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 re-expressing $v$ 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 . We can total order bases $A$ and $B$. Then a change of basis (from $A$ to $B$) has the matrix representation

$${[I]}_B^A,$$

where $I:V\to V$ is the identity operator. ${[I]}_B^A$ is called a change of basis matrix. By applying ${[I]}_B^A$ to a vector $v$ expressed in terms of $A$, we get $v$ expressed in terms of $B$:

$${[v]}_B={[I]}_B^A{[v]}_A,$$

where ${[v]}_A$ and ${[v]}_B$ are $v$ expressed in the two bases $A$ and $B$ respectively.

Since $I$ is obviously invertible, ${[I]}_B^A$ is invertible also, whose inverse is ${[I]}_A^B$. Furthermore, ${[I]}_A=I_n$ for any basis $A$. Here, $I_n$ is the identity matrix.

Examples.

1. Let $V={ℝ}^3$ and the following two sets

   $$A=\{\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right),\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right),\left(\begin{array}{c}\hfill 2\hfill \\ \hfill 3\hfill \\ \hfill 0\hfill \end{array}\right)\}\text{ and }B=\{\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right),\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right),\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)\}$$

   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

   Notice that the columns of ${[I]}_B^A$ 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:

   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

   $${[w]}_A={[I]}_A^B{[w]}_B.$$

   Now, ${[w]}_B$ is just $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)$, so ${[w]}_A$ is nothing more than the first column of ${[I]}_A^B$, which is just the inverse of the matrix ${[I]}_B^A$, so

   Therefore, ${[w]}_A=\left(\begin{array}{c}\hfill 1/3\hfill \\ \hfill -2/3\hfill \\ \hfill 1/3\hfill \end{array}\right)$. A quick verification shows that this is indeed the case:

   $$\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)=(1/3)\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right)+(-2/3)\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)+(1/3)\left(\begin{array}{c}\hfill 2\hfill \\ \hfill 3\hfill \\ \hfill 0\hfill \end{array}\right).$$

3. Now let $C$ be the set $\{\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right),\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right),\left(\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)\}$. It is easy to check that $C$ forms a basis for ${ℝ}^3$ (determinant is non-zero). Order $C$ in the obvious manner. What is the change of basis matrix ${[I]}_A^C$? One way is to express each element of $C$ in terms of the elements of $A$. Another way is to use the formula ${[I]}_A^C={[I]}_A^B{[I]}_B^C$. Applying the first example, we see that ${[I]}_B^C$ is just the matrix whose columns are elements of $C$. As a result:

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 $A$, ${[I]}_E^A$ 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 one-to-one fashion) a change of basis from the basis $S_A$ whose elements are columns of $A$ to $E$, the standard basis: $A={[I]}_E^{S_A}$.

• 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 ${[I]}_B^A$, which is ${[I]}_B^E{[I]}_E^A={({[I]}_E^B)}^{-1}{[I]}_E^A.$

• Let $A^{\prime }$ be a re-ordering of the ordered basis $A$, where each $v_i^{\prime }\in A^{\prime }$ is just $v_{\pi (i)}$ for some permutation in $S_n$. Then ${[I]}_A^{A^{\prime }}$ 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 ${[T]}_B^A$. Under changes of basis from $A$ to $A^{\prime }$, and $B$ to $B^{\prime }$, we have

  $${[T]}_{B^{\prime }}^{A^{\prime }}={[ITI]}_{B^{\prime }}^{A^{\prime }}={[I]}_{B^{\prime }}^B{[T]}_B^A{[I]}_A^{A^{\prime }}.$$

• 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 ${[I]}_A^{A^{\prime }}$. 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.