Change of basis isomorphism So I am reading the following article on wikipedia:

My confusion stems from line 5 under the "Change of Basis" heading. They propose that $\phi_2^{-1}$$\phi_1$ is a change of coordinates automorphism.
However, $\phi_1$ is an isomorphism and takes the standard basis $e_i$ and changes it to the basis $\alpha_i$ and then $\phi_2^{-1}$ takes the standard basis $e_i$ and changes it to the basis $\alpha_i'$. So it's inverse suggests it would take coordinates in the $\alpha_i'$ basis and convert it back to the canonical basis. Therefore, when applied in that order I am confused that it does the necessary transformation given that we first convert to $\alpha_i$ and then the application of $\phi_2^{-1}$ seems redundant as it is being fed the wrong basis.
I hope that makes sense. If someone could explain this to me I would be very grateful.
Thanks,
 A: $\phi_1$ and $\phi_2$ are the maps from $\Bbb R^n \to V$ defined by $$\phi_1(x_1, \ldots, x_n) = \sum_{i=1}^n x_i\alpha_i\\\phi_2(y_1, \ldots, y_n) = \sum_{i=1}^n y_i\beta_i$$
These are called coordinate isomorphisms because each imposes a coordinate system on $V$. Every vector $v \in V$ is identified by $\phi_1$ with a unique set of coordinates $(x_1, \ldots, x_n)$ such that $v = \sum_i x_i\alpha_i$, and similarly, $\phi_2$ identifies the same vector $v$ with a different set of coordinates $(y_1, \ldots, y_n)$.
So I can locate $v$ using its $\phi_1$ coordinates, or I can locate it using its $\phi_2$ coordinates. But what if I want to compare these two coordinate systems? How are they related to each other? That is what the change of basis matrix is about:
We have $\phi_1(x_1, \ldots, x_n) = v$ and $\phi_2(y_1, \ldots, y_n) = v$, or since these are invertible, $$(y_1, \ldots, y_n) = \phi_2^{-1}(v) = \phi_2^{-1}(\phi_1(x_1, \ldots, x_n))$$
So $\phi_2^{-1}\circ\phi_1$ carries the $\phi_1$ coordinates of $v$ to the $\phi_2$ coordinates of $v$. Since both $\phi_1$ and $\phi_2$ are linear, this is a linear transformation from $\Bbb R^n$ to itself. a.k.a, an $n\times n$ matrix. The matrix tells you how to convert $\phi_1$ coordinates into $\phi_2$ coordinates.
For example, Let $V = \{f: \Bbb R \to \Bbb R : t\mapsto at + b \mid a,b \in \Bbb R\}$ be the space of all affine functions on $\Bbb R$. A couple of bases for $V$ are $$\{\alpha_1 : t \mapsto 2t+1, \alpha_2 : t \mapsto t+2\}\\\{\beta_1 : t \mapsto -t, \alpha_2 : t \mapsto t-1\}$$
Then $$\phi_1(x_1, x_2) = \bigg(t\mapsto (2x_1 + x_2)t + (x_1 + 2x_2)\bigg)\\\phi_2(y_1, y_2) = \bigg(t \mapsto (y_2 - y_1)t - y_2\bigg)$$
It is easy to calculate $\phi_2^{-1}(t \mapsto at + b) = (-a-b,-b)$, so
$$\phi_2^{-1}\circ\phi_1 = \begin{bmatrix}-3&&-3\\-2&&-1\end{bmatrix}$$
So if I know that some $v \in V$ is given in $\alpha$ coordinates by $v = -2\alpha_1 + 3\alpha_2$, to find its expression in $\beta$ coordinates, I can just apply the change of basis matrix to calculate it directly:
$$\begin{bmatrix}-3&&-3\\-2&&-1\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix} = \begin{bmatrix}-3\\1\end{bmatrix}$$ So $$v = -3\beta_1 + \beta_2$$
