$[v{]}_{\mathcal{C}}=P_{\mathcal{C}\leftarrow \mathcal{B}}[v{]}_{\mathcal{B}}$ and $P_{\mathcal{B}\leftarrow \mathcal{C}}=P_{\mathcal{C}\leftarrow \mathcal{B}}^{-1}$

For ordered bases $\mathcal{B},\mathcal{C}$ of a finite-dimensional vector space $V$ and every $v\in V$,

$[v{]}_{\mathcal{C}}=P_{\mathcal{C}\leftarrow \mathcal{B}}[v{]}_{\mathcal{B}}\mathrm{.}$

Moreover,

$P_{\mathcal{B}\leftarrow \mathcal{C}}=P_{\mathcal{C}\leftarrow \mathcal{B}}^{-1}\mathrm{.}$

Let $n=\dim V$, and write ${\kappa }_{\mathcal{B}}:V\to F^n$ for $v\mapsto [v{]}_{\mathcal{B}}$, with ${\kappa }_{\mathcal{C}}$ defined similarly. Here $F^n$ is identified with the coordinate-column space $M_{n\times 1}(F)$. The coordinate-change identity in step 1.1 is the commutativity of this square; the bottom arrow denotes multiplication by the displayed matrix.

The reverse coordinate conversion is multiplication by $P_{\mathcal{B}\leftarrow \mathcal{C}}$. Step 3.1 gives both compositions:

$P_{\mathcal{B}\leftarrow \mathcal{C}}{P}_{\mathcal{C}\leftarrow \mathcal{B}}=I_n,P_{\mathcal{C}\leftarrow \mathcal{B}}{P}_{\mathcal{B}\leftarrow \mathcal{C}}=I_n\mathrm{.}$

Thus the two bottom coordinate spaces are inverse identifications of the same vector space, including when $n=0$. The proof above is independent of the figure.