Matrix representation equivalence introductory (very basic and probably trivial) I am studying representation theory. I updated the question m since the photo seems to have poor quality. Please take a look at the link (the last two slides).
I understand why the matrix representations differ by $P^{-1}R_gP=R’_g$ at the top of the page (we used change base formula).
However, I don’t really get the Isomorphism part, why is it not
$T(g_1v)=g_2T(v)$? I have seen other sources writing it that way I believe (not sure).
My main struggle is why will the associated matrix representations be equal (mentioned at the bottom). Why is $R’_g=R_g$  Did we not agree to use the change of base formula given at the top? My professor said that this is a really simple observation that follows from the isomorphism defined.
Sorry I am really confused. The book notation might be confusing so please take a look at the website at the last two slides if you have time)
.
Thank you 

I found a related question by my classmate here:
Equivalence in representation theory and isomorphism between equivalent represenation basis
More information can be found on here: the last two slides is what I don’t get. Please explain to me the last two slides (especially the proof given at the very last slide. I don’t get how our isomorphism equals one with this basis. Please be detailed.
http://www.maths.usyd.edu.au/u/UG/SM/MATH3966/boblec21.pdf
 A: The answer to your question has to do with what bases for $V$ and $V'$ ones uses for the computation. Really, when one talks about a matrix representation, one means a representation $V$ of $G$, together with a basis $\mathbf{B}$ of $V$. This extra data of a basis allows you to turn each element $g \in G$ (thought of as a linear operator on $V$) into a matrix. Specifically, if the basis is $b_1,\dots,b_n$ then $g b_i  = \sum c_{ij}b_j$ for some unique scalars $c_{ij}$. The matrix of $g$ is then the matrix of the $c_{ij}$. 
When they write $\mathbf{B}$ and $\mathbf{B'}$ are "corresponding bases" what is meant is that having fixed $\mathbf{B}$, $\mathbf{B'}$ is defined to be $T \mathbf{B}$.
Knowing this, let us compute the matrix of $g \in G$ on $V'$, using the basis $\mathbf{B'}$. We know in advance what the matrix of $g$ is on $V$ using the basis $\mathbf{B}$.
We have
$$
g b_i = \sum_j c_{ij} b_j \implies
T(g b_i) = \sum_j c_{ij} T(b_j) 
$$
Now use the fact that $T$ is an isomorphism of representations of $G$ to commute the $g$ past the $T$ and get
$$
g T(b_i) = \sum_j c_{ij} T(b_j)
$$
Finally, remember that the basis $\mathbf{B'}$ is defined to be the transformed basis $T \mathbf{B}$. Hence the matrix of $g$ will be the same on the pair $(V,\mathbf{B})$ as on $(V',\mathbf{B'})$.
