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Euclidean vector spaces

Aaron Greicius

Section 5.5 Change of basis

Coordinate vectors and matrix representations work in tandem to model vectors in an abstract vector space \(V\) as column vectors in \(\R^n\text{,}\) and linear transformations \(T\colon V\rightarrow W\) as \(m\times n\) matrices. In both cases the model depends on our choice of basis. In this section we investigate how different basis choices affect these various models. Specifically, we consider the two questions below.
  1. Given \(V\) and two ordered bases \(B\) and \(B'\text{,}\) what is the algebraic relation between \([\boldv]_B\) and \([\boldv]_{B'}\text{?}\)
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  2. Given \(T\colon V\rightarrow V\) and two ordered bases \(B\) and \(B'\text{,}\) what is the relation between \([T]_{B}\) and \([T]_{B'}\text{?}\)
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We will tackle each question in turn. Both answers rely on something called a change of basis matrix \(\underset{B\rightarrow B'}{P}\text{.}\)
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Subsection 5.5.1 Change of basis matrix

We define change of basis matrices via a column-by-column formula and motivate the definition retroactively with Theorem 5.5.2.
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Definition 5.5.1. Change of basis matrix.

Let \(B=(\boldv_1, \boldv_2, \dots, \boldv_n)\) and \(B'\) be two ordered bases for the vector space \(V\text{.}\) The change of basis from \(B\) to \(B'\) is the \(n\times n\) matrix \(\underset{B\rightarrow B'}{P}\) defined as
\begin{equation} \underset{B\rightarrow B'}{P}= \begin{bmatrix} \vert \amp \vert \amp \amp \vert \\ \phantom{v}[\boldv_1]_{B'} \amp \phantom{v}[\boldv_2]_{B'}\amp \dots \amp \phantom{v}[\boldv_n]_{B}\\ \vert \amp \vert \amp \amp \vert \end{bmatrix}\text{.}\tag{5.13} \end{equation}
In other words, the \(j\)-th column of \(\underset{B\rightarrow B'}{P}\) is obtained by computing the coordinate vector of the \(j\)-th element of the original basis \(B\) with respect to the new basis \(B'\text{.}\)
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Proof.

  1. Let \(B=(\boldv_1, \boldv_2,\dots, \boldv_n)\text{.}\) From formula (5.10) applied to \(I_v\text{,}\) we see that the \(j\)-th column of \([I_V]_B^{B'}\) is
    \begin{equation*} [\boldv_j]_{B'} \end{equation*}
    for all \(1\leq j\leq n\text{.}\) Using formula (5.13) from Definition 5.5.1 this is precisely the \(j\)-th column of \(\underset{B\rightarrow B'}{P}\) for all \(1\leq j\leq n\text{.}\) We conclude that
    \begin{equation*} [I_V]_B^{B'} = \underset{B\rightarrow B'}{P}\text{.} \end{equation*}
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  2. This follows from (1) and Theorem 5.4.6:
    \begin{align*} \underset{B\rightarrow B'}{P}[\boldv]_B \amp = [\id_V]_{B}^{B'}[\boldv]_B \\ \amp = [\id_V(\boldv)]_{B'} \amp (\knowl{./knowl/xref/th_matrixrep.html}{\text{Theorem 5.4.6}})\\ \amp =[\boldv]_{B'} \text{.} \end{align*}
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  3. By (2) of Theorem 5.4.6 (the uniqueness claim), if \(A\) satisfies \(A[\boldv]_B=[\boldv]_{B'}\) for all \(\boldv\in \R^n\text{,}\) then \(A=[\id_V]_{B}^{B'}\text{.}\) Since \([\id_V]_{B}^{B'}=\underset{B\rightarrow B'}{P}\text{,}\) we conclude \(A=\underset{B\rightarrow B'}{P}\text{.}\)
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Example 5.5.3. Change of basis.

Let \(V=\R^2\text{,}\) \(B=(\boldv_1=(1,1),\boldv_2=(1,-1))\text{,}\) \(B'=(\boldw_1=(1,2), \boldw_2=(2,-1))\text{.}\)
  1. Compute \(\underset{B\rightarrow B'}{P}\text{.}\)
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  2. Let \(\boldx=(4,-2)\text{.}\) Compute \([\boldx]_{B'}\) using the change of basis formula (5.14).
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Solution.
  1. Using Definition 5.5.1, we have
    \begin{align*} \underset{B\rightarrow B'}{P}\amp = \begin{bmatrix}\vert \amp \vert \\ \hspace{7pt}[\boldv_1]_{B'} \amp \hspace{7pt}[\boldv_2]_{B'}\\ \vert \amp \vert \end{bmatrix}\\ \amp = \begin{bmatrix}\frac{3}{5}\amp -\frac{1}{5}\\ \frac{1}{5}\amp \frac{3}{5} \end{bmatrix} \text{.} \end{align*}
    The two coordinate vector computations \([(1,1)]_{B'}=(3/5, 1/5)\) and \([(1,-1)]_{B'}=(-1/5,3/5)\) were done as usual using Procedure 5.3.4: that is, by setting up in turn the vector equations
    \begin{align*} (1,1) \amp = c_1(1,2)+c_2(2,-1) \amp (1,-1)\amp =c_1(1,2)+c_2(2,-1) \end{align*}
    and solving for \(c_1,c_2\) using Gaussian elimination.
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  2. The usual application of Procedure 5.3.4 produces the coordinate vector \([\boldx]_{B}=(1, 3)\text{.}\) We leave the details to you. To compute \([\boldv]_{B'}\text{,}\) we use the change of basis formula (5.14):
    \begin{align*} [\boldx]_{B'} \amp =\underset{B\rightarrow B'}{P}[\boldx]_B \\ \amp = \begin{bmatrix}\frac{3}{5}\amp -\frac{1}{5}\\ \frac{1}{5}\amp \frac{3}{5} \end{bmatrix} \begin{amatrix}[c]1 \\ 3 \end{amatrix}\\ \amp = \begin{amatrix}[r]0\\ 2 \end{amatrix}\text{.} \end{align*}
    This should come as now surprise since
    \begin{equation*} \boldx=(4,-2)=2(2,-1)=0\boldw_1+2\boldw_2\text{.} \end{equation*}
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Example 5.5.4. Change of basis.

Let \(B=(\bolde_1,\bolde_2,\bolde_3)\) be the standard basis of \(\R^3\) and consider the nonstandard basis \(B'=(\boldx_1=(1,-4,4),\boldx_2=(0,1,-2), \boldx_3=(0,0,1))\text{.}\)
  1. Compute \(\underset{B\rightarrow B'}{P}\text{.}\)
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  2. Compute \([(1,1,1)]_{B'}\) using (5.14).
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Solution.
  1. We have
    \begin{align*} \underset{B\rightarrow B'}{P} \amp = \begin{bmatrix} \vert \amp \vert \amp \vert \\ [\bolde_1]_{B'}\amp [\bolde_2]_{B'}\amp [\bolde_3]_{B'} \\ \vert \amp \vert \amp \vert \end{bmatrix} \\ \amp = \begin{bmatrix} 1\amp 0\amp 0 \\ 4\amp 1\amp 0\\ 4\amp 2\amp 1 \end{bmatrix} \text{.} \end{align*}
    The first two coordinate vector computations are nontrivial, but can be computed as usual using Procedure 5.3.4; you can verify for yourself that
    \begin{align*} (1,0,0)\amp =1(1,-4,4)+4(0,1,-2)+4(0,0,1)\\ (0,1,0) \amp =0(1,-4,4)+1(0,1,-2)+2(0,0,1)\text{.} \end{align*}
    The third coordinate vector computation, \([(0,0,1)]_{B'}\) is easier: since \((0,0,1)=\boldv_3\text{,}\) we have
    \begin{equation*} (0,0,1)=0\boldx_1+0\boldx_2+1\boldx_3\text{,} \end{equation*}
    and thus \([(0,0,1)]_{B'}=(0,0,1)\text{.}\)
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  2. Since \(B\) is the standard basis, we see easily that \([(1,1,1)]_{B}=(1,1,1)\text{.}\) Using (5.14) we have
    \begin{align*} [(1,1,1)]_{B'} \amp = \underset{B\rightarrow B'}{P}[(1,1,1)]_{B}\\ \amp = \underset{B\rightarrow B'}{P}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}\\ \amp =\begin{bmatrix} 1\\ 5\\ 7 \end{bmatrix} \text{.} \end{align*}
    Verify for yourself that we do indeed have
    \begin{equation*} (1,1,1)=1(1,-4,4)+5(0,1,-2)+7(0,0,1)\text{.} \end{equation*}
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The following properties are often useful when computing various change of basis matrices.
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Proof.

  1. Let \(B=(\boldv_1, \boldv_2,\dots, \boldv_n)\text{.}\) From formula (5.10) applied to \(\id_v\text{,}\) we see that the \(j\)-th column of \([\id_V]_B^{B'}\) is
    \begin{equation*} [\boldv_j]_{B'} \end{equation*}
    for all \(1\leq j\leq n\text{.}\) Using formula (5.13) from Definition 5.5.1 this is precisely the \(j\)-th column of \(\underset{B\rightarrow B'}{P}\) for all \(1\leq j\leq n\text{.}\) We conclude that
    \begin{equation*} [I_V]_B^{B'} = \underset{B\rightarrow B'}{P}\text{.} \end{equation*}
    Choosing \(B'=B\text{,}\) we see that
    \begin{align*} \underset{B\rightarrow B}{P} \amp =[\id_V]_{B}^{B} \\ \amp = I\text{.} \end{align*}
    The last equality above follows since the \(j\)-th column of \([\id_V]_{B}^{B}\) is
    \begin{equation*} [\id_V(\bolde_j)]_{B}=[\bolde_j]_{B}=\bolde_j\text{,} \end{equation*}
    since \(B\) is the standard basis.
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  2. This follows from (1) and Theorem 5.4.6:
    \begin{align*} \underset{B\rightarrow B'}{P}[\boldv]_B \amp = [\id_V]_{B}^{B'}[\boldv]_B \\ \amp = [\id_V(\boldv)]_{B'} \amp (\knowl{./knowl/xref/th_matrixrep.html}{\text{Theorem 5.4.6}})\\ \amp =[\boldv]_{B'} \text{.} \end{align*}
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  3. By (2) of Theorem 5.4.6 (the uniqueness claim), if \(A\) satisfies \(A[\boldv]_B=[\boldv]_{B'}\) for all \(\boldv\in \R^n\text{,}\) then \(A=[\id_V]_{B}^{B'}\text{.}\) Since \([\id_V]_{B}^{B'}=\underset{B\rightarrow B'}{P}\text{,}\) we conclude \(A=\underset{B\rightarrow B'}{P}\text{.}\)
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Example 5.5.6. \(V=\R^n\text{,}\) \(B\) standard basis.

Consider the special situation where \(V=\R^n\text{,}\) \(B\) is the standard basis, and \(B'=\{\boldv_1,\dots,\boldv_n\}\) is some nonstandard basis. In this case we have
\begin{align*} \underset{B'\rightarrow B}{P}\amp =\begin{bmatrix}\vert\amp\vert\amp \amp \vert \\ [\boldv_1]_B\amp [\boldv_2]_B \amp \cdots\amp [\boldv_n]_B\\ \vert\amp \vert\amp \amp \vert \end{bmatrix}\\ \amp = \begin{bmatrix}\vert\amp\vert\amp \amp \vert \\ \boldv_1\amp \boldv_2\amp\cdots\amp \boldv_n\\ \vert\amp \vert\amp \amp \vert \end{bmatrix} \amp (B \text{ standard basis})\text{.} \end{align*}
In other words, \(\underset{B'\rightarrow B}{P}\) is the matrix whose \(j\)-th column is just the \(j\)-th element of \(B'\text{.}\) Thus, in this situation we can compute \(\underset{B'\rightarrow B}{P}\) by placing the elements of \(B'\) as columns of a matrix, and then use (2) of Theorem 5.5.5 to compute \(\underset{B\rightarrow B'}{P}=\left(\underset{B'\rightarrow B}{P}\right)^{-1}\text{.}\)
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Example 5.5.7. Change of basis, \(B\) standard.

Let \(V=\R^2\text{,}\) \(B=((1,0),(0,1))\text{,}\) \(B'=\{(1,\sqrt{3}),(-\sqrt{3},1)\}\text{.}\) Compute \(\underset{B\rightarrow B'}{P}\) and \(\underset{B'\rightarrow B}{P}\text{.}\)
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Solution.
According to Example 5.5.6 we have
\begin{equation*} \underset{B'\rightarrow B}{P}=\begin{bmatrix}1\amp -\sqrt{3}\\ \sqrt{3}\amp 1 \end{bmatrix}\text{.} \end{equation*}
We then compute
\begin{equation*} \underset{B\rightarrow B'}{P}=(\underset{B'\rightarrow B}{P})^{-1}=\left(\begin{bmatrix}1\amp -\sqrt{3}\\ \sqrt{3}\amp 1 \end{bmatrix} \right)^{-1}=\frac{1}{4}\begin{bmatrix}1\amp \sqrt{3}\\ -\sqrt{3}\amp 1 \end{bmatrix}\text{.} \end{equation*}
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Remark 5.5.8. \(B\) standard basis of \(V\).

The observation from Example 5.5.6 applies more generally when \(B\) is the standard basis of the given vector space \(V\) and \(B'=(\boldv_1, \boldv_2, \dots, \boldv_n)\) is nonstandard. In this case computing \(\underset{B'\rightarrow B}{P}\) will be easy as the coordinate vectors \([\boldv_j]_{B}\) can be produced by inspection. See Example 5.5.9.
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Example 5.5.9. Change of basis, \(B\) standard.

Let \(V=M_{22}\text{,}\) \(B=(E_{11}, E_{12}, E_{21}, E_{22})\) (standard basis) and \(B'=(A_1,A_2,A_3,A_4)\text{,}\) where
\begin{equation*} A_1=\begin{amatrix}[rr] 1\amp 1\\ 1\amp 1 \end{amatrix}, A_2=\begin{amatrix}[rr] 1\amp -1\\ 1\amp -1 \end{amatrix}, A_3=\begin{amatrix}[rr] 1\amp 1\\ -1\amp -1 \end{amatrix}, A_4=\begin{amatrix}[rr] -1\amp 1\\ 1\amp -1 \end{amatrix}\text{.} \end{equation*}
Compute \(\underset{B'\rightarrow B}{P}\text{.}\)
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Solution.
We have
\begin{align*} \underset{B'\rightarrow B}{P}\amp = \begin{bmatrix} \vert\amp \vert\amp \vert\amp \vert\\ [A_1]_{B}\amp [A_2]_B\amp [A_3]_B\amp [A_4]_B\\ \vert\amp \vert\amp \vert\amp \vert \end{bmatrix}\\ \amp = \begin{amatrix}[rrrr] 1\amp 1\amp 1\amp -1\\ 1\amp -1\amp 1\amp 1\\ 1\amp 1\amp -1\amp 1\\ 1\amp -1\amp -1\amp -1 \end{amatrix} \text{.} \end{align*}
Here the coordinate vectors \([A_i]_B\) are easily computed by inspection since \(B\) is the standard basis.
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It turns out that \(\underset{B\rightarrow B'}{P}=(\underset{B'\rightarrow B}{P})^{-1}\) is not so difficult to compute in this case since the columns \(\boldc_j\) of \(\underset{B'\rightarrow B}{P}\) satisfy
\begin{equation*} \boldc_i\cdot\boldc_j=\begin{cases} 4\amp \text{if } i=j\\ 0\amp \text{if } i\ne j \end{cases}\text{.} \end{equation*}
From this observation and Theorem 3.1.25 it is easy to see that
\begin{equation*} \underset{B\rightarrow B'}{P}=(\underset{B'\rightarrow B}{P})^{-1}=\frac{1}{4} \begin{amatrix}[rrrr] 1\amp 1\amp 1\amp 1\\ 1\amp -1\amp 1\amp -1\\ 1\amp 1\amp -1\amp -1\\ -1\amp -1\amp 1\amp -1 \end{amatrix}\text{.} \end{equation*}
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Example 5.5.10. Video example: change of basis matrix.

Figure 5.5.11. Video: change of basis matrix
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Before connecting change of basis matrices with matrix representations of linear transformations, it is worth gathering some of the different techniques for computing change of basis matrices we have discussed so far.
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Subsection 5.5.2 Change of basis for transformations

We now investigate how our choice of basis affects matrix representations of linear transformations. We will only consider the special case where \(T\colon V\rightarrow V\) and we are comparing matrix representations \([T]_B\) and \([T]_{B'}\) for two different ordered bases of \(V\text{.}\)
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Proof.

First observe that (5.17) follows from (5.16) and (2) of Theorem 5.5.5. Next, to prove (5.16), it suffices by (2) of Theorem 5.4.6 to show that the matrix \(A=\underset{B\rightarrow B'}{P}\, [T]_B\, \underset{B'\rightarrow B}{P}\) satisfies
\begin{equation*} A[\boldv]_{B'}=[T(\boldv)]_{B'} \end{equation*}
for all \(\boldv\in V\text{.}\) To this end, given any \(\boldv\in V\text{,}\) we have
\begin{align*} A[\boldv]_{B'}=\underset{B\rightarrow B'}{P}\, [T]_B\, \underset{B'\rightarrow B}{P}[\boldv]_{B'} \amp= \underset{B\rightarrow B'}{P}\, [T]_B [\boldv]_B \amp (\knowl{./knowl/xref/th_change_of_basis_coordinates.html}{\text{Theorem 5.5.2}})\\ \amp= \underset{B\rightarrow B'}{P}[T(\boldv)]_{B} \amp (\knowl{./knowl/xref/th_matrixrep.html}{\text{Theorem 5.4.6}}, (1)) \\ \amp = [\boldv]_{B'} \amp (\knowl{./knowl/xref/th_change_of_basis_coordinates.html}{\text{Theorem 5.5.2}})\text{.} \end{align*}
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Remark 5.5.14. Getting change of basis formulas correct.

It is easy to get the various details of the change of basis formula wrong. Here is a potential way to keep things organized in your mind.
  1. We wish to relate \([T]_{B'}\) and \([T]_B\) with an equation of the form \([T]_{B'}=*[T]_B*\text{,}\) where the asterisks are to be replaced with change of basis matrices or their inverses. Think of the three matrices on the right-hand side of this equation as a sequence of three things done to coordinate vectors, reading from right to left.
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  2. \([T]_{B'}\) takes as inputs \(B'\)-coordinates of vectors, and outputs \(B'\)-coordinates. Thus the same should be true for \(*[T]_B*\text{.}\)
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  3. Since \([T]_B\) takes as inputs \(B\)-coordinates, we must first convert from \(B'\)-coordinates to \(B\)-coordinates. So we should have \([T]_{B'}=*[T]_B\underset{B'\rightarrow B}{P}\text{.}\)
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  4. Since \([T]_B\) outputs \(B\)-coordinates, we need to then convert back to \(B'\)-coordinates. Thus \([T]_{B'}=\underset{B\rightarrow B'}{P}[T]_B\underset{B'\rightarrow B}{P}\text{.}\)
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  5. If desired you may replace \(\underset{B\rightarrow B'}{P}\) with \(\underset{B'\rightarrow B}{P}^{-1}\text{.}\)
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Example 5.5.15. Matrix representations and change of basis.

The function
\begin{align*} T\colon \R^3 \amp \rightarrow \R^3\\ (x,y,z)\amp\mapsto (2x-y-z, -x+2y-z, -x-y+2z) \end{align*}
is linear.
  1. Compute \([T]_B\text{,}\) where \(B=(\bolde_1,\bolde_2,\bolde_3)\) is the the standard basis of \(\R^3\text{.}\)
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  2. Consider the nonstandard basis \(B'=((1,-1,0),(1,1,-2),(1,1,1))\) of \(\R^3\text{.}\) Compute \([T]_{B'}\) directly using Definition 5.4.1.
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  3. Now compute \([T]_{B'}\) using \([T]_B\) and the change of basis formula (5.17).
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Solution.
  1. According to Theorem 5.4.3, since \(B\) is the standard basis of \(\R^3\text{,}\) \([T]_B\) is just the standard matrix of \(T\text{.}\) Thus we can use Procedure 5.1.29 to compute
    \begin{equation*} [T]_B=\begin{amatrix}[rrr] 2\amp -1 \amp -1 \\ -1\amp 2 \amp -1 \\ -1 \amp -1 \amp 2 \end{amatrix}\text{.} \end{equation*}
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  2. We compute \([T]_{B'}\) directly using Definition 5.4.1:
    \begin{align*} [T]_{B'} \amp = \begin{bmatrix} \vert \amp \vert \amp \vert \\ [T(1,-1,0)]_{B'} \amp [T(1,1,-2)]_{B'}\amp [T(1,1,1)]_{B'}\\ \vert \amp \vert \amp \vert \end{bmatrix}\\ \amp = \begin{bmatrix} \vert \amp \vert \amp \vert \\ [(3,-3,0)]_{B'} \amp [(3,3,-6)]_{B'}\amp [(0,0,0)]_{B'}\\ \vert \amp \vert \amp \vert \end{bmatrix}\\ \amp = \begin{bmatrix} 3\amp 0 \amp 0 \\ 0\amp 3\amp 0 \\ 0 \amp 0 \amp 0\end{bmatrix}\text{.} \end{align*}
    The coordinate vector computations in the last equality were all done by inspection:
    \begin{align*} (3,-3,0) \amp =3(1,-1,0)+0(1,1,-2)+0(1,1,1)\\ (3,3,-6) \amp =0(1,-1,0)+3(1,1,-2)+0(1,1,1)\\ (0,0,0) \amp =0(1,-1,0)+0(1,1,-2)+0(1,1,1)\text{.} \end{align*}
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  3. To use the change of basis formula (5.17), we need to compute \(\underset{B'\rightarrow B}{P}\) and \(\left( \underset{B'\rightarrow B}{P}\right)^{-1}\text{.}\) Since \(B\) is the standard basis of \(\R^3\text{,}\) using Remark 5.5.8 we see that we can build \(\underset{B'\rightarrow B}{P}\) simply by placing the elements of \(B'\) in as its columns:
    \begin{equation*} \underset{B'\rightarrow B}{P}=\begin{amatrix}[rrr]1\amp 1\amp 1\\ -1\amp 1\amp 1\\ 0\amp -2\amp 1\end{amatrix}\text{.} \end{equation*}
    We then compute the inverse
    \begin{equation*} \left( \underset{B'\rightarrow B}{P}\right)^{-1}=\begin{amatrix}[rrr] \frac{1}{2} \amp -\frac{1}{2} \amp 0 \\ \frac{1}{6} \amp \frac{1}{6} \amp -\frac{1}{3} \\ \frac{1}{3} \amp \frac{1}{3} \amp \frac{1}{3} \end{amatrix}\text{.} \end{equation*}
    Lastly, we compute
    \begin{align*} [T]_{B'} \amp = \left( \underset{B'\rightarrow B}{P}\right)^{-1}\,[T]_B\, \underset{B'\rightarrow B}{P}\\ \amp =\begin{amatrix}[rrr] \frac{1}{2} \amp -\frac{1}{2} \amp 0 \\ \frac{1}{6} \amp \frac{1}{6} \amp -\frac{1}{3} \\ \frac{1}{3} \amp \frac{1}{3} \amp \frac{1}{3} \end{amatrix} \begin{amatrix}[rrr] 2\amp -1 \amp -1 \\ -1\amp 2 \amp -1 \\ -1 \amp -1 \amp 2 \end{amatrix} \begin{amatrix}[rrr]1\amp 1\amp 1\\ -1\amp 1\amp 1\\ 0\amp -2\amp 1\end{amatrix}\\ \amp = \begin{bmatrix} 3\amp 0 \amp 0 \\ 0\amp 3\amp 0 \\ 0 \amp 0 \amp 0\end{bmatrix}\text{,} \end{align*}
    as we expected.
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Consider the special case where \(T\colon \R^n\rightarrow \R^n\text{:}\) that is, when \(V=\R^n\) is a space of \(n\)-tuples. We know from Theorem 5.1.26 that \(T=T_A\) for a unique \(n\times n\) matrix \(A\text{:}\) the standard matrix of \(T\text{.}\) The standard matrix of \(T\) is useful, as it provides a convenient matrix formula for \(T\text{.}\) To compute \(A\) directly using the recipe in 5.1.26, we must compute \(T(\bolde_j)\) for each of the standard basis elements \(\bolde_j\text{.}\) For many naturally occurring transformations \(T\text{,}\) this is often not so easy to do. Theorem 5.5.13 provides an indirect method in such cases, as we now explain.
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According to Theorem 5.4.3 we have \(A=[T]_B\text{:}\) i.e., the standard matrix of \(T\) is none other than the matrix representing \(T\) with respect to the standard basis. This connection allows us to compute \(A=[T]_B\) by first computing \([T]_{B'}\) for some more convenient basis \(B'\text{,}\) and then using the change of basis formula.
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Procedure 5.5.16 is a powerful technique for computing matrix formulas for many interesting geometric linear transformations of \(\R^n\text{:}\) e.g., rotations, reflections, and orthogonal projections. Often the very definition of such transformations will suggest a more convenient nonstandard basis \(B'\text{:}\) one that reflects the geometry involved. The next example illustrates this nicely.
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Example 5.5.17. Reflection in \(\R^2\).

Let \(T\colon \R^2\rightarrow \R^2\) be reflection through the line \(\ell\) with equation \(y=2x\text{.}\) We showed in Theorem 5.1.38 that such a reflection is a linear transformation.
  1. Using the geometric definition of reflection (see Definition 5.1.37), come up with a basis \(B'=(\boldx_1,\boldx_2)\) such that \(T(\boldx_1)=\boldx_1\) and \(T(\boldx_2)=-\boldx_2\text{:}\) i.e. the reflection \(T\) “fixes” \(\boldx_1\) and “flips” \(\boldx_2\text{.}\) You might use the interactive in Figure 5.1.40 to see how you should choose \(B'\text{.}\)
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  2. Show that
    \begin{align*} [T]_{B'} \amp = \begin{amatrix}[rr] 1\amp 0 \\ 0\amp -1\end{amatrix}\text{.} \end{align*}
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  3. Use the change of basis formula (5.17) to compute \([T]_{B}\text{,}\) the standard matrix of \(T\text{.}\)
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  4. Explain how your expression for \([T]_B\) is consistent with the matrix formula given in Theorem 5.1.38.
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Solution.
  1. Geometrically, reflection through the line \(\ell\) defined by \(y=2x\) fixes any vector that points along \(\ell\) and flips any vector that points perpendicularly to \(\ell\text{.}\) The vector \(\boldx_1=(1,2)\) points along \(\ell\text{;}\) and the vector \(\boldx_2=(2,-1)\) points perpendicularly to \((1,2)\) (and hence \(\ell\)). It follows that \(B'=(\boldx_1,\boldx_2)\) is a basis satisfying \(T(\boldx_1)=\boldx_1\) and \(T(\boldx_2)=-\boldx_2\text{.}\)
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  2. We compute
    \begin{align*} [T]_{B'} \amp = \begin{bmatrix}\vert \amp \vert \\ [T(\boldx_1)]_{B'}\amp [T(\boldx_2)]_{B'}\\ \vert \amp \vert \end{bmatrix}\\ \amp = \begin{bmatrix}\vert \amp \vert \\ [\boldx_1]_{B'}\amp [-\boldx_2]_{B'}\\ \vert \amp \vert \end{bmatrix} \amp (T(\boldx_1)=\boldx_1, T(\boldx_2)=-\boldx_2)\\ \amp =\begin{amatrix}[rr] 1\amp 0\\ 0 \amp -1\end{amatrix}\text{.} \end{align*}
    The coordinate vectors in the last equality were computed by inspection:
    \begin{align*} \boldx_1 \amp =1\boldx_1+0\boldx_2\\ -\boldx_2 \amp =0\boldx_1+(-1)\boldx_2\text{.} \end{align*}
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  3. Using Remark 5.5.8, we have
    \begin{equation*} \underset{B'\rightarrow B}{P}=\begin{amatrix}[rr] 1\amp 2 \\ 2\amp -1 \end{amatrix}\text{,} \end{equation*}
    and thus
    \begin{equation*} \underset{B\rightarrow B'}{P}=\left(\underset{B'\rightarrow B}{P}\right)^{-1}=\frac{1}{5}\begin{amatrix}[rr]1\amp 2 \\2\amp -1\end{amatrix}\text{.} \end{equation*}
    We conclude that
    \begin{align*} [T]_B \amp =\underset{B'\rightarrow B}{P}\, [T]_{B'}\, \underset{B\rightarrow B'}{P}\\ \amp = \begin{amatrix}[rr] 1\amp 2 \\ 2\amp -1 \end{amatrix} \begin{amatrix}[rr] 1\amp 0\\ 0 \amp -1\end{amatrix} \left( \frac{1}{5}\begin{amatrix}[rr]1\amp 2 \\2\amp -1\end{amatrix} \right)\\ \amp = \frac{1}{5}\begin{amatrix}[rr] 1\amp 2 \\ 2\amp -1 \end{amatrix} \begin{amatrix}[rr] 1\amp 0\\ 0 \amp -1\end{amatrix} \begin{amatrix}[rr]1\amp 2 \\2\amp -1\end{amatrix} \\ \amp =\frac{1}{5}\begin{amatrix}[rr]-3\amp 4\\ 4\amp 3\end{amatrix} \text{.} \end{align*}
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  4. According to Theorem 5.1.38, we should have
    \begin{equation*} [T]_B=\begin{amatrix}[rr]\cos 2\alpha \amp \sin 2\alpha \\ \sin 2\alpha \amp -\cos 2\alpha \end{amatrix}\text{,} \end{equation*}
    where \(\alpha\) is the angle that \(\ell\) makes with the positive \(x\)-axis. Since \((1,2)\) lies along \(\ell\text{,}\) it follows by drawing the relevant right triangle that
    \begin{align*} \cos \alpha\amp =\frac{1}{\sqrt{5}} \amp \sin \alpha=\frac{2}{\sqrt{5}} \text{,} \end{align*}
    and hence that
    \begin{align*} \cos 2\alpha \amp = \cos^2\alpha-\sin^2\alpha\\ \amp = \frac{1}{5}-\frac{4}{5}=-\frac{3}{5}\\ \sin 2\alpha \amp =2\sin\alpha\cos\alpha=\frac{4}{5}\text{.} \end{align*}
    This shows
    \begin{equation*} \begin{amatrix}[rr]\cos 2\alpha \amp \sin 2\alpha \\ \sin 2\alpha \amp -\cos 2\alpha \end{amatrix}= \frac{1}{5}\begin{amatrix}[rr]-3\amp 4\\ 4\amp 3\end{amatrix} \end{equation*}
    in this case, as expected.
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Example 5.5.18. Video example: change of basis for transformations.

Figure 5.5.19. Video: change of basis for transformations
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Example 5.5.20. Video example: change of basis and reflection.

Figure 5.5.21. Video: computing reflection via change of basis
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Subsection 5.5.3 Similarity and the holy commutative tent of linear algebra

Theorem 5.5.13 supplies an algebraic answer to the question: What is the relation between two matrix representations \(A=[T]_B\) and \(A'=[T]_{B'}\text{?}\) Letting \(P=\underset{B'\rightarrow B}{P}\text{,}\) equation (5.17) becomes \(A'=P^{-1}AP\text{.}\) Matrices satisfying such a relation are said to be similar.
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Definition 5.5.22. Similar matrices.

Matrices \(A, A'\in M_{nn}\) are similar if there is an invertible matrix \(P\) such that \(A'=P^{-1}AP\text{.}\)
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So any two matrix representations of a linear transformation \(T\colon V\rightarrow V\) are similar in the technical sense of Definition 5.5.22. In fact, a converse of sorts is also true, as articulated in the theorem below.
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Proof.

The discussion above shows that if \(A=[T]_B\) and \(A'=[T]_{B'}\text{,}\) then \(A'=P^{-1}AP\text{,}\) where \(P=\underset{B'\rightarrow B}{P}\text{;}\) thus \(A\) and \(A'\) are similar in this case.
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Now assume that \(A\) and \(A'\) are similar. By definition this means there is an invertible matrix \(P\) such that \(A'=P^{-1}AP\text{.}\) Define \(T\colon \R^n\rightarrow \R^n\) as the matrix transformation \(T=T_A\text{.}\) According to Theorem 5.4.3 we have \(A=[T]_B\) where \(B\) is the standard basis of \(\R^n\text{.}\) Next, letting \(B'\) be the ordered basis whose \(j\)-th element is the \(j\)-th column of \(P\text{,}\) we have \(P=\underset{B'\rightarrow B}{P}\) (Example 5.5.6), and hence
\begin{equation*} A'=P^{-1}AP=\underset{B\rightarrow B'}{P}\, [T]_B\, \underset{B'\rightarrow B}{P}=[T]_{B'}\text{,} \end{equation*}
as desired.
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We will see in Section 6.1 that similar matrices are indeed similar algebraically speaking: i.e., they share many of the same properties. Theorem 5.5.23 provides the theoretical foundation to understand why this should be so: if \(A\) and \(A'\) are similar, then they are two matrix representations of a common linear transformation \(T\text{;}\) their many shared properties are simply inherited from the single overlying linear transformation that they both represent! This circle of ideas is neatly encompassed by Figure 5.5.24.
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Figure 5.5.24. The holy commutative tent of linear algebra. Here we have \(P=\underset{B'\rightarrow B}{P}\) and \(A'=P^{-1}AP\text{.}\)
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Perhaps a little exegesis is in order here. Think of the map \(T\colon V\rightarrow V\) as a linear transformation up in abstract heaven; and think of the two matrices \(A=[T]_B\) and \(A'=[T]_{B'}\) as two earthly shadows of \(T\text{.}\) OK, this gets at the holy bit somewhat, but why commutative? Each face of the tent is a commutative diagram, as we now explain.
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Slanted sides of the tent.

The commutativity of the two slanted sides of the tent is a consequence of Theorem 5.4.9:
\begin{align*} [T(\boldv)]_B[\boldv]_B \amp = [T(\boldv)]_B \amp [T]_{B'}[\boldv]_{B'}\amp =[T(\boldv)]_{B'}\text{.} \end{align*}
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Triangular ends of the tent.

Let \(P=\underset{B'\rightarrow B}{P}\text{,}\) so that \(P^{-1}=\underset{B\rightarrow B'}{P}\text{.}\) The commutativity of the two triangular ends of the tent are consequences of Theorem 5.5.2:
\begin{align*} P[\boldv]_{B'} \amp=[\boldv]_B \amp P^{-1}[\boldv]_B\amp=[\boldv]_{B'} \text{.} \end{align*}
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Base of tent.

Lastly the commutativity of the base of the tent is a consequence of Theorem 5.5.13:
\begin{equation*} [T]_{B'}=\underset{B\rightarrow B'}{P}[T]_B\underset{B'\rightarrow B}{P}, \end{equation*}
or equivalently,
\begin{equation*} A'=P^{-1}AP\text{.} \end{equation*}
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In summary, the holy commutative tent conveys a close connection between the three maps
\begin{equation*} \R^n\xrightarrow{A}\R^n, \R^n\xrightarrow{A'}\R^n, V\xrightarrow{T}V\text{.} \end{equation*}
Since the base of the tent is commutative, and since the maps given by \(P\) and \(P^{-1}\) are invertible, we can translate back and forth between the matrices \(A\) and \(A'\text{.}\) Furthermore, since the two slanted sides of the tent are commutative, and since the coordinate vector transformations are invertible, we can translate up and down between our two matrix representations \(A\) and \(A'\) and the overlying linear transformation \(T\text{.}\) There is one true \(T\text{!}\)
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Exercises 5.5.4 Exercises

WeBWork Exercises

1.
Consider the ordered bases \(B=((-1,5),(1,-6))\) and \(C=((-3,-2),(-1,-1))\) for the vector space \({\mathbb R}^2\text{.}\)
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a. Find the transition matrix from \(C\) to the standard ordered basis \(E=((1,0),(0,1))\text{.}\)
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\(T_C^E =\) (2 × 2 array)
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b. Find the transition matrix from \(B\) to \(E\text{.}\)
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\(T_B^E =\) (2 × 2 array)
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c. Find the transition matrix from \(E\) to \(B\text{.}\)
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\(T_E^B =\) (2 × 2 array)
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d. Find the transition matrix from \(C\) to \(B\text{.}\)
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\(T_C^B =\) (2 × 2 array)
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e. Find the coordinates of \(u= (-1,-3)\) in the ordered basis \(B\text{.}\) Note that \([u]_B = T_E^B [u]_E\text{.}\)
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\([u]_B =\) (2 × 1 array)
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f. Find the coordinates of \(v\) in the ordered basis \(B\) if the coordinate vector of \(v\) in \(C\) is \([v]_C = (-1,2)\text{.}\)
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\([v]_B =\) (2 × 1 array)
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2.
Consider the ordered bases \(B=(\left[\begin{array}{cc} 1 \amp 0\cr 0 \amp -1 \end{array}\right],\left[\begin{array}{cc} 0 \amp 0\cr 2 \amp 0 \end{array}\right])\) and \(C=(\left[\begin{array}{cc} 3 \amp 0\cr 1 \amp -3 \end{array}\right],\left[\begin{array}{cc} 2 \amp 0\cr 1 \amp -2 \end{array}\right])\) for the vector space \(V\) of lower triangular \(2\times 2\) matrices with zero trace.
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a. Find the transition matrix from \(C\) to \(B\text{.}\)
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\(T_C^B =\) (2 × 2 array)
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b. Find the coordinates of \(M\) in the ordered basis \(B\) if the coordinate vector of \(M\) in \(C\) is \([M]_C = \left[\begin{array}{c} 3\cr -3 \end{array}\right]\text{.}\)
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\([M]_B =\) (2 × 1 array)
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c. Find \(M\text{.}\)
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\(M =\) (2 × 2 array)
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3.
Let \(f : \mathbb{R}^{2} \to \mathbb{R}^{3}\) be the linear transformation defined by
\begin{equation*} f(\vec{x}) = \left[\begin{array}{cc} 3 \amp -1\cr -3 \amp -2\cr 2 \amp -4 \end{array}\right] \vec{x}. \end{equation*}
Let
\begin{equation*} \begin{array}{lcl} \mathcal{B} \amp = \amp \lbrace \left\lt -1,1\right>, \left\lt 3,-4\right> \rbrace, \\ \mathcal{C} \amp = \amp \lbrace \left\lt 1,-1,1\right>, \left\lt 1,0,1\right>, \left\lt -3,1,-2\right> \rbrace, \end{array} \end{equation*}
be bases for \(\mathbb{R}^{2}\) and \(\mathbb{R}^{3}\text{,}\) respectively. Find the matrix \(\lbrack f \rbrack_{\mathcal{B}}^{\mathcal{C}}\) for \(f\) relative to the basis \(\mathcal{B}\) in the domain and \(\mathcal{C}\) in the codomain.
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\(\lbrack f \rbrack_{\mathcal{B}}^{\mathcal{C}} =\) (3 × 2 array)
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4.
Let \(\mathcal{P}_{n}\) be the vector space of all polynomials of degree \(n\) or less in the variable \(x\text{.}\) Let \(D : \mathcal{P}_{3} \to \mathcal{P}_{2}\) be the linear transformation defined by \(D(p(x)) = p'(x)\text{.}\) That is, \(D\) is the derivative operator. Let
\begin{equation*} \begin{array}{lcl} \mathcal{B} \amp = \amp \lbrace 1,x,x^2,x^3 \rbrace, \\ \mathcal{C} \amp = \amp \lbrace 1,x,x^2 \rbrace, \end{array} \end{equation*}
be ordered bases for \(\mathcal{P}_{3}\) and \(\mathcal{P}_{2}\text{,}\) respectively. Find the matrix \(\lbrack D \rbrack_{\mathcal{B}}^{\mathcal{C}}\) for \(D\) relative to the basis \(\mathcal{B}\) in the domain and \(\mathcal{C}\) in the codomain.
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\(\lbrack D \rbrack_{\mathcal{B}}^{\mathcal{C}} =\) (3 × 4 array)
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5.
Let \(\mathcal{P}_{n}\) be the vector space of all polynomials of degree \(n\) or less in the variable \(x\text{.}\) Let \(D : \mathcal{P}_{3} \to \mathcal{P}_{2}\) be the linear transformation defined by \(D(p(x)) = p'(x)\text{.}\) That is, \(D\) is the derivative operator. Let
\begin{equation*} \begin{array}{lcl} \mathcal{B} \amp = \amp \lbrace 1, x, x^{2}, x^{3} \rbrace, \\ \mathcal{C} \amp = \amp \lbrace 2+x-x^{2}, 2+2x-x^{2}, -3-3x+2x^{2} \rbrace, \end{array} \end{equation*}
be ordered bases for \(\mathcal{P}_{3}\) and \(\mathcal{P}_{2}\text{,}\) respectively. Find the matrix \(\lbrack D \rbrack_{\mathcal{B}}^{\mathcal{C}}\) for \(D\) relative to the basis \(\mathcal{B}\) in the domain and \(\mathcal{C}\) in the codomain.
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\(\lbrack D \rbrack_{\mathcal{B}}^{\mathcal{C}} =\) (3 × 4 array)
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Change of basis matrix.

In each exercise a vector space \(V\) is given along with two ordered bases \(B\) and \(B'\text{.}\) Compute \(\underset{B\rightarrow B'}{P}\) and \(\underset{B'\rightarrow B}{P}\text{.}\)
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6.
\(V=\R^2\text{,}\) \(B=(\bolde_1, \bolde_2)\text{,}\) \(B'=\left((3,4),(1,1)\right)\)
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7.
\(V=\R^2\text{,}\) \(B=\left((1,2), (2,1)\right)\text{,}\) \(B'=\left((-1,1),(1,1)\right)\)
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8.
\(V=\R^3\text{,}\) \(B=(\bolde_1, \bolde_2, \bolde_3)\text{,}\) \(B'=\left( (1,0,0),(1,1,0),(1,1,1) \right)\)
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9.
\(V=P_2\text{,}\) \(B=(x^2, x, 1)\text{,}\) \(B'=((x-2)^2, (x-2), 1)\)
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10.

Let \(V=\R^3\text{,}\) \(B=(\bolde_1, \bolde_2, \bolde_3)\text{,}\) \(B'=\left( (1,0,0),(1,1,0),(1,1,1) \right)\text{,}\) as in Exercise 5.5.4.8.
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  1. Compute \([(1,2,3)]_{B'}\) directly using Definition 5.3.3.
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  2. Compute \([(1,2,3)]_{B'}\) using the change of basis matrix \(\underset{B\rightarrow B'}{P}\) and (5.14).
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11.

Let \(V=P_2\text{,}\) \(B=(x^2, x, 1)\text{,}\) \(B'=((x-2)^2, (x-2), 1)\text{,}\) as in Exercise 5.5.4.9.
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  1. Compute \([x^2-x+4]_{B'}\) directly using Definition 5.3.3.
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  2. Compute \([x^2-x+4]_{B'}\) using the change of basis matrix \(\underset{B\rightarrow B'}{P}\) and the change of basis formula (5.14).
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12.

Let \(B\) be the standard basis of \(\R^2\text{.}\) Find the ordered basis \(B'\) for which the change of basis matrix \(\underset{B\rightarrow B'}{P}\) is given by
\begin{equation*} \underset{B\rightarrow B'}{P}=\begin{amatrix}[rr]5\amp 1\\ -3\amp 2 \end{amatrix}\text{.} \end{equation*}
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13.

Let \(B\) be the standard basis of \(P_2\text{.}\) Find the ordered basis \(B'\) for which the change of basis matrix \(\underset{B'\rightarrow B}{P}\) is given by
\begin{equation*} \underset{B'\rightarrow B}{P}=\begin{amatrix}[rrr]1\amp 1\amp 2\\ -3\amp 2\amp 0\\ 0\amp -1\amp 1 \end{amatrix}\text{.} \end{equation*}
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14.

Suppose \(B=(\boldv_1, \boldv_2)\) and \(B'=(\boldw_1, \boldw_2)\) are two bases for the space \(V\) related by the change of basis matrix
\begin{equation*} \underset{B\rightarrow B'}{P}=\begin{amatrix}[rr]1\amp -2 \\ 3\amp 1 \end{amatrix}\text{.} \end{equation*}
  1. Let \(\boldv=-3\boldv_1+2\boldv_2\text{.}\) Compute \([\boldv]_B\) and \([\boldv]_{B'}\text{.}\)
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  2. Let \(\boldw=\boldw_1+2\boldw_2\text{.}\) Compute \([\boldw]_B\) and \([\boldw]_{B'}\text{.}\)
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15.

Let \(B\text{,}\) \(B'\text{,}\) and \(B''\) be three ordered bases of the vector space \(V\text{.}\)
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  1. Show that
    \begin{equation} \underset{B\rightarrow B''}{P}=\underset{B'\rightarrow B''}{P}\underset{B\rightarrow B'}{P}\text{.}\tag{5.18} \end{equation}
    To do so, set \(A=\underset{B'\rightarrow B''}{P}\) and \(B=\underset{B\rightarrow B'}{P}\) and show that the matrix \(AB\) satisfies the defining property of \(\underset{B\rightarrow B''}{P}\text{:}\) i.e.,
    \begin{equation*} AB[\boldv]_{B}=[\boldv]_{B''} \end{equation*}
    for all \(\boldv\in V\text{.}\)
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  2. Using (a), show that
    \begin{equation*} \underset{B'\rightarrow B''}{P}=\underset{B\rightarrow B''}{P}\underset{B'\rightarrow B}{P}\text{.} \end{equation*}
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Change of basis methods.

In each exercise a vector space \(V\) is given along with two ordered bases \(B'\) and \(B''\text{.}\)
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  1. Compute \(\underset{B'\rightarrow B''}{P}\) directly using Definition 5.5.1
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  2. Let \(B\) be the standard basis for \(V\text{.}\) Compute \(\underset{B'\rightarrow B''}{P}\) using formula (5.18) from Exercise 5.5.4.15.
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16.
\(V=\R^2\text{,}\) \(B'=\left((1,1),(1,-1)\right)\text{,}\) \(B''=\left((2,1),(-1,2)\right)\)
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17.
\(V=P_2\text{,}\) \(B'=(x^2+1, x+1, 1)\text{,}\) \(B''=(x^2+x+1, x^2+x, x^2)\)
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18.

Let \(T\colon \R^3\rightarrow \R^3\) be the linear transformation defined as \(T(x,y,z)=(x+2y+z, -y, x+7z)\text{.}\) Let \(B\) be the standard basis of \(\R^3\text{,}\) and let \(B'=\left((1,0,0), (1,1,0), (1,1,1)\right)\text{.}\)
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  1. Compute \([T]_B\text{.}\)
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  2. Compute \([T]_{B'}\) using Theorem 5.5.13.
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19.

Let \(T\colon P_1\rightarrow P_1\) be the linear transformation defined as \(T(p(x))=(x+1)p'(x)\text{.}\) Let \(B\) be the standard basis of \(P_1\text{,}\) and let \(B'=\left(2x+1, x-1\right)\text{.}\)
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  1. Compute \([T]_B\text{.}\)
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  2. Compute \([T]_{B'}\) using Theorem 5.5.13.
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20. Reflection in \(\R^2\).

Let \(\boldv=(a,b)\in \R^2\) be nonzero and define \(\ell=\Span\{\boldv\}\text{,}\) the line passing through the origin with direction vector \(\boldv\text{.}\) Let \(T\colon \R^2\rightarrow \R^2\) be reflection through \(\ell\text{.}\) (See Definition 5.1.37.) In this exercise we will use a change of basis argument to find a formula for the standard matrix of \(T\text{:}\) i.e., the matrix \(A\) satisfying \(T(\boldx)=A\boldx\) for all \(\boldx\in \R^2\text{.}\) Our answer will be expressed in terms of \(a\) and \(b\text{.}\)
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  1. Pick a basis \(B'=\{\boldv_1, \boldv_2\}\) where \(\boldv_1\) points along \(\ell\) and \(\boldv_2\) is orthogonal to \(\boldv_1\text{.}\) (Both vectors will be expressed in terms of \(a\) and \(b\text{.}\)) Compute \([T]_{B'}\text{.}\)
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  2. Let \(B\) be the standard basis of \(\R^2\text{.}\) Use Theorem 5.5.13 to compute \(A=[T]_{B}\text{.}\)
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  3. How do we know that \(A\) is the standard matrix of \(T\text{?}\)
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  4. Explain why your matrix \(A\text{,}\) expressed in terms of \(a\) and \(b\) for \(T\) agrees with the matrix formula provided in Theorem 5.1.38, which is expressed in terms of the angle \(\alpha\) that \(\ell\) makes with the \(x\)-axis.
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