I'm am learning linear algebra and I am very confused as to what my textbook is saying about the change of basis operator.
The discussion is limited to a single finite-dimensional space $V$ and to linear operators on $V$. Begin by examining how the coordinates of $\mathbf{v}\in V$ change as the basis for $V$ change as the basis for $V$ changes. Consider two different bases.
$B=\lbrace \mathbf{x}_1,\mathbf{x}_2,...,\mathbf{x}_n \rbrace$ and $B'=\lbrace \mathbf{y}_1,\mathbf{y}_2,...,\mathbf{y}_n \rbrace$.
It's convenient to regard $B$ as an old basis for $V$ and $B'$ as a new basis for $V$. Throughout this section $\mathbf{T}$ will denote the linear operation such that
$\mathbf{T} (\mathbf{y}_i)=\mathbf{x}_i$ for $i=1,2,...,n$.
$\mathbf{T}$ is called the change of basis operator because it maps the new basis vectors in $B'$ to the old basis vectors in $B$. Notice that $[\mathbf{T}]_B=[\mathbf{T}]_{B'}=[\mathbf{I}]_{BB'}$. To see this, observe that $\mathbf{x}_i=\displaystyle\sum_{j=1}^{n} \alpha_j \mathbf{y}_j \implies \mathbf{T}(\mathbf{x}_i)=\displaystyle\sum_{j=1}^{n} \alpha_j \mathbf{T}(\mathbf{y}_j)=\displaystyle\sum_{j=1}^{n} \alpha_j \mathbf{x}_j$,
which means $[\mathbf{x}_i]_{B'}=[\mathbf{T}(\mathbf{x_i})]_B$.
I am very unclear on what this last line means and what it has to do with the line above or anything that comes after. The rest goes
So
$[\mathbf{T}]_B = \begin{pmatrix} [\mathbf{T}(\mathbf{x}_1)]_{B} & [\mathbf{T}(\mathbf{x}_2)]_{B} & \cdots & [\mathbf{T}(\mathbf{x}_1)]_{B} \end{pmatrix}=$ $\begin{pmatrix} [\mathbf{x}_1]_{B'} & [\mathbf{x}_2]_{B'} & \cdots & [\mathbf{x}_1]_{B'} \end{pmatrix}= [\mathbf{T}]_{B'}$
What is going on here?