Intuitive understanding of the $BAB^{-1}$ formula for changing basis in linear transformations. I would please like to have a better understanding on why we use the formula $BAB^{-1}$ when chaning basis on linear transformations. (I have not a complete understanding on this so there are probably errors in my question, please point them out so I can learn more...)
For example:
In $\mathbb{R^3}$
Lets say we have a basis $e = {(e_1, e_2, e_3)}$ and another basis $f = {(f_1, f_2, f_3)}$. We have a change of basis matrix from $f$ to $e$ called $B$ that expresses what $f_1, f_2, f_3$ is in the basis $e$.
We have defined a linear transformation $R_e$ in the basis $e$ with the matrix $A$.
Let's say we have three vectors expressed in the base $f$: $P = \left(
\begin{array}{c}
s\\
t\\
u\\
\end{array}
\right)$
Now we want to apply the linear transformation $R_e = A$ to these vectors:
First we have to change the basis of $P$ which gives us: $BP$. $BP$ is now $P$ expressed in the basis $e$. To apply the linear transformation we now only has to multiply $BP$ with $A$: $ABP$. But according to the formula it should be $BAB^{-1}P$. Where is my thinking incorrect?
 A: Suppose that you have a closed box in Sweden which you wish to be open in Sweden. You only know how to open the box if you are in China. Then 
(open box in Sweden) = (go from Sweden to China) THEN (open box in China) THEN (go from China to Sweden)
(note that function composition is written funny in math compared to the English word "THEN.")
A: First off, with matrices being applied on the left, the correct chancge of basis formula is $B^{-1}AB$ (the formula $BAB^{-1}$ would be correct for matrices applying to the right).
The way to understand this is as follows. Matrix$\times$vector multiplication models applying a linear transformation only when bases at entry and exit are specified, and the initial vector is expressed in coordinates on the entry basis, while the product expresses the result on the exit basis. This is natural for linear maps between different spaces, but also applies to maps from a space to itself. However in the latter case it is usually confusing to use different bases at entry and exit. Therefore in that situation using different bases at entry and exit is best reserved for the situation where changing basis is all one want to do, in other words the linear map is just the identity. Now if one has the matrix $A$ of a linear map $\phi:V\to V$ with respect to a basis $e$, but want to apply it to a vector expressed in coordinates with respect to another basis $f$, a $3$-step procedure applies: first apply the identity but using $f$ as entry basis and $e$ as exit basis, then apply $\phi$ which can be done using $A$, giving an expression onthe same basis $e$, and finally convert the coordinates back to $f$ by applying the identity with $e$ as entry basis and $f$ as exit basis.
It remains to identify the matrices for the identity with different entry and exit basis. It turns out that $B$ converts coordinates w.r.t. $f$ into coordinates w.r.t. $e$ (despite it being called change of basis from $e$ to$~f$!). So the procedure is: multiply by $B$ then by $A$ and finally by $B^{-1}$.
A: Preliminary remark: You should replace the sentence 
– Let's say we have three vectors expressed in the base $f\!: \ P =\left(\matrix{s \cr t\cr u\cr}\right)$ 
by 
– Let's say we have a point $P$ expressed in the base $f\!: \ [P]_f =[s,t,u]^\top$.
Now to your question: The two matrices $A$ and $B$ play a completely different rôle in this game. The matrix $A$ is the important one. It encodes how the "geometric points" $P$ will be moved around in space by the linear transformation ${\cal A}$ in question. Unfortunately this encoding refers  to a particular basis $(e_i)_{1\leq i\leq 3}$ and is valid only when the points $P$ as well as their images $P':={\cal A}(P)$ are coordinatized with respect to this basis. If it so happens that you want to use some other basis $(f_j)_{1\leq j\leq3}$ and corresponding coordinates then some computational overhead will arise. That's where the matrix $B$ comes in. Its columns are the $e$-coordinates of the new basis vectors $f_j$. The rules of linear algebra tell us that the $e$-coordinates of any point $P$ are computed from its $f$-coordinates by means of
$$[P]_e=B\,[P]_f\ .$$ Then we get the $e$-coordinates of the image point $P'={\cal A}(P)$ by applying the matrix $A$ to $[P]_e$:
$$[P']_e=A\,[P]_e=AB\,[P]_f\ .$$
But in the end we want the $f$-coordinates of the image point $P'$, and it is again the matrix $B$ that does the proper bookkeeping:
$$[P']_f=B^{-1}\,[P']_e=B^{-1}AB\,[P]_f\ .\tag{1}$$
Looking at $(1)$ we see that when the world is described in $f$-coordinates  the matrix $\tilde A$ that encodes the map ${\cal A}$ is given by
$$\tilde A=B^{-1} A B\ .$$
When we have to compute the $f$-coordinates of a million image points $P'={\cal A}(P)$ it pays to compute the matrix $\tilde A$ once and for all, so that we then have the simple formula
$$[P']_f=\tilde A\,[P]_f\ .$$
A: Suppose $g :span(u_1,...,u_n)\to span(u_1,...,u_n)$ define like $g(x)=Ax$ and $f:span(u_1,...,u_n)\to span(v_1,...,v_n)$ definie by $f(x)=Bx$. Then, you have that $f^{-1}:span(v_1,...,v_n)\to span(u_1,...,u_n)$ and thus 
$$f\circ g\circ f^{-1}: span(v_1,...,v_n)\to span(v_1,...,v_n)$$ and is define by $f\circ g\circ f^{-1}(x)=B^{-1}A B x$ that give the matrix $A$ in $span(v_1,...,v_n)$.
