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Given an arbitrary 3x3 matrix M and an angle θ I can use this to, we'll say rotate about my space's Z-axis, so I'd create the rotation matrix:

$$ R_z = \begin{bmatrix} \cos\theta&-\sin\theta&0\\ \sin\theta&\cos\theta&0\\ 0&0&1 \end{bmatrix}$$

What I'm struggling with is the description of these 2 multiplications:

  1. M*Rz
  2. Rz*M

I believe that 1 describes M rotated about my space's Z-axis and 2 describes rotating M about it's Z-axis. Is this correct? Can someone help me with understanding this?

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  • $\begingroup$ I assume * represents matrix multiplication (instead of complex conjugation for example)? $\endgroup$ – YiFan Sep 11 at 13:42
  • $\begingroup$ @YiFan Yes, would it be more clear if I just wrote MR? I can change it if you think so? $\endgroup$ – Jonathan Mee Sep 11 at 13:46
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In general, for $n\times n$ matrices $A,B$, the matrix product $AB$ represents the composite map obtained by applying $B$ first then applying $A$. This is simply because for any vector $v$, $ABv=A(Bv)$, so multiplication by $B$ is done first.

I believe that $1$ describes $M$ rotated about my space's $Z$-axis and $2$ describes rotating $M$ about it's $Z$-axis.

No, this is not correct the way it is written. What does it even mean for a matrix to be "rotated" about an axis? It is however true that $MR_z$ corresponds to the linear transformation which first rotates the space about the $z$-axis, and then applies the linear transformation defiend by $M$. On the other hand, $R_zM$ corresponds to applying $M$ first, and then rotating about the $z$-axis. In general, these two are not the same, of course.

It would also be correct to say that $R_zM$ has column vectors which are rotated copies of the column vectors of $M$. That is, if $$M=\begin{bmatrix}\vert&\vert&\vert\\v_1&v_2&v_3\\\vert&\vert&\vert\end{bmatrix},$$ then $$R_zM=\begin{bmatrix}\vert&\vert&\vert\\R_zv_1&R_zv_2&R_zv_3\\\vert&\vert&\vert\end{bmatrix}.$$ Of course, you will observe that $R_zv_i$ is just $R_z$ applied to $v_i$, which corresponds to $v_i$ rotated about the $z$-axis (for each $i=1,2,3$). Furthermore, $MR_z$ has row vectors which are rotated copies of the row vectors of $M$. That is, if $$M=\begin{bmatrix}-&u_1&-\\-&u_2&-\\-&u_3&-\end{bmatrix}$$ then $$MR_z=\begin{bmatrix}-&u_1R_z&-\\-&u_2R_z&-\\-&u_3R_z&-\end{bmatrix}.$$ Of course, you will again recognise that $u_iR_z$ is just a rotated copy of the $u_i$, for each $i=1,2,3$.

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