Proof of the matrix of change coordinates If i have the reference frames  $R$ and $R '$, and I call $M(R',R)$ the matrix that changes coordinates from  $R'$ to $R$. 
Let's say that $R=\{p,e_1,\dots,e_n\}$, $R'=\{p,e'_1,\dots,e'_n\}$ and $R''=\{p,e''_1,\dots,e''_n\}$ are the reference frames from $A$ I have to prove that $M(R,R')=M(R'',R') \cdot M(R,R'')$
I tried to do it in a long way, putting all the members of the matrix and doing the product but I ended doing a mess.
Does anyone know any easier way to do it?
Sorry if I didn't use the correct way to write the mathematic things, I'll try to do it beter next time.
 A: Let be the same point $A$ espressed in different coordinate systems:
$A\; is\;\{x_1,x_2,...x_n\}$ $\; and\; is\;\{x'_1,x'_2,...x'_n\}$ $\; and\;is\;\{x''_1,x''_2,...x''_n\}$
The matrix that transforms from $R$ to $R'$ is $M(R,R')$,  and let's note any matrix $M(K, J)$ as "from $K$ to $J$". Example with coordinates in "row":
 $\{x'_1,x'_2,...x'_n\}=\{x_1,x_2,...x_n\}·M(R,R')$ $\;\;\;$(1)
So we can also say
 $\{x''_1,x''_2,...x''_n\}=\{x_1,x_2,...x_n\}·M(R,R'')$ $\;\;\;$(2)
 $\{x'_1,x'_2,...x'_n\}=\{x''_1,x''_2,...x''_n\}·M(R'',R')$ $\;\;\;$(3)
Replacing (2) in (3) we get:
 $\{x'_1,x'_2,...x'_n\}=\{x_1,x_2,...x_n\}·M(R,R'')·M(R'',R')$ $\;\;\;$(4)
Compare (1) and (4) and you get
$M(R,R')=M(R,R'')·M(R'',R')$ 
Now, if you work with vectors in column, instead of row, you must transpose all equations, and remember "the transpose of a product is the swapped product of transposed":
 $\{x'_1,x'_2,...x'_n\}^T=M(R,R')^T·\{x_1,x_2,...x_n\}^T$ $\;\;\;$(5)
and, by rewritting (2)(3) you can arrive to $M(R,R')^T=M(R'',R')^T·M(R,R'')^T$
Notice that if you define $M$ as "row ordered" you use (1). But if the same $M$ notation means "column ordered" then (1) is of type $\;X'=M·X\;$ and the transpose operations are not needed and you can do the prove by simple (2)(3) rewritting in proper order.
