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Is there a reason that the transformation matrix is on the left?

For example: $$ \begin{bmatrix}2&1\\-1&1\end{bmatrix}\begin{bmatrix}x&z\\y&v\end{bmatrix} $$ I understand that when multiplying matrices order is important, but is there any other reason other than that that the transformation matrix has to come first? Is it because there is no other way to 'write' the transformation matrix that allows it to be on the right?

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  • $\begingroup$ The matrix represents a linear map $f$, and usually we write $f(x)$, and not $x(f)$. But there are books writing $xA$ instead of $Ax$. $\endgroup$ – Dietrich Burde Aug 18 '17 at 15:26
  • $\begingroup$ Read up on matrix multiplication. $\endgroup$ – mathreadler Aug 18 '17 at 15:31
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Read up on matrix multiplication.

Yes you can have it on the right, if the vector you are transforming is a row vector:

$$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}2&-1\\1&1\end{bmatrix}$$

But as you can see here $-1$ and $1$ switch places as we need to do this transpose operator we describe below and which is on wikipedia.

If it is a more complicated transformation, say given by the matrix $\bf M$, then a column vector being transformed (multiplied by $\bf M$) from the left is the same as a row vector being transformed (multiplied by ${\bf M}^T$ from the right: $$({\bf Mv})^T = {\bf v}^T {\bf M}^T$$ This is due to a famous algebra rule of transpose (the $(\cdot)^T$ operator).

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  • $\begingroup$ I just realized for a transformation of \begin{bmatrix}-1&0\\0&-1\end{bmatrix} that it does not matter what order it comes in, but what about more complicated transformations? Is there a reason that it has to be on the left? I have updated the question. $\endgroup$ – Kyzen Aug 18 '17 at 15:47
  • $\begingroup$ So just to confirm, for a given matrix A, in order for it to be transformed by a a matrix M on the right, A has to be transposed? Would this work if A is a 2x2 matrix, or any matrix that can be multiplied regardless of order? $\endgroup$ – Kyzen Aug 18 '17 at 16:04
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    $\begingroup$ Yes the rule of transpose $({\bf Mv})^T = {\bf v}^T{\bf M}^T$ is general regardless of the dimensions of first and last matrix, as long as the matrix multiplication is well defined ( inner dimensions must agree ). $\endgroup$ – mathreadler Aug 18 '17 at 16:09
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We have $$\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-x\\-y\end{bmatrix}$$ whereas $$\begin{bmatrix}x\\y\end{bmatrix}\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$$ is undefined because it is a $2\times 1$ matrix multiplying a $2\times 2$ matrix. Matrix multiplication requires the number of columns of the left matrix to equal the number of rows of the right matrix. If the transformation matrix were taking other square matrices then this could be acceptable. e.g. if the transformation were something like $$\begin{bmatrix}w&x\\y&z\end{bmatrix}\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$$ then that would be allowed. In general though we have the transformation matrix on the left of the multiplication.

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  • $\begingroup$ It seems that using such a simple transformation and a 2x1 matrix that my question has been misunderstood. What I'm really trying to ask is for complicated transformation with matrices that can be multiplied regardless of order, why is the transformation matrix on the left. I have updated the question. $\endgroup$ – Kyzen Aug 18 '17 at 15:50
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    $\begingroup$ One could define a transformation with the transformation matrix on the right, as long as the multiplication is valid. I believe that the reason that the transformation matrices are typically on the left is because it is a natural convention. We read left to right, so this seems a natural way to perform the operation. This question addresses some similar questions: math.stackexchange.com/questions/61200/… $\endgroup$ – Dave Aug 18 '17 at 15:56

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