Need help with a better understanding of change of basis matrix and corresponding theorems I'll try to summarize here what I understand so far about the concepts of change of basis matrix etc.
Let $\beta$ and $\gamma$ be two different ordered bases for the vectorspace $V$, and let $P =[Id_V]_{\beta}^{\gamma}$ be the matrixrepresentation of the identical transformation, with respect to $\beta$ and $\gamma$. It can be easily shown that $P$ is invertible; the inverse is $[Id_V]_{\gamma}^{\beta}$. When an arbitrary vector $x \in V$ has coordinate vector $[x]_{\beta}$ with respect to $\beta$, then its coordinate vector with respect to $\gamma$ is given as \begin{align*} [x]_{\gamma} = P [x]_{\beta}. \end{align*} The matrix $P = [Id_V]_{\beta}^{\gamma}$ is called the change of basis matrix from $\beta$ to $\gamma$.
Example:
If we work with $\mathbb{R}^2$, then two possible bases are $\gamma = \left\{(1,1), (1,-1)\right\}$ and $\beta = \left\{(2,4),(3,1)\right\}$. Now, since  $(2,4) = 3(1,1) - 1(1,-1)$ and $(3,1)=2(1,1)+1(1,-1)$ we have that \begin{align*} P = [Id_V]_{\beta}^{\gamma} = \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix}, \end{align*} whereby $P$ the $\beta$-coordinates in $\gamma$-coordinates changes. Let $x = (7,9) \in \mathbb{R}^2$ be an arbitrary vector that has a coordinate vector $[x]_{\beta} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ with respect to the basis $\beta$. Then its coordinate vector with respect to $\gamma$ is given as \begin{align*} \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 8 \\ -1 \end{pmatrix}. \end{align*} We notice that in the columns of this matrix $P$ are the coordinates of vectors $\beta$ with respect to the basis $\gamma$.
So far so good, now I'm having trouble relating this information to the next two theorems.
Theorem 1.: Let $T$ be a linear operator over a finite dimensional vectorspace $V$, and let $\beta$ and $\gamma$ be two ordered bases for $V$. Let $[T]_{\beta}$ and $[T]_{\gamma}$ be two matrixrepresentations of this transformation. If $P$ is the change of basis matrix from $\beta$ to $\gamma$, then we have that \begin{align*} [T]_{\beta} = P^{-1} [T]_{\gamma} P. \end{align*} 
Proof: Let $Id: V \rightarrow V$ be the identical transformation. Then we have that $T = Id \circ T = T \circ Id$. Now, since $P = [Id]_{\beta}^{\gamma}$, we have that \begin{align*} P[T]_{\beta} = [Id]_{\beta}^{\gamma} [T]_{\beta}^{\beta} = [Id \circ T]_{\beta}^{\gamma} = [T \circ Id]_{\beta}^{\gamma} = [T]_{\gamma}^{\gamma} [Id]_{\beta}^{\gamma} = [T]_{\gamma} P, \end{align*} from where it follows that $[T]_{\beta} = P^{-1} [T]_{\gamma} P$. Q.E.D.
I'm not sure what's the connection between this theorem and the information above about coordinate vectors. 
I could also make this theorem more general, expanding it to a linear map (not necessarily an operator). 
Theorem 2.: Let $T : V \rightarrow W$ be a linear map, and let $\beta, \beta'$ be two different ordered bases for $V$, and $\gamma, \gamma'$ two different ordered bases for $W$. Let $A = [T]_{\beta}^{\gamma}$ and $B = [T]_{\beta'}^{\gamma'}$ be two matrixrepresentations of $T$. If $P$ is the change of basis matrix from $\beta$ to $\beta'$, and $Q$ is the change of basis matrix from $\gamma$ to $\gamma'$, then the relationship between $A$ and $B$ is given as \begin{align*} B = QAP^{-1}. \end{align*}
Would this be correct? Also, what do $P$ and $Q$ represent now? Are they still representing the identical transformation? If this is correct, could someone give me an example?
Thanks in advance for clarifying up any misconceptions I might have.
 A: Let $T$ be reflection over $y=x$. It is a linear transformation. We have
$$T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\x\end{pmatrix}$$
Let $\beta=\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\}, \gamma=\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}1\\1\end{pmatrix}\}$ be two basis. We can find $P =[Id_V]_{\beta}^{\gamma}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$
The matrices of the linear transformation $T$ in terms of two basis are
$$[T]_{\beta}=\begin{pmatrix}0&1\\1&0\end{pmatrix}, [T]_{\gamma}=\begin{pmatrix}-1&0\\1&1\end{pmatrix}$$
Theorem 1 says
$$\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}^{-1}\begin{pmatrix}-1&0\\1&1\end{pmatrix}\begin{pmatrix}1&-1\\0&1\end{pmatrix}$$
which can be easily verified.
An example for the second theorem can be constructed from a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^1$. For example, projection onto $x$-axis. $P$ and $Q$ are still the identical transformation. This time we have 
$$T\begin{pmatrix}x\\y\end{pmatrix}=x$$
Let $\beta=\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\}, \beta'=\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}1\\1\end{pmatrix}\}$ be two basis for $\mathbb{R}^2$. We still have $P =[Id_V]_{\beta}^{\beta'}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$.
Let $\gamma=1, \gamma'=-1$. Then $Q=[Id_V]_{\gamma}^{\gamma'}=-1$. 
$$[T]_{\beta}^{\gamma}=\begin{pmatrix}1&0\end{pmatrix}, [T]_{\beta'}^{\gamma'}=\begin{pmatrix}-1&-1\end{pmatrix}$$
To get $[T]_{\beta'}^{\gamma'}$, compute transformation of $T$ on the basis $\beta'$, and write the results in terms of $\gamma'$: 
$$T\begin{pmatrix}1\\0\end{pmatrix}=1=-1\cdot -1\\
T\begin{pmatrix}1\\1\end{pmatrix}=1=-1\cdot -1$$
It can be found that
$$\begin{pmatrix}-1&-1\end{pmatrix}=Q\begin{pmatrix}1&0\end{pmatrix}P^{-1}$$
