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: $$v=\sum_{v_i\in A} r_iv_i$$ 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 $n<\infty$ . 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 $$A=\Bigg\lbrace \begin{pmatrix} 1\\-1\\2 \end{pmatrix}, \begin{pmatrix} 0\\1\\1 \end{pmatrix}, \begin{pmatrix} 2\\3\\0 \end{pmatrix} \Bigg\rbrace \mbox{ and }B=\Bigg\lbrace \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \begin{pmatrix} 0\\1\\0 \end{pmatrix}, \begin{pmatrix} 0\\0\\1 \end{pmatrix} \Bigg\rbrace$$ 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 $\Bigg\lbrace \begin{pmatrix} 1\\0\\2 \end{pmatrix}, \begin{pmatrix} 0\\1\\1 \end{pmatrix}, \begin{pmatrix} 2\\1\\0 \end{pmatrix} \Bigg\rbrace$ . It is easy to check that $C$ forms a basis for $\mathbb{R}^3$ (determinant is non-zero). 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 one-to-one fashion) a change of basis from the basis $S_A$ whose elements are columns of $A$ to $E$ , the standard basis: $A=[I]^{S_A}_E$ .
• 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 $A'$ be a re-ordering of the ordered basis $A$ , where each $v'_i\in A'$ is just $v_{\pi(i)}$ for some permutation in $S_n$ . Then $[I]^{A'}_A$ 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 $A'$ , and $B$ to $B'$ , we have $$[T]^{A'}_{B'}=[ITI]^{A'}_{B'}=[I]^B_{B'}[T]^A_B[I]^{A'}_A.$$
• If $T$ is a linear operator on $V$ , then setting $V=W$ , $A=B$ and $A'=B'$ from above, we have that $$[T]_{A'}=P^{-1}[T]_A P,$$ where $P$ is the change of basis matrix $[I]^{A'}_A$ . This shows that $[T]_A$ and $[T]_{A'}$ are similar matrices. In other words, under a change of basis, the linear transformation $T$ is basically the same.

Bibliography

1

Friedberg, Insell, Spence. Linear Algebra. Prentice-Hall Inc., 1997.