Change of basis

Let \(V\) be a vector space with a basis \(v_1, \dots, v_n\). For brevity we write \(\underline v\) for this basis. Recall from Proposition 3.64 that then any vector \(x \in V\) can be written uniquely as

\[ x = \alpha_1 v_1 + \dots + \alpha_n v_n \]

and we regard the coefficients \(\alpha_1, \dots, \alpha_n\) as the coordinates of \(x\) with respect to the basis \(\underline v\). We indicate this notationally by writing \((\alpha_1, \dots, \alpha_n)_{\underline v}\).

If we take, instead, another basis \(\underline w\) consisting of vectors \(w_1, \dots, w_n \in V\) then

\[ x = \beta_1 w_1 + \dots + \beta_n w_n, \]

giving different coordinates of \(x\) with respect to the basis \(\underline w\). Our goal in this section is to answer the natural question how to pass from the coordinates \((\alpha_1, \dots, \alpha_n)_{\underline v}\) to \((\beta_1, \dots, \beta_n)_{\underline w}\).

Example 4.84

Consider the identity map \({\mathrm {id}}_V : V \to V\) (Example 4.7). We fix a basis \(\underline v = \{v_1, \dots, v_n\}\) of \(V\) and determine the matrix of \({\mathrm {id}}_V\) with respect to this basis both in the domain and in the codomain. We have

\[ \begin{align*} {\mathrm {id}}_V(v_1) & = v_1 = 1 \cdot v_1 + 0 v_2 + \dots + 0 v_n \\ \vdots \\ {\mathrm {id}}_V(v_n) & = v_1 = 0 \cdot v_1 + \dots + 0 v_{n-1} + 1 v_n. \end{align*} \]

These coefficients in \({\mathrm {id}}_V(v_i)\) form the \(i\)-th column of the matrix, which therefore is equal to

\[ {\mathrm M}_{{\mathrm {id}}_V, \underline v, \underline v} = \left ( \begin{array}{cccc} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & \dots & 0 & 1 \end{array} \right ) = {\mathrm {id}}_n. \]

Example 4.85 (Related exercises: Exercise 3.12)

We now consider still the identity map \({\mathrm {id}}_V\), but with a basis \(\underline v = \{v_1, \dots, v_n\}\) on the domain and another basis \(\underline w = \{w_1, \dots, w_n\}\) on the codomain. The matrix of \({\mathrm {id}}_V\) with respect to these bases is found by expressing

\[ {\mathrm {id}}_V(v_1) = v_1 = a_{11} w_1 + \dots + a_{n1} w_n \]

i.e., we express \(v_1\) in coordinates with respect to the basis \(\underline w\). More generally, for all \(i \le n\):

\[ {\mathrm {id}}_V(v_i) = v_1 = a_{1i} w_1 + \dots + a_{ni} w_n. \]

The matrix of \({\mathrm {id}}_V\) with respect to the bases \(\underline v\) (domain) and \(\underline w\) (codomain) is then

\[ {\mathrm M}_{{\mathrm {id}}_V, \underline v, \underline w} = \left ( \begin{array}{cccc} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{array} \right ). \]

We refer to this matrix as the base change matrix from \(\underline v\) to \(\underline w\).

Example 4.86

Here is a concrete example of the above situation. Let \(V = {\bf R}^2\), \(\underline v = \{e_1, e_2\} = \{(1,0), (0,1)\}\) be the standard basis and \(\underline w = \{w_1, w_2\} = \{(1,1), (1,3)\}\) be another basis. We compute the matrix following the above lines:

\[ {\mathrm {id}}_{{\bf R}^2}(1,0) = (1,0) = a_{11} (1,1) + a_{21} (1,3) \]

has the solution \(a_{11}=\frac 32\) and \(a_{21} = -\frac 12\).

\[ {\mathrm {id}}_{{\bf R}^2}(0,1) = (0,1) = a_{12} (1,1) + a_{22}(1,3) \]

has the solution \(a_{12} = -\frac 12\), \(a_{22}=\frac 12\), so the matrix reads

\[ {\mathrm M}_{{\mathrm {id}}_{{\bf R}^2}, \underline v, \underline w} = \left ( \begin{array}{cc} \frac 32 & -\frac 12 \\ -\frac 12 & \frac 12 \end{array} \right ). \]

Thus, for example, the vector \((3,2) = 3e_1 + 2e_2\) can be expressed as

\[ {\mathrm M}_{{\mathrm {id}}_{{\bf R}^2}, \underline v, \underline w} {\left ( \begin{array}{c} 3 \\ 2 \end{array} \right )}_{\underline v} = \left ( \begin{array}{cc} \frac 32 & -\frac 12 \\ -\frac 12 & \frac 12 \end{array} \right ) {\left ( \begin{array}{c} 3 \\ 2 \end{array} \right )}_{\underline v} = {\left ( \begin{array}{c} \frac 72 \\ -\frac 12 \end{array} \right )}_{\underline w}, \]

i.e.,

\[ (3, 2) = \frac 72 w_1 - \frac 12 w_2. \]

By Proposition 4.52, the composition of linear maps corresponds to the product of matrices. We apply this to the composition

\[ {\mathrm {id}}_V \circ {\mathrm {id}}_V = {\mathrm {id}}_V. \]

We choose the following bases on \(V\), indicated by a subscript:

\[ V_{\underline v} \xrightarrow{{\mathrm {id}}_V} V_{\underline w} \xrightarrow{{\mathrm {id}}_V} V_{\underline v}. \]

We consider the associated matrices \({\mathrm M}_{{\mathrm {id}}_V, \underline v, \underline w}\) for the first map, and \({\mathrm M}_{{\mathrm {id}}_V, \underline w, \underline v}\) for the second one. We have

\[ {\mathrm M}_{{\mathrm {id}}_V, \underline w, \underline v} \cdot {\mathrm M}_{{\mathrm {id}}_V, \underline v, \underline w} = {\mathrm M}_{{\mathrm {id}}, \underline v, \underline v} = {\mathrm {id}}_n. \]

In other words,

\[ {\mathrm M}_{{\mathrm {id}}_V, \underline w, \underline v} = ({\mathrm M}_{{\mathrm {id}}_V, \underline v, \underline w})^{-1}. \]

The above observations lead to the following.

Method 4.87

Let \(f : V \to V\) be a linear map represented by a matrix \(E \in {\mathrm {Mat}}_{n \times n}\) with respect to a fixed basis \(\underline v\) on the domain and codomain. The matrix of \(f\) with respect to another basis \(\underline w\) (again on the domain and the codomain) is

\[ {\mathrm M}_{f, \underline w, \underline w} = A E A^{-1}, \]

where \(A\) is the matrix describing the change of basis from \(\underline v\) to \(\underline w\), i.e., \(A = {\mathrm M}_{{\mathrm {id}}_V, \underline v, \underline w}\) (cf. Example 4.85).

Proof. Indeed, \(E\) is the matrix for \(V_{\underline v} \xrightarrow{f} V_{\underline v}\), which we indicate by writing

\[ V_{\underline v} \xrightarrow[E]{f} V_{\underline v}. \]

The composition

\[ V \xrightarrow{{\mathrm {id}}_V} V \xrightarrow{f} V \xrightarrow{{\mathrm {id}}_V} V \]

is again \(f\), but as above we now put a different basis on \(V\):

\[ V_{\underline w} \xrightarrow[{\mathrm M}_{{\mathrm {id}}_V, \underline w, \underline v}]{{\mathrm {id}}_V} V_{\underline v} \xrightarrow[E]{f} V_{\underline v} \xrightarrow[{\mathrm M}_{{\mathrm {id}}_V, \underline v, \underline w}] {{\mathrm {id}}_V} V_{\underline w}. \]

According to the above, this simplifies to

\[ V_{\underline w} \xrightarrow[A^{-1}]{{\mathrm {id}}_V} V_{\underline v} \xrightarrow[E]{f} V_{\underline v} \xrightarrow[A] {{\mathrm {id}}_V} V_{\underline w}. \]

◻

Example 4.88

Consider the linear map \(f : {\bf R}^2 \to {\bf R}^2\) given in the standard basis \(\underline e = \{e_1, e_2\}\) by multiplication with the matrix \(E = \left ( \begin{array}{cc} 2 & 0 \\ -1 & 3 \end{array} \right )\). We compute the matrix associated to \(f\) with respect to the basis \(\underline w = \{w_1, w_2\} = \{(1,1), (1,-2)\}\) (on the domain and codomain). According to the above, we need to compute the base change matrix \(A = {\mathrm M}_{{\mathrm {id}}, \underline e, \underline w}\) and its inverse \(A^{-1} = {\mathrm M}_{{\mathrm {id}}, \underline w, \underline e}\). The matrix \(A^{-1}\) can be computed more easily than \(A\), because its entries are given by coefficients in the following linear combinations:

\[ {\mathrm {id}}(w_1) = w_1 = (1,1) = 1 \cdot e_1 + 1 \cdot e_2, \]

\[ {\mathrm {id}}(w_2) = w_2 = (1,-2) = 1 \cdot e_1 - 2 \cdot e_2. \]

Thus \(A^{-1} = \left ( \begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array} \right )\). We compute \(A = (A^{-1})^{-1}\) using the method described in Theorem 4.81:

\[ \begin{align*} (A^{-1} | {\mathrm {id}}) & = \left ( \begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ 1 & -2 & 0 & 1 \end{array} \right ) \leadsto \left ( \begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ 0 & -3 & -1 & 1 \end{array} \right ) \\ & \leadsto \left ( \begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ 0 & 1 & \frac 13 & -\frac 13 \end{array} \right ) \leadsto \left ( \begin{array}{cc|cc} 1 & 0 & \frac 23 & \frac 13 \\ 0 & 1 & \frac 13 & -\frac 13 \end{array} \right ) \\ & = ({\mathrm {id}} | (A^{-1})^{-1}) = ({\mathrm {id}} | A). \end{align*} \]

This leads to

\[ {\mathrm M}_{f, \underline w, \underline w } = A E A^{-1} = \left ( \begin{array}{cc} \frac 23 & \frac 13 \\ \frac 13 & -\frac 13 \end{array} \right ) \cdot \left ( \begin{array}{cc} 2 & 0 \\ -1 & 3 \end{array} \right ) \cdot \left ( \begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array} \right ) = \left ( \begin{array}{cc} 2 & -1 \\ 0 & 3 \end{array} \right ). \]