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change of basis
Let $V$ be a vector space. Given a basis $A$ for $V$, each vector $v\in V$ can be uniquely expressed in terms of the base elements $v_{i}\in A$ as follows:
$v=\sum_{{v_{i}\in A}}r_{i}v_{i}$ 
where the sum is taken over a finite number of elements in $A$. Suppose now that $B$ is another basis for $V$. By a change of basis from $A$ to $B$ we mean reexpressing $v$ in terms of base elements $w_{i}\in B$.
Formally, we can think of a change of basis as the identity function (viewed as a linear operator) on a vector space $V$, such that elements in the domain are expressed in terms of $A$ and elements in the range are expressed in terms of $B$.
Note that, by the very design of a basis, a change of basis in a vector space is always possible.
Now, if $V$ has dimension $n<\infty$. We can total order bases $A$ and $B$. Then a change of basis (from $A$ to $B$) has the matrix representation
$[I]^{A}_{B},$ 
where $I:V\to V$ is the identity operator. $[I]^{A}_{B}$ is called a change of basis matrix. By applying $[I]^{A}_{B}$ to a vector $v$ expressed in terms of $A$, we get $v$ expressed in terms of $B$:
$[v]_{B}=[I]^{A}_{B}[v]_{A},$ 
where $[v]_{A}$ and $[v]_{B}$ are $v$ expressed in the two bases $A$ and $B$ respectively.
Since $I$ is obviously invertible, $[I]^{A}_{B}$ is invertible also, whose inverse is $[I]^{B}_{A}$. Furthermore, $[I]_{A}=I_{n}$ for any basis $A$. Here, $I_{n}$ is the identity matrix.
Examples.
1. Let $V=\mathbb{R}^{3}$ and the following two sets
$A=\Bigg\{\begin{pmatrix}1\\ 1\\ 2\end{pmatrix},\begin{pmatrix}0\\ 1\\ 1\end{pmatrix},\begin{pmatrix}2\\ 3\\ 0\end{pmatrix}\Bigg\}\mbox{ and }B=\Bigg\{\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}\Bigg\}$ be the two ordered bases for $V$, ordered in the way the elements are arranged in the set. For each $v_{i}\in A$, $I(v_{i})=v_{i}=[v_{i}]_{{E_{3}}}$, we see that
$[I]^{A}_{B}=\begin{pmatrix}1&0&2\\ 1&1&3\\ 2&1&0\end{pmatrix}.$ Notice that the columns of $[I]^{A}_{B}$ are exactly the elements of $A$. Indeed, each element of $A$ is already written in terms of the standard basis elements (in $B$). For example, let $v$ be the first basis element in $A$. Let us see what $[v]_{A}$ is, when expressed using base elements in $B$, the standard ordered basis:
$[v]_{B}=[I]^{A}_{B}[v]_{A}=\begin{pmatrix}1&0&2\\ 1&1&3\\ 2&1&0\end{pmatrix}[v]_{A}=\begin{pmatrix}1&0&2\\ 1&1&3\\ 2&1&0\end{pmatrix}\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}=\begin{pmatrix}1\\ 1\\ 2\end{pmatrix},$ exactly as we have expected.
2. Conversely, let $w$ be the first basis element in $B$. What is $w$ when expressed in terms of basis elements of $A$? In other words, we need to find
$[w]_{A}=[I]^{B}_{A}[w]_{B}.$ Now, $[w]_{B}$ is just $\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}$, so $[w]_{A}$ is nothing more than the first column of $[I]^{B}_{A}$, which is just the inverse of the matrix $[I]^{A}_{B}$, so
$[I]^{B}_{A}={([I]^{A}_{B})}^{{1}}=\begin{pmatrix}1&0&2\\ 1&1&3\\ 2&1&0\end{pmatrix}^{{1}}=\begin{pmatrix}1/3&2/9&2/9\\ 2/3&4/9&5/9\\ 1/3&1/9&1/9\end{pmatrix}.$ Therefore, $[w]_{A}=\begin{pmatrix}1/3\\ 2/3\\ 1/3\end{pmatrix}$. A quick verification shows that this is indeed the case:
$\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}=(1/3)\begin{pmatrix}1\\ 1\\ 2\end{pmatrix}+(2/3)\begin{pmatrix}0\\ 1\\ 1\end{pmatrix}+(1/3)\begin{pmatrix}2\\ 3\\ 0\end{pmatrix}.$ 3. Now let $C$ be the set $\Bigg\{\begin{pmatrix}1\\ 0\\ 2\end{pmatrix},\begin{pmatrix}0\\ 1\\ 1\end{pmatrix},\begin{pmatrix}2\\ 1\\ 0\end{pmatrix}\Bigg\}$. It is easy to check that $C$ forms a basis for $\mathbb{R}^{3}$ (determinant is nonzero). Order $C$ in the obvious manner. What is the change of basis matrix $[I]^{C}_{A}$? One way is to express each element of $C$ in terms of the elements of $A$. Another way is to use the formula $[I]^{C}_{A}=[I]^{B}_{A}[I]^{C}_{B}$. Applying the first example, we see that $[I]^{C}_{B}$ is just the matrix whose columns are elements of $C$. As a result:
$[I]^{C}_{A}=[I]^{B}_{A}[I]^{C}_{B}=\begin{pmatrix}1/3&2/9&2/9\\ 2/3&4/9&5/9\\ 1/3&1/9&1/9\end{pmatrix}\begin{pmatrix}1&0&2\\ 0&1&1\\ 2&1&0\end{pmatrix}=\begin{pmatrix}7/9&0&4/9\\ 4/9&1&8/9\\ 1/9&0&7/9\end{pmatrix}.$
Remarks. Let us summarize what we have learned from the examples above, as well as list some additional facts. Let $V$ be a finite dimensional vector space of dimension $n$.

If $E$ is the standard basis (ordered), then for any ordered basis $A$, $[I]^{A}_{E}$ is the matrix whose columns are exactly the basis elements in $A$ (assuming these elements have already been expressed in terms of $E$) such that the $i$column corresponds to the $i$th element in the ordered set $A$.

This also means that every invertible matrix $A$ corresponds to (in a onetoone fashion) a change of basis from the basis $S_{A}$ whose elements are columns of $A$ to $E$, the standard basis: $A=[I]^{{S_{A}}}_{E}$.

Continue to assume that $E$ is the standard basis. Let $A,B$ be any ordered bases for $V$. Using the above property, we can easily compute $[I]^{A}_{B}$, which is $[I]^{E}_{B}[I]^{A}_{E}=([I]^{B}_{E})^{{1}}[I]^{A}_{E}.$

Let $A^{{\prime}}$ be a reordering of the ordered basis $A$, where each $v^{{\prime}}_{i}\in A^{{\prime}}$ is just $v_{{\pi(i)}}$ for some permutation in $S_{n}$. Then $[I]^{{A^{{\prime}}}}_{A}$ is the permutation matrix corresponding to the permutation $\pi$.

Suppose $T$ is a linear transformation from $V$ to $W$ (both finite dimensional). Under a bases $A\subset V$ and $B\subset W$, $T$ has matrix representation $[T]^{A}_{B}$. Under changes of basis from $A$ to $A^{{\prime}}$, and $B$ to $B^{{\prime}}$, we have
$[T]^{{A^{{\prime}}}}_{{B^{{\prime}}}}=[ITI]^{{A^{{\prime}}}}_{{B^{{\prime}}}}=% [I]^{B}_{{B^{{\prime}}}}[T]^{A}_{B}[I]^{{A^{{\prime}}}}_{A}.$ 
If $T$ is a linear operator on $V$, then setting $V=W$, $A=B$ and $A^{{\prime}}=B^{{\prime}}$ from above, we have that
$[T]_{{A^{{\prime}}}}=P^{{1}}[T]_{A}P,$ where $P$ is the change of basis matrix $[I]^{{A^{{\prime}}}}_{A}$. This shows that $[T]_{A}$ and $[T]_{{A^{{\prime}}}}$ are similar matrices. In other words, under a change of basis, the linear transformation $T$ is basically the same.
References
 1 Friedberg, Insell, Spence. Linear Algebra. PrenticeHall Inc., 1997.
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