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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?

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$(x,\ y,\ z,)M^T=(x+2y,\ 2x+3y-4z, \ 4x-4y)$

and

$ M \begin{pmatrix}x\\y\\z\end{pmatrix}=(x+2y,\ 2x+3y-4z, \ 4x-4y)^T$,

since $( M \begin{pmatrix}x\\y\\z\end{pmatrix})^T=(x,\ y,\ z,)M^T.$

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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.)

And just to be clear, the convention I'm referring to is a combined effect of how matrix multiplication is defined and how the matrix of a linear map, with respect to given bases, is defined.

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