§ 12. The Change of Coordinate Matrix¶

The Change of Coordinate Matrix¶

Theorem 12.1 : Change of coordinate matrix is invertible.

Let \(\beta\) and \(\beta^{\prime}\) be two ordered bases for a finite-dimensional vector space \(V\), and let \(Q=[I_{V}]_{\beta^{\prime}}^{\beta}\). Then

(a) \(Q\) is invertible.
(b) For any \(v \in V\), \([v]_{\beta}=Q[v]_{\beta^{\prime}}\).

Proof

(a) Since \(I_{V}\) is invertible, \(Q\) is invertible by Theorem 11.6.
(b) For any \(v \in V\),

\[ [v]_{\beta}=\left[I_{V}(v)\right]_{\beta}=[I_{V}]_{\beta^{\prime}}^{\beta}[v]_{\beta^{\prime}}=Q[v]_{\beta^{\prime}} \]

by Theorem 10.14.

Definition 12.2 : Change of Coordinate Matrix

The matrix \(Q=[I_{V}]_{\beta^{\prime}}^{\beta}\) defined in Theorem 12.1 is called a change of coordinate matrix. Because of part (b) of the theorem, we say that \(Q\) changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates. Observe that if \(\beta=\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}\) and \(\beta^{\prime}=\left\{x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{n}^{\prime}\right\}\), then

\[ x_{j}^{\prime}=\sum_{i=1}^{n} Q_{i j} x_{i} \]

for \(j=1,2, \ldots, n\); that is, the \(j\)th column of \(Q\) is \(\left[x_{j}^{\prime}\right]_{\beta}\).

Example 12.3 : Computing a change of coordinate matrix

In \(\mathbb{R}^{2}\), let \(\beta=\{(1,1),(1,-1)\}\) and \(\beta^{\prime}=\{(2,4),(3,1)\}\). Since

\[ (2,4)=3(1,1)-1(1,-1) \quad \text { and } \quad(3,1)=2(1,1)+1(1,-1), \]

the matrix that changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates is

\[ Q=\left(\begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array}\right) . \]

Thus, for instance,

\[ [(2,4)]_{\beta}=Q[(2,4)]_{\beta^{\prime}}=Q\binom{1}{0}=\binom{3}{-1} . \]

Theorem 12.4 : Inverse of change of coordinate matrix changes coordinates back.

Let \(\beta\) and \(\beta^{\prime}\) be ordered bases for a finite-dimensional vector space \(V\). Let \(Q\) be the change of coordinate matrix that changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates, that is, \([v]_{\beta}=Q[v]_{\beta^{\prime}}\) for all \(v \in V\). Then \(Q^{-1}\) changes \(\beta\)-coordinates into \(\beta^{\prime}\)-coordinates; that is, \([v]_{\beta^{\prime}}=Q^{-1}[v]_{\beta}\) for all \(v \in V\). (See Exercise 12.11.)

Proof

Since \(Q\) is invertible by Theorem 12.1, the matrix \(Q^{-1}\) exists. Let \(v \in V\). From \([v]_{\beta}=Q[v]_{\beta^{\prime}}\), multiplying both sides by \(Q^{-1}\) yields \(Q^{-1}[v]_{\beta}=[v]_{\beta^{\prime}}\).

The Change of Basis Formula for a Linear Operator¶

Definition 12.5 : Linear Operator

Linear transformations that map a vector space \(V\) into itself is called a linear operator on \(V\).

Theorem 12.6 : Change of basis formula for a linear operator

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(\beta\) and \(\beta^{\prime}\) be ordered bases for \(V\). Suppose that \(Q\) is the change of coordinate matrix that changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates. Then

\[ [T]_{\beta^{\prime}}=Q^{-1}[T]_{\beta} Q . \]

Proof

Let \(I\) be the identity transformation on \(V\). Then \(T=IT=TI\); hence, by Theorem 10.6,

\[ Q[T]_{\beta^{\prime}}=[I]_{\beta^{\prime}}^{\beta}[T]_{\beta^{\prime}}=[IT]_{\beta^{\prime}}^{\beta}=[TI]_{\beta^{\prime}}^{\beta}=[T]_{\beta}[I]_{\beta^{\prime}}^{\beta}=[T]_{\beta} Q . \]

Therefore \([T]_{\beta^{\prime}}=Q^{-1}[T]_{\beta} Q\).

Theorem 12.6 can be generalized to allow \(T: V \rightarrow W\), where \(V\) is distinct from \(W\). In this case, we can change bases in \(V\) as well as in \(W\) (see Exercise 12.8).

Example 12.7 : Computing the change of basis formula for a linear operator

Let \(T\) be the linear operator on \(\mathbb{R}^{2}\) defined by

\[ T\binom{a}{b}=\binom{3 a-b}{a+3 b} \]

and let \(\beta\) and \(\beta^{\prime}\) be the ordered bases in Example 12.3. The reader should verify that

\[ [T]_{\beta}=\left(\begin{array}{rr} 3 & 1 \\ -1 & 3 \end{array}\right) . \]

In Example 12.3, we saw that the change of coordinate matrix that changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates is

\[ Q=\left(\begin{array}{rr} 3 & 2 \\ -1 & 1 \end{array}\right), \]

and it is easily verified that

\[ Q^{-1}=\frac{1}{5}\left(\begin{array}{rr} 1 & -2 \\ 1 & 3 \end{array}\right) . \]

Hence, by Theorem 12.6,

\[ [T]_{\beta^{\prime}}=Q^{-1}[T]_{\beta} Q=\left(\begin{array}{rr} 4 & 1 \\ -2 & 2 \end{array}\right) . \]

To show that this is the correct matrix, we can verify that the image under \(T\) of each vector of \(\beta^{\prime}\) is the linear combination of the vectors of \(\beta^{\prime}\) with the entries of the corresponding column as its coefficients. For example, the image of the second vector in \(\beta^{\prime}\) is

\[ T\binom{3}{1}=\binom{8}{6}=1\binom{2}{4}+2\binom{3}{1} . \]

Notice that the coefficients of the linear combination are the entries of the second column of \([T]_{\beta^{\prime}}\).

Example 12.8 : Using a change of basis to compute \(T(a,b)\)

Recall the reflection about the \(x\)-axis in Example 8.4. The rule \((x, y) \rightarrow(x,-y)\) is easy to obtain. We now derive the less obvious rule for the reflection \(T\) about the line \(y=2 x\). We wish to find an expression for \(T(a, b)\) for any \((a, b)\) in \(\mathbb{R}^{2}\). Since \(T\) is linear, it is completely determined by its values on a basis for \(\mathbb{R}^{2}\). Clearly, \(T(1,2)=(1,2)\) and \(T(-2,1)=-(-2,1)=(2,-1)\). Therefore if we let

\[ \beta^{\prime}=\left\{\binom{1}{2},\binom{-2}{1}\right\}, \]

then \(\beta^{\prime}\) is an ordered basis for \(\mathbb{R}^{2}\) and

\[ [T]_{\beta^{\prime}}=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right) . \]

Let \(\beta\) be the standard ordered basis for \(\mathbb{R}^{2}\), and let \(Q\) be the matrix that changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates. Then

\[ Q=\left(\begin{array}{rr} 1 & -2 \\ 2 & 1 \end{array}\right) \]

and \([T]_{\beta^{\prime}}=Q^{-1}[T]_{\beta} Q\). We can solve this equation for \([T]_{\beta}\) to obtain that \([T]_{\beta}=Q[T]_{\beta^{\prime}} Q^{-1}\). Because

\[ Q^{-1}=\frac{1}{5}\left(\begin{array}{rr} 1 & 2 \\ -2 & 1 \end{array}\right), \]

the reader can verify that

\[ [T]_{\beta}=\frac{1}{5}\left(\begin{array}{rr} -3 & 4 \\ 4 & 3 \end{array}\right) . \]

Since \(\beta\) is the standard ordered basis, it follows that \(T\) is left-multiplication by \([T]_{\beta}\). Thus for any \((a, b)\) in \(\mathbb{R}^{2}\), we have

\[ T\binom{a}{b}=\frac{1}{5}\left(\begin{array}{rr} -3 & 4 \\ 4 & 3 \end{array}\right)\binom{a}{b}=\frac{1}{5}\binom{-3 a+4 b}{4 a+3 b} . \]

Corollary 12.9 : Matrix representation of \(L_A\) in an arbitrary basis

Let \(A \in \mathrm{M}_{n \times n}(F)\), and let \(\gamma\) be an ordered basis for \(F^{n}\). Then \([L_{A}]_{\gamma}=Q^{-1} A Q\), where \(Q\) is the \(n \times n\) matrix whose \(j\)th column is the \(j\)th vector of \(\gamma\).

Example 12.10 : Computing \([L_A]_{\gamma}\)

Let

\[ A=\left(\begin{array}{rrr} 2 & 1 & 0 \\ 1 & 1 & 3 \\ 0 & -1 & 0 \end{array}\right) \]

and let

\[ \gamma=\left\{\left(\begin{array}{r} -1 \\ 0 \\ 0 \end{array}\right),\left(\begin{array}{l} 2 \\ 1 \\ 0 \end{array}\right),\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)\right\} \]

which is an ordered basis for \(\mathbb{R}^{3}\). Let \(Q\) be the \(3 \times 3\) matrix whose \(j\)th column is the \(j\)th vector of \(\gamma\). Then

\[ Q=\left(\begin{array}{rrr} -1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right) \quad \text { and } \quad Q^{-1}=\left(\begin{array}{rrr} -1 & 2 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{array}\right) \]

So by the preceding corollary,

\[ [L_{A}]_{\gamma}=Q^{-1} A Q=\left(\begin{array}{rrr} 0 & 2 & 8 \\ -1 & 4 & 6 \\ 0 & -1 & -1 \end{array}\right) . \]

Similarity¶

Definition 12.11 : Similarity

Let \(A\) and \(B\) be matrices in \(\mathrm{M}_{n \times n}(F)\). We say that \(B\) is similar to \(A\) if there exists an invertible matrix \(Q\) such that \(B=Q^{-1} A Q\).

Theorem 12.12 : Similarity is an equivalence relation.

The relation of similarity is an equivalence relation on \(\mathrm{M}_{n \times n}(F)\).

Proof

Reflexive.
For any \(A \in \mathrm{M}_{n \times n}(F)\), we have \(A=I_{n}^{-1} A I_{n}\), so \(A\) is similar to \(A\).
Symmetric.
Suppose that \(B\) is similar to \(A\). Then \(B=Q^{-1} A Q\) for some invertible \(Q\). Multiplying on the left by \(Q\) and on the right by \(Q^{-1}\) gives \(A=Q B Q^{-1}\), hence \(A\) is similar to \(B\).
Transitive.
Suppose that \(B\) is similar to \(A\) and that \(C\) is similar to \(B\). Then \(B=Q^{-1} A Q\) and \(C=P^{-1} B P\) for some invertible matrices \(P\) and \(Q\). Thus

\[ C=P^{-1} Q^{-1} A(Q P). \]

Since \((QP)(P^{-1}Q^{-1})=I_{n}\) and \((P^{-1}Q^{-1})(QP)=I_{n}\), the matrix \(QP\) is invertible with inverse \(P^{-1}Q^{-1}\). Hence \(C=(Q P)^{-1} A(Q P)\), so \(C\) is similar to \(A\).

So we need only say that \(A\) and \(B\) are similar.

Corollary 12.13 : Matrix representations of a linear operator are similar.

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(\beta\) and \(\beta^{\prime}\) be any ordered bases for \(V\). Then \([T]_{\beta^{\prime}}\) is similar to \([T]_{\beta}\).

Proof

By Theorem 12.6, \([T]_{\beta^{\prime}}=Q^{-1}[T]_{\beta} Q\) for the change of coordinate matrix \(Q\). By Definition 12.11, this means \([T]_{\beta^{\prime}}\) is similar to \([T]_{\beta}\).

Exercise¶

Exercise 12.8

Prove the following generalization of Theorem 12.6. Let \(T: V \rightarrow W\) be a linear transformation from a finite-dimensional vector space \(V\) to a finite-dimensional vector space \(W\). Let \(\beta\) and \(\beta^{\prime}\) be ordered bases for \(V\), and let \(\gamma\) and \(\gamma^{\prime}\) be ordered bases for \(W\). Then \([T]_{\beta^{\prime}}^{\gamma^{\prime}}=P^{-1}[T]_{\beta}^{\gamma} Q\), where \(Q\) is the matrix that changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates and \(P\) is the matrix that changes \(\gamma^{\prime}\)-coordinates into \(\gamma\)-coordinates.

Exercise 12.10

Prove that if \(A\) and \(B\) are similar \(n \times n\) matrices, then \(\operatorname{tr}(A)=\operatorname{tr}(B)\).

Hint: Use Exercise 10.13.

Exercise 12.11

Let \(V\) be a finite-dimensional vector space with ordered bases \(\alpha, \beta\), and \(\gamma\).

(a) Prove that if \(Q\) and \(R\) are the change of coordinate matrices that change \(\alpha\)-coordinates into \(\beta\)-coordinates and \(\beta\)-coordinates into \(\gamma\)-coordinates, respectively, then \(RQ\) is the change of coordinate matrix that changes \(\alpha\)-coordinates into \(\gamma\)-coordinates.
(b) Prove that if \(Q\) changes \(\alpha\)-coordinates into \(\beta\)-coordinates, then \(Q^{-1}\) changes \(\beta\)-coordinates into \(\alpha\)-coordinates.

Exercise 12.13

Let \(V\) be a finite-dimensional vector space over a field \(F\), and let \(\beta=\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}\) be an ordered basis for \(V\). Let \(Q\) be an \(n \times n\) invertible matrix with entries from \(F\). Define

\[ x_{j}^{\prime}=\sum_{i=1}^{n} Q_{i j} x_{i} \quad \text { for } 1 \leq j \leq n \]

and set \(\beta^{\prime}=\left\{x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{n}^{\prime}\right\}\). Prove that \(\beta^{\prime}\) is a basis for \(V\) and hence that \(Q\) is the change of coordinate matrix changing \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates.

Exercise 12.14

Prove the converse of Exercise 12.8: If \(A\) and \(B\) are each \(m \times n\) matrices with entries from a field \(F\), and if there exist invertible \(m \times m\) and \(n \times n\) matrices \(P\) and \(Q\), respectively, such that \(B=P^{-1} A Q\), then there exist an \(n\)-dimensional vector space \(V\) and an \(m\)-dimensional vector space \(W\) (both over \(F\)), ordered bases \(\beta\) and \(\beta^{\prime}\) for \(V\) and \(\gamma\) and \(\gamma^{\prime}\) for \(W\), and a linear transformation \(T: V \rightarrow W\) such that

\[ A=[T]_{\beta}^{\gamma} \quad \text { and } \quad B=[T]_{\beta^{\prime}}^{\gamma^{\prime}} . \]

Hints: Let \(V=F^{n}\), \(W=F^{m}\), \(T=L_{A}\), and \(\beta\) and \(\gamma\) be the standard ordered bases for \(F^{n}\) and \(F^{m}\), respectively. Now apply the results of Exercise 12.13 to obtain ordered bases \(\beta^{\prime}\) and \(\gamma^{\prime}\) from \(\beta\) and \(\gamma\) via \(Q\) and \(P\), respectively.