Matrix of linear operator in different bases [PMA Rudin] 
$\mathbf{9.35}\quad$ Theorem $\ \; $ If $[A]$ and $[B]$ are $n$ by $n$ matrices, then $$\det([B][A])=\det[B]\det[A].$$ 
$\mathbf{9.36}\quad$ Theorem $\ \; $ A linear operator $A$ on $R^n$ is invertible if and only if $\det[A]\neq 0$.
$\qquad\mathbf{9.37}\quad$ Remark $\ \; $ Suppose $\{\mathbf e_1,\ldots,\mathbf e_n\}$ and $\{\mathbf u_1,\ldots,\mathbf u_n\}$ are bases in $R^n$. Every linear operator $A$ on $R^n$ determines matrices $[A]$ and $[A]_U$, with entries $a_{ij}$ and $\alpha_{ij}$, given by $$A\mathbf e_j=\sum_i a_{ij}\mathbf e_i,\qquad\; A\mathbf{u}_j=\sum_i \alpha_{ij}\mathbf{u}_i.$$ If $\color{red}{\underset{\Rule{8.5em}{0.1em}{0.09em}}{\color{black}{\mathbf u_j=B\mathbf e_j=\sum b_{ij}\mathbf e_i}}}$, then $A\mathbf u_j$ is equal to $$\sum_k\alpha_{kj}B\mathbf{e}_k=\sum_k \alpha_{kj}\sum_i b_{ik}\mathbf{e}_i=\sum_i\left(\displaystyle\sum_k b_{ik}\alpha_{kj}\right)\mathbf e_i,$$ and also to $$AB\mathbf e_j=A\sum_k b_{kj}\mathbf e_k=\sum_i\left(\displaystyle\sum_k a_{ik}b_{kj}\right)\mathbf{e}_i.$$ Thus $\sum b_{ik}\alpha_{kj}=\sum a_{ik}b_{kj}$, or $$\tag{91} [B][A]_U=[A][B].$$ $\color{red}{\underset{\Rule{8.5em}{0.1em}{0.09em}}{{\color{black}{\text{Since $B$ is invertible,}}}}}$ $\det[B]\neq 0$. Hence $(91)$, combined with Theorem $9.35$, shows that $$\tag{92}\det[A]_U=\det[A].$$ $\qquad$ The determinant of the matrix of a linear operator does therefore not depend on the basis which is used to construct the matrix. It is thus meaningful to speak of the determinant of a linear operator, without having any basis in mind.

This excerpt from Rudin's Principles of Mathematical Analysis and I have couple of questions. 
1) What is $B$? He wrote that $\mathbf{u}_j=B\mathbf{e}_j=\sum \limits_{i=1}^{n}b_{ij}\mathbf{e}_i$. I guess that $B$ is some  linear operator that translates each $\mathbf{e}_j$ to $\mathbf{u}_j$. Am I right? 
If yes why such operator exists?
2) Why $B$ is invertible? Why it's so obvious?
Can anyone answer to my questions please. 
 A: Geometrically we have just one operator, namely $A$. This is the map that moves the points of our space $V:={\mathbb R}^n$ around. In order to deal with this $A$ in a computational way we have to choose a basis for $V$. At the outset we would use the standard basis $(e_i)_{1\leq i\leq n}$ of ${\mathbb R}^n$, and with respect to this basis the operator $A$ assumes the matrix $[a_{ij}]$.
If we choose another basis $(u_j)_{1\leq j\leq n}$ then each $u_j$ is a linear combination of the $e_i$, hence $u_j=\sum_i b_{ij}e_i$ $\>(1\leq j\leq n)$ for a certain matrix $B$. Don't think of $B$ as an operator. It's just an $(n\times n)$-array of numbers encoding the numerical relations between the two sets of basis vectors.  
Since in the same way each $e_i$ is a linear combination of the $u_k$ we have as well $e_i=\sum_k c_{ki} u_k$ $\>(1\leq i\leq n)$ for a certain matrix $C$. It is then easy to see that in fact $BC=I$, which proves that $B$ is invertible.
A: The first part is a notational device, to be able to describe a change of basis operator $B$, by defining its action on the (standard?) basis $\{e_1,\ldots,e_n\}$:
$$ Be_j = u_j, \;\; j=1,\ldots,n $$
Then we can related the matrix representation of any other linear operator $A$ on $\mathbb{R}^n$ with respect to the standard basis and the representation of the same operator with respect to (nonstandard) basis $U$.
The second part of the question asks why we know that $B$ is invertible.  This is because its action on one (ordered) basis gives the other ordered basis.  That is, the inverse of $B$ is exactly the linear operator whose action on a basis is as follows:
$$  (B^{-1})u_j = e_j, \;\; j = 1,\ldots,n $$
Since this definition of a linear operator by its "action on a basis" is well-defined, it remains only to verify that in fact $(B^{-1})$ composed with $B$ is the identity operator.
Added: 
W. Rudin concisely treats the definition of linear operators, at least in finite dimensional spaces, through "action on a basis" in a couple of paragraphs at the top of page $207$ in PMA 3rd edition.
Assuming no familiarity of the "action on a basis" principle for defining (uniquely) a linear operator, we can write out just why this principle is justified in a particular case of change of basis.
That is, supposing $\{e_1,\ldots,e_n\}$ and $\{u_1,\ldots,u_n\}$ are bases, we can define a linear operator $B$ such that:
$$ Be_j = u_j, \;\; j=1,\ldots,n $$
by using the ability of basis elements to span a vector space, and the linear independence property to show uniqueness (well-definition).
So given any $v\in \mathbb{R}^n$, we can express $v$ as a linear combination of the basis vectors $e_j$ in exactly one way:
$$ v = \sum_{j=1}^n c_j e_j $$
where the scalar coefficients $c_j$ are uniquely determined by $v$.
Now define $B:\mathbb{R}^n \to \mathbb{R}^n$ so that:
$$ Bv = \sum_{j=1}^n c_j u_j $$
There are details to verify, that $B$ as thus defined is indeed a linear operator.  One way to visualize this is by identifying $B$ as acting by matrix multiplication on the "standard" coordinates of $\mathbb{R}^n$ with respect to the basis $\{e_1,\ldots,e_n\}$, namely (treating input as column vectors):
$$ B \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} = \begin{bmatrix} u_1 & u_2 & \dots & u_n \end{bmatrix}
\begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} $$
where the multiplying matrix contains the ordered basis elements $u_j$ as columns.
Since $e_j$ in these coordinates is simply the column vector with a $1$ in the $j$th entry and zeros elsewhere, multiplying $e_j$ by that matrix picks out the $j$th column, namely $u_j$.  Thus $Be_j = u_j$, as promised.
