I have a question on one of the steps in the following proof.
To prove: if two matrices A,B are similar then they are the matrices of the linear transformation for some matrix with respect to two different bases
Proof:
Suppose A,B are n x n matrices over a field K and that they are similar via $B=P^{-1}AP$.
We define a linear transformation $T_A:K^n\to K^n, T_A(v)=Av$.
We now define C to be the standard basis for $K^n$ :$\{e_1,...,e_n\}$ (where e_i denotes the n-column vector with a 1 in the ith entry and zeroes in all the other entries.)
and D to the standard basis translated by P: $D=\{Pe_1,...,Pe_n\} $
$A=\space _C[T_A]_C$ $\space$by how we've defined our linear transformation.
Note that $P=P_{D\to C} $ (How do we know this????) (I have my own reasoning but it seems a bit lengthy)
then $B=P^{-1}AP=P^{-1}_C[T_A]_CP=(P_{D\to C})^{-1}\space_C[T_A]_C P_{D\to C}$
$= _D[T_A]_D$
My question is : how do we know that $P=P_{D\to C} $
I take it that $P_{D\rightarrow C}$ means the change of basis matrix from $D$ to $C$. By definition, the change of basis matrix is the transformation matrix of the identity respective to those bases. Again by definition, the $i$-th column of $P_{D\rightarrow C}$ is the coordinate vector of $d_i$, the $i$-th element of $D$, with respect to the basis $C$. By the way we defined $D$, $d_i=Pe_i$, but $Pe_i$ is the $i$-th column of $P$. The coordinate mapping respective to the standard basis is the identity and hence the $i$-th column of $P_{D\rightarrow C}$ is the $i$-th column of $P$. Thus, $P_{D\rightarrow C}=P$.
More symbolically, $$P_{D\rightarrow C}\ e_i=[d_i]_C=[Pe_i]_C=Pe_i\ \Rightarrow\ P_{D\rightarrow C}=P.$$