I have asked a similar question relating to this theorem.
Suppose $P$ is an $n\times n$ invertible matrix over $F.$ Let $V$ be an $n$-dimensional vector space over $F,$ and let $\scr B$ be an ordered basis of $V.$ Then there is a unique ordered basis $\scr \overline{B}$ of $V$ such that
\begin{align} \tag{1}[\alpha]_{\scr B} = P[\alpha]_{\scr \overline{B}} \end{align} for every vector $\alpha \in V$
$\Big($ $[\alpha]_{\scr B}$ is the coordinate matrix of $\alpha$ relative to the ordered basis $\scr B.$ $\Big)$
Attempt : Let $\scr B$ consist of the vectors $\alpha_1,\dots,\alpha_n.$ Let $\scr \overline{B}= \{\overline{\alpha_1},\dots ,\overline{\alpha_n}\}$ be an ordered set in which each element is defined by \begin{align} \overline{\alpha_j}=\sum_{i=1}^n P_{ij}\alpha_i, 1\le j\le n \end{align} Thus we need only show that the vectors $\overline{\alpha_j},$ defined by these equations form a basis. Let $Q=P^{-1}.$ Then \begin{align} \sum_j Q_{jk}\overline{\alpha_{j}}&=\sum_j Q_{jk}\sum_i P_{ij}\alpha_i\\ &=\sum_j \sum_iP_{ij}Q_{jk}\alpha_i\\ &=\sum_i\Big( \sum_jP_{ij}Q_{jk}\Big)\alpha_i\\ &=\alpha_k \end{align}
Thus the subspace spanned by the set $\scr \overline{B}$ contains $\scr B$ and hence spans $V.$ Thus $(1)$ is valid due to a "previous" theorem.
"Previous Theorem": Let $W$ be an $n$-dimensional vector space over the field $F,$ and let $\scr D$ and $\scr \overline{D}$ be two ordered bases of $V.$ Then there is a unique, necessarily invertible, $n\times n$ matrix $A$ with entries in $F$ such that $[\alpha]_{\scr D} = A[\alpha]_{\scr \overline{D}}$
Is my proof valid?