Theorem 5.2.6. Let $B = \{b_1, b_2, \dots, b_n\}$ and $D$ be two ordered bases of $V$.
Then
$P_{D \leftarrow B} = [C_D(b_1) \quad \dots \quad C_D(b_n)],$
and satisfies $C_D(\mathbf{v}) = P_{D \leftarrow B} C_B(\mathbf{v})$ for all $\mathbf{v} \in V$.
The matrix $P_{D \leftarrow B}$ is invertible, and $(P_{D \leftarrow B})^{-1} = P_{B \leftarrow D}$. Moreover, if $E$ is a third
ordered basis, then
$P_{E \leftarrow D} P_{D \leftarrow B} = P_{E \leftarrow B}$.
