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.

Title	change of basis
Canonical name	ChangeOfBasis
Date of creation	2013-03-22 17:30:18
Last modified on	2013-03-22 17:30:18
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	20
Author	CWoo (3771)
Entry type	Definition
Classification	msc 15A04
Synonym	change of coordinates
Synonym	change of bases
Synonym	basis change
Synonym	base change
Synonym	base change matrix
Defines	change of basis matrix