expressed as rows?

$$(x,\ y,\ z,)M^T=(x+2y,\ 2x+3y-4z, \ 4x-4y)$$
$$M \begin{pmatrix}x\\y\\z\end{pmatrix}=(x+2y,\ 2x+3y-4z, \ 4x-4y)^T$$,
since $$( M \begin{pmatrix}x\\y\\z\end{pmatrix})^T=(x,\ y,\ z,)M^T.$$
By convention, an $$m\times n$$ matrix $$A$$ over a field $$K$$ coincides with the matrix (with respect to standard bases) of the linear map $$K^n\to K^m$$ that is defined by left-multiplication by $$A$$, that is the map $$v\mapsto A\cdot v$$ . It is not the matrix of any map $$K^m\to K^n$$ (if $$n\neq m$$), and since right-multiplication by$$~A$$, that is $$w\mapsto w\cdot A$$ (viewing $$w$$ as a row vector) is such a map, $$A$$ is in particular not the matrix of right-multiplication by$$~A$$. (The transpose of $$A$$ is.)