Here is an extract from my notes on how I think about the change of basis matrices
I am not sure that my interpretation is correct; that is how I have defined $P$ and $Q$ as the matrix representation of the identity with different bases. Could someone let me know if this is correct and if not, why it is incorrect?
On
All this manner of stating things is horribly confusing in my opinion. Forgetting bases to begin with, every $n\times m$ matrix$~M$ corresponds directly and bijectively to a linear map $K^m\to K^n$ (where $K$ is the field the we are considering vector spaces over; read $\Bbb R^n$ for $K^n$ if you are considering real vector spaces), namely $v\mapsto M\cdot v$, the map of multiplying (on the left) by $M$. Composition of such maps corresponds to matrix multiplication (with the rightmost matrix corresponding to the first map applied).
Bases are used to bring a general vector space $V$ (in other words, one does not specify what kind of value its vectors $v\in V$ are) of dimension$~n$ into correspondence with the special vector space $K^n$ (whose elements are $n$-tuples of scalars in$~K$). If $\def\B{\mathcal B}\def\b{\mathbf b}\B=[\b_1,\ldots,\b_n]$ is a list of vectors in$~V$, one always has a linear map $g_\B:K^n\to V$ defined by forming linear combinations: $(x_1,\ldots,x_n)\mapsto x_1\b_1+x_2\b_2+\cdots+x_n\b_n$, and $\B$ is a basis precisely when this linear map is a bijection $K^n\to V$. A mental image that can be useful is that $\B$ serves as a (multiple) unit of measure that allows all elements of$~V$ to be expressed numerically, via the inverse bijection $f_\B:V\to K^n$, which is the coordinate map associated to the basis$~\B$.
A basic question when more than one basis of$~V$ is considered is that of coordinate change. If after $\B$ we consider a new bases $\def\CC{\mathcal C}\def\cc{\mathbf c}\CC=[\cc_1,\ldots,\cc_n]$, then what is the relation between $f_\B(v)$ and $f_\CC(v)$ for arbitrary vectors $v\in V$? To answer this, one first needs to know how the vectors of $\CC$ are situated relative to the basis$~\B$, and this information can be given in the form of an $n\times n$ matrix$~P$, called the change of basis matrix from $\B$ to$~\CC$, whose columns are $f_\B(\cc_1),\ldots,f_\B(\cc_n)$, that is, column $j$ contains the coordinates with respect to$~\B$ of the vector $\cc_j\in V$. Now the relation between $f_\B(v)$ and $f_\CC(v)$ can be given first for the cases where $v=\cc_j$ for some$~j$, and then extended by linearity to arbitrary vectors. Now since $\cc_j=0\cc_1+0\cc_2+\cdots+1\cc_j+\cdots+0\cc_n$ one has $f_\CC(\cc_j)=(\delta_{i,j})_{i=1}^n$, the $j$-th column of the $n\times n$ identity matrix (also called $j$-th standard basis vector of $K^n$), while we saw that $f_\B(\cc_j)$ is the $j$-th column of$~P$, so that $f_\B(\cc_j)=P\cdot f_\CC(\cc_j)$ for all$~j$. By linearity if the maps $f_\B$, $f_\CC$, and of multiplication by$~P$, it follows that $f_\B(v)=P\cdot f_\CC(v)$ for all$~v\in V$. In other words, multiplication by the change of basis matrix$~P$ transforms the coordinates $f_\CC(v)$ of $v$ with respect to the new basis$~\CC$ into its coordinates with respect to the old basis$~\B$: it equals the composition $f_\B\circ g_\CC$. To go from the old basis to the new basis, one must instead perform the inverse $f_\CC\circ g_\B$, which is given by multiplication by$~P^{-1}$.
Now given vector spaces $V,W$ and bases $\B_1$ of $V$ en $B_2$ of $W$, any linear map $\phi:V\to W$ can be described by a matrix $\def\Mat{\operatorname{Mat}}A={}_{\B_2}\Mat_{\B_1}(\phi)$, which is that of the composite map $\def\from{\leftarrow}f_{\B_2}\circ\phi\circ g_{\B_1}:K^n\from W\from V\from K^m$ (written right-to-left as that is how composite maps apply). If we want to compute from this the matrix ${}_{\B_2}\Mat_{\CC_1}(\phi)$ where in $V$ one uses a new basis$~\CC_1$, one can get this by right-composition with $f_{\B_1}\circ g_{\CC_1}$, whose matrix is the change of basis matrix $P$ from $\B_1$ to $\CC_1$, so on must replace $A$ by $A\cdot P$. Similarly changing to ${}_{\CC_2}\Mat_{\B_1}(\phi)$ where in $W$ one uses a new basis$~\CC_2$ is done by left-composition with $g_{\CC_2}\circ f_{\B_2}$, whose matrix is the inverse of the change of basis matrix $Q$ from $\B_2$ to $\CC_2$, so on must replace $A$ by $Q^{-1}A$. Doing both changes at one we must replace $A$ by $Q^{-1}AP$.
I think I've probably interchanged $m$ and $n$ with respect to your presentation, and I have no idea how my $P$ and $Q$ relate to yours. In any case I think it is a fundamentally wrong approach when discussing change of basis to invoke any non-identity linear maps $V\to V$ at all, like the one sending one basis to another; the only meaningful maps in this context are the ones like $f_{\B_2}\circ g_{\B_1}:K^n\to K^n$ that interpret coordinates for $\B_1$ and then express the resulting vector in coordinates for$~B_2$. Also the use of the $\sum$ symbol is more confusing than helpful (the only summations needed are implicit in the definition of matrix multiplication).
If you are looking at a matrix-transform of a vector, you can consider it a dot product of the vector and columns of the matrix, the result is a vector instead of a numeric value.