I wish to prove formally that the change of basis matrix $M$ is given by,
$$M=[I]_{\mathscr{B_{1}}}^{\mathscr{B_{2}}}$$
I wish to tidy up my proof, and would like someone to verify my steps.
Proof.
Let $\mathscr{B}_{1}=\{u_{1},u_{2},\ldots,u_{j},\ldots,u_{n}\}$ and $\mathscr{B}_{2}=\{w_{1},w_{2},\ldots,w_{j},\ldots,w_{n}\}$ be the two bases of the finite dimensional vector space $V$.
Suppose, the co-ordinates of $w_{j}$ with respect to the basis $\mathscr{B}_{1}$ are $\{a_{1j},a_{2j},\ldots,a_{nj}\}$. That is,
$$w_{j}=\sum_{i=1}^{n}a_{ij}u_{i}$$
We also know that, for any transformation, the below relationship holds
$$[T(x)]_{\mathscr{B}_{2}}=[T]_{\mathscr{B}_{1}}^{\mathscr{B}_{2}}\cdot[x]_{\mathscr{B}_{1}}$$
So,
$$[I(w_{j})]_{\mathscr{B}_{1}}=[I]_{\mathscr{B}_{2}}^{\mathscr{B}_{1}}\cdot[w_{j}]_{\mathscr{B}_{2}}$$
But, $I(w_{j})=w_{j}$.
And $[w_{j}]_{\mathscr{B}_{1}}=\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\{a_{nj}}\end{bmatrix}$. And $[w_{j}]_{\mathscr{B}_{2}}$ is like a standard basis vector with $j$th co-ordinate $1$.
The right hand side is the $j$th column of $[I]_{\mathscr{B}_{2}}^{\mathscr{B}_{1}}$.
My challenge here is, that, I wish to derive an expression for $M=[I]_{\mathscr{B}_{1}}^{\mathscr{B}_{2}}$ and not $[I]_{\mathscr{B}_{2}}^{\mathscr{B}_{1}}$ . I know that, $[I]_{\mathscr{B}_{1}}^{\mathscr{B}_{2}}$ must represent the co-ordinates of the basis vectors of ${\mathscr{B}_{1}}$ relative to ${\mathscr{B}_{2}}$.