Similar matrices proof question I am reading about similar matrices. 

Here is the theorem: suppose V is a finite dimensional vector space with $B_1$ and $B_2$ as ordered basis and let T be a linear operator on V. Then the matrices $[T]_{B_1}$ and $[T]_{B_2}$ are similar.

Proof: use $\beta = \{u_1, \dotsc, u_m\}$ and $\gamma = \{v_1, \dotsc, v_n\}$ as the bases. For any linear transformation $T$, we can write $T(u_j)$ uniquely in terms of $\gamma$ as $T(u_j) = \sum_{i = 1}^n a_{ij}v_i$. We can arrange $a_{ij}$ into a unique matrix and denote it $[T]_{\beta}^{\gamma}$ or just $[T]_\beta$ if both bases are $\beta$. Similarly, if $v = a_i u_i$ we can denote $[v]_\beta$ to be the column matrix of vector $v$ with respect to basis $\beta$ consisting of $a_i$.
Going through all the details, it can be proven that $[ST]_\beta^\gamma = [S]_\alpha^\gamma [T]_\beta^\alpha$ and $[Tv]_\gamma = [T]_\beta^\gamma [v]_\beta$. 

Question 1:  $[Tv]_\gamma = [T]_\beta^\gamma [v]_\beta$, why don't we write as for $[Tv]^{\gamma}_{\beta}$? By the principle of composite mapping? That's my only question here.

In particular we can consider the linear operators $T$ and $I$(identity operator) on a $n$ dimensional vector space $V$. Let $Q = [I]_\beta^\gamma$. Then, for any $v \in V$, we have
$$
[Iv]_\gamma = [I]_\beta^\gamma [v]_\beta \implies [v]_\gamma = Q[v]_\beta. \tag{1}
$$
Also,
$$
[Tv]_\gamma = [I]_\beta^\gamma [Tv]_\beta \implies [T]_\gamma [v]_\gamma = Q[T]_\beta [v]_\beta.
$$
Using $(1)$, this becomes
$$
[T]_\gamma Q [v]_\beta = Q[T]_\beta [v]_\beta \implies [T]_\gamma Q = Q[T]_\beta \implies [T]_\beta = Q^{-1} [T]_\gamma Q.
$$
Note that $Q$ is invertible since $I$ is invertible. The last equation show the transformation matrices with respect to two bases are similar.
 A: There are two pieces of notation here, and it's important not to confuse them:


*

*If $V$ is a vector space with basis $\beta = (v_1, \ldots, v_n)$, and $v \in V$, we denote by $[v]_\beta$ the coordinate column vector. That is, we set
$$[v]_\beta = \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix},$$
where $a_1, \ldots, a_n$ are the unique scalars such that
$$v = a_1 v_1 + \ldots + a_n v_n.$$

*Suppose, in addition to $V$ and $\beta$ as above, we have a vector space $W$ (over the same scalar field), with basis $\gamma = (w_1, \ldots, w_m)$, and $T : V \to W$ is linear. Then we denote by $[T]_\beta^\gamma$ the matrix for $T$ from basis $\beta$ to basis $\gamma$. Specifically, $[T]_\beta^\gamma$ is the unique $m \times n$ matrix over the common scalar field such that
$$[Tv]_\gamma = [T]_\beta^\gamma [v]_\beta$$
for all $v \in V$. This matrix can be explicitly calculated like so:
$$[T]_\beta^\gamma = \left[\begin{array}{c|c}[Tv_1]_\gamma & [Tv_2]_\gamma & \cdots & [Tv_n]_\gamma\end{array}\right].$$
Confusingly, when $V = W$ (i.e. when $T$ is an operator on $V$) and $\beta = \gamma$, rather than writing $[T]_\beta^\beta$, we shorten it to $[T]_\beta$, understanding implicitly that we are mapping to the same basis $\beta$. This is not the same as writing $[v]_\beta$ when $v \in V$; $[T]_\beta$ is an $n\times n$ matrix, not an $n \times 1$ column vector like $[v]_\beta$. It's up to the reader to know from context whether, say, $X$ is a vector in $V$ or an operator on $V$, and hence whether $[X]_\beta$ is a column vector or a matrix.
We also denote the application of a linear map, and the composition of linear maps similarly. If $T : V \to W$ and $S : W \to X$ are linear, then we denote the composite map $S \circ T$ by simply $ST$. Similarly, if $v \in V$, we often denote the application of $T$ to vector $v$ by $Tv$ (instead of the usual $T(v)$). Again, it's up to the reader to tell, from context, whether $TX$ is a composition of maps $T$ and $X$, or the application of $T$ to the vector $X$.
So, to answer your question directly, $[Tv]_\beta^\gamma$ doesn't make sense, as $v$ is a vector in $V$, not an operator on $V$. If it were the latter, then $Tv$ would be composition of $T$ and $v$, making an operator $Tv$, and the notation $[Tv]_\beta^\gamma$ would sensibly define an $n \times n$ matrix. However, because $v$ is a vector in $V$, $Tv$ is another vector in $V$, and the notation is undefined in this case.
