I have a question regarding a change of basis and the pseudoinverse.
Let $\mathit{B}_{1}$ and $\mathit{B}_{2}$ be basis for $\mathbb{R}^2$ with square $2 \times 2$ $\mathit{A}$ matrix describing $\mathit{B}_{2}$ in terms of $\mathit{B}_{1}$. Some transformation $\mathit{T}_{2}$ described in $\mathit{B}_{2}$ maps vector $\mathbf{x}_{2}$ described in $\mathit{B}_{2}$ to vector $\mathbf{y}_{2}$ described in $\mathit{B}_{2}$. If I wanted to apply transformation $\mathit{T}_{2}$ to a vector $\mathbf{x}_{1}$ described in $\mathit{B}_{1}$, I believe:
$$ \begin{equation} \mathit{T}_{1} = \mathit{A}\mathit{T}_{2}\mathit{A}^{-1} \end{equation} \tag{1} \label{eq1} $$
$$ \begin{equation} \mathbf{y}_{1} = \mathit{T}_{1} \mathbf{x}_{1}. \end{equation} \tag{2} \label{eq2} $$
Now let $\mathit{B}_{3}$ be a basis for $\mathbb{R}^3$ with rectangular $2 \times 3$ matrix $\mathit{C}$ describing $\mathit{B}_{3}$ in terms of $\mathit{B}_{1}$. Some transformation $\mathit{M}_{3}$ described in $\mathit{B}_{3}$ maps vector $\mathbf{x}_{3}$ described in $\mathit{B}_{3}$ to vector $\mathbf{w}_{3}$ described in $\mathit{B}_{3}$. If I wanted to apply transformation $\mathit{M}_{3}$ to $\mathbf{x}_{1}$, could I use the right pseudoinverse to apply a "pseudo-change of basis" (I'm not sure what the correct term is here):
$$ \begin{equation} \mathit{M}_{1} = \mathit{C}\mathit{M}_{3}\mathit{C}^{~+} \end{equation} \tag{3} \label{eq3} $$
$$ \begin{equation} \mathbf{w}_{1} = \mathit{M}_{1} \mathbf{x}_{1} \end{equation} \tag{4} \label{eq4} $$
where
$$ \begin{equation} \mathit{C}^{~+} = \mathit{C}^{\mathsf{~T}} \left(\mathit{C}\mathit{C}^{\mathsf{~T}} \right)^{-1} ? \end{equation} \tag{5} \label{eq} $$
Further, if $\mathit{M}_{3}$ simply scaled the components of $\mathbf{x}_{3}$, such as $\mathit{M}_{3} = diag\{0.5 ~~ 1 ~~ 2 \}$, what would $\mathit{M}_{1}$ do to $\mathbf{x}_{1}$?
Sorry if the above math is confusing. I'm an engineering student trying my best. Thanks!