Is my reasoning on Transformation Matrices right?

Question

Let $A=\{v_1,\cdots,v_n\}$ and $B=\{w_1,\cdots,w_n\}$ be two basis of $V$. Then we define $[v]_A = \{\alpha_1,\cdots,\alpha_n\}^T$ as the coordinate tuple of $v$ respect to $A$. We want to obtain the map of the coordinate tuple respect to $A$ to coordinate tuple respect to $B$.

In terms of Linear Algebra this is what I know so far:

$$ \begin{cases} A\cdot [v]_A = \overrightarrow{v} \\ B.[v]_B = \overrightarrow{v} \end{cases} $$

Here we are trying to express $\overrightarrow{v}$ as a linear combination in basis $A$ and $B$. Since we know $[v]_A$ we obtain the coordinate vector of $\overrightarrow{v}$ in basis $B$ this is $[v]_B$:

$$[v]_B = B^{-1}\cdot A\cdot [v]_A$$

But I'm trying to expand my knowledge to Linear Maps. What I've got is the following:

$$T : V \to K^n$$ where $T(\overrightarrow{v}) = [v]_B$ and $v \in V, [v]_B \in K^n$

$$T(\overrightarrow{v}) = T(\alpha_1 \cdot v_1 + \cdots + \alpha_n \cdot v_n) = T(\alpha_1 \cdot v_1) + \cdots + T(\alpha_n \cdot v_n) = \alpha_1 \cdot T(v_1) + \cdots + \alpha_n \cdot T(v_n)$$

Then we define the transformation operation as:

$$[v]_B = M.[v]_A$$

where $M = [T(v_1),\cdots,T(v_n)]$ is the Transformation Matrix.

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Questions:

1. Can we say in this case that $M = B^{-1} \cdot A$?

In other means: $$[v]_B = \underbrace{B^{-1}\cdot A}_{\text{M?}} \cdot [v]_A$$

2. Is my linear map well-defined according to the operations described above?

Thanks in advance.

Answer 1

I think I got it and i can answer myself:

1. Can we say in this case that $M = B^{-1} \cdot A$?

In other means: $$[v]_B = \underbrace{B^{-1}\cdot A}_{\text{M?}} \cdot [v]_A$$

$T(\overrightarrow v)$ is the Linear Transformation that maps $\overrightarrow v$ to coordinate tuple on basis $B$

As described before this is: $$T(\overrightarrow v) = [v]_B$$

Since $\overrightarrow v$ is expressed as linear combination on basis $A$, the linear transformation acts on each vector of basis $A$, yielding the associated Transformation Matrix $M$ of the linear transformation.

The column vectors of $M$ are the vectors of basis $A$ transformed to coordinate tuple on basis $B$. Since $T(\overrightarrow{v_i}) = B^{-1}\cdot \overrightarrow{v_i}$

Then $[v]_B = M\cdot [v]_A$ as we already know that $[v]_B = B^{-1}\cdot A \cdot [v]_A$ the proof is concluded.