Change of Basis

Section 5.6 Change of Basis

Every basis for a vector space gives a different angle on that space—we get a different coordinate system for each basis. Since any finite-dimensional vector space has many bases, in this section we explain how to move between bases.

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Subsection 5.6.1 The Change-of-Basis Matrix

We will first describe a situation in which this technique will be useful. Consider the following two bases for \(\rr^2\text{:}\) \(\mcb = \{\bfv_1, \bfv_2 \}\) and \(\mcc = \{ \bfw_1, \bfw_2 \}\text{,}\) where

\begin{equation*} \bfv_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \hspace{6pt} \bfv_2 = \begin{bmatrix} 0 \\ 2 \end{bmatrix}, \hspace{6pt} \bfw_1 = \begin{bmatrix} -2 \\ 1 \end{bmatrix}, \hspace{6pt} \bfw_2 = \begin{bmatrix} -1 \\ 3 \end{bmatrix}\text{.} \end{equation*}

We can verify that \(\mcb\) and \(\mcc\) are bases for \(\rr^2\) since they are both linearly independent sets of two vectors in a two-dimensional space.

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If we have a vector \(\bfv \in \rr^2\text{,}\) it is straightforward to calculate both \([\bfv]_{\mcb}\) and \([\bfv]_{\mcc}\text{.}\) The question for us is this: How do these two coordinate vectors relate to each other? Specifically, how might we calculate one coordinate vector from the other one?

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It turns out that we already have the necessary machinery for this calculation. We summarize the process in the following proposition.

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Proposition 5.6.1.

Let \(\mcb\) and \(\mcc\) be two bases for a finite-dimensional vector space \(V\text{.}\) Then, for any \(\bfv \in V\text{,}\) we have

\begin{equation*} [\bfv]_{\mcc} = [I]_{\mcb,\mcc} [\bfv]_{\mcb}\text{,} \end{equation*}

where \(I: V\to V\) is the identity transformation.

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Proof.

This is a simple application of Proposition 5.5.9 to the identity transformation \(I\text{:}\)

\begin{equation*} [\bfv]_{\mcc} = [I(\bfv)]_{\mcc} = [I]_{\mcb,\mcc} [\bfv]_{\mcb}\text{.} \end{equation*}

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Definition 5.6.2.

If \(\mcb\) and \(\mcc\) are two bases for a finite-dimensional vector space \(V\text{,}\) then the matrix \(P_{\mcb,\mcc} = [I]_{\mcb,\mcc}\) is called the change-of-basis matrix from \(\mcb\) to \(\mcc\text{.}\)

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Example 5.6.3.

We will continue the example begun earlier in this section. If \(\mcb = \{\bfv_1, \bfv_2 \}\) and \(\mcc = \{ \bfw_1, \bfw_2 \}\text{,}\) then we can calculate \(P_{\mcb, \mcc}\) by determining the coordinate vectors \([\bfv_1]_{\mcc}\) and \([\bfv_2]_{\mcc}\text{.}\) We need only to row-reduce two matrices:

\begin{equation*} \begin{bmatrix} -2 \amp -1 \amp 1 \\ 1 \amp 3 \amp -1 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp -2/5 \\ 0 \amp 1 \amp -1/5 \end{bmatrix}, \hspace{6pt} \begin{bmatrix} -2 \amp -1 \amp 0 \\ 1 \amp 3 \amp 2 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp -2/5 \\ 0 \amp 1 \amp 4/5 \end{bmatrix}\text{.} \end{equation*}

From these calculations, we can see how to write the \(\mcb\)-basis vectors in terms of the vectors in \(\mcc\text{,}\) and these form the columns of our change-of-basis matrix:

\begin{equation*} P_{\mcb,\mcc} = \begin{bmatrix} -2/5 \amp -2/5 \\ -1/5 \amp 4/5 \end{bmatrix}\text{.} \end{equation*}

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We now consider a vector \(\bfv = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\) in \(\rr^2\text{.}\) We can calculate \([\bfv]_{\mcb}\) in this way:

\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 2 \\ -1 \amp 2 \amp 3 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp 2 \\ 0 \amp 1 \amp 5/2 \end{bmatrix} \text{.} \end{equation*}

Now that we have \([\bfv]_{\mcb} = \begin{bmatrix} 2 \\ 5/2 \end{bmatrix}\text{,}\) we can use the change-of-basis matrix to find \([\bfv]_{\mcc}\text{:}\)

\begin{equation*} [\bfv]_{\mcc} = P_{\mcb,\mcc} [\bfv]_{\mcb} = \begin{bmatrix} -2/5 \amp -2/5 \\ -1/5 \amp 4/5 \end{bmatrix} \begin{bmatrix} 2 \\ 5/2 \end{bmatrix} = \begin{bmatrix} -9/5 \\ 8/5 \end{bmatrix} \text{.} \end{equation*}

We can verify that this is the correct \(\mcc\)-coordinate vector for \(\bfv\) by calculating it directly:

\begin{equation*} \begin{bmatrix} -2 \amp -1 \amp 2 \\ 1 \amp 3 \amp 3 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp -9/5 \\ 0 \amp 1 \amp 8/5 \end{bmatrix}\text{.} \end{equation*}

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Example 5.6.4.

We consider the vector space \(P_2\text{.}\) Let \(\mcc\) be the standard basis for \(P_2\) and let \(\mcb\) be the set \(\{p_1, p_2, p_3 \}\text{,}\) where

\begin{equation*} p_1 = 2+t, \hspace{6pt} p_2 = -1-t+2t^2, \hspace{6pt} p_3 = -2t+3t^2\text{.} \end{equation*}

In order to find the change-of-basis matrix, we need to write the coordinate vectors of the basis vectors of \(\mcb\) with respect to \(\mcc\text{.}\) But since \(\mcc\) is the standard basis of \(P_2\text{,}\) this is an easy task to complete. Here is the change-of-basis matrix:

\begin{equation*} P_{\mcb, \mcc} = \begin{bmatrix} 2 \amp -1 \amp 0 \\ 1 \amp -1 \amp -2 \\ 0 \amp 2 \amp 3 \end{bmatrix}\text{.} \end{equation*}

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What we saw in Example 5.6.4 is an indication that some change-of-basis matrices are easier to calculate than others. In particular, when the standard basis is the target (not the source) basis, the result is almost immediate.

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Lemma 5.6.5.

Let \(\mce\) be the standard basis of \(\ff^n\text{,}\) and let \(\mcb\) be any other basis of \(\ff^n\text{.}\) Then the columns of \(P_{\mcb, \mce}\) are the vectors of \(\mcb\text{,}\) in order.

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Proof.

If \(\mcb = \{\bfv_1, \ldots, \bfv_n \}\text{,}\) then column \(j\) of \(P_{\mcb, \mce}\) is \([\bfv_j]_{\mce}\text{.}\) But since \(\mce\) is the standard basis, then \([\bfv_j]_{\mce} = \bfv_j\text{.}\)

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The next lemma also shows that the change-of-basis matrices from one basis to another and back again have the inverse relationship we might expect.

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Lemma 5.6.6.

Let \(\mcb\) and \(\mcc\) be two bases for a finite-dimensional vector space \(V\text{.}\) Then the relationship between the two change-of-basis matrices is

\begin{equation*} P_{\mcc, \mcb} = \left( P_{\mcb, \mcc} \right)^{-1}\text{.} \end{equation*}

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Proof.

Since \(P_{\mcb,\mcc} = [I]_{\mcb,\mcc}\text{,}\) by Theorem 5.5.15 we have

\begin{equation*} P_{\mcb,\mcc}[I]_{\mcc,\mcb} = [I]_{\mcb,\mcc}[I]_{\mcc,\mcb} = [I]_{\mcc,\mcc} = I_n\text{.} \end{equation*}

Since \(P_{\mcb,\mcc}\) is square, this proves that \([I]_{\mcc,\mcb} = (P_{\mcb,\mcc})^{-1}\text{.}\)

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Example 5.6.7.

We consider two bases for \(\ff_5^3\text{:}\) the standard basis \(\mce\) and \(\mcb = \{\bfv_1, \bfv_2, \bfv_3 \}\text{,}\) where

\begin{equation*} \bfv_1 = \begin{bmatrix} 4 \\ 2 \\ 2 \end{bmatrix}, \hspace{6pt} \bfv_2 = \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}, \hspace{6pt} \bfv_3 = \begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix}\text{.} \end{equation*}

(The reader should verify that \(\mcb\) is a basis for \(\ff_5^3\text{.}\))

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Lemma 5.6.5 tells us that the change-of-basis matrix \(P_{\mcb,\mce}\) is easy to write down:

\begin{equation*} P_{\mcb, \mce} = \begin{bmatrix} 4 \amp 3 \amp 0 \\ 2 \amp 1 \amp 4 \\ 2 \amp 0 \amp 1 \end{bmatrix} \text{.} \end{equation*}

Then Lemma 5.6.6 says that \(P_{\mce,\mcb} = (P_{\mcb,\mce})^{-1}\text{,}\) so we can find that matrix without too much difficulty as well:

\begin{equation*} P_{\mce,\mcb} = \begin{bmatrix} 3 \amp 1 \amp 1 \\ 3 \amp 2 \amp 2 \\ 4 \amp 3 \amp 4 \end{bmatrix}\text{.} \end{equation*}

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The final results of this section deal with linear transformations. This theorem relates the coordinate matrix for a linear transformation to the situation in which we want to change bases in the domain and codomain.

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Theorem 5.6.8.

Let \(V\) and \(W\) be finite-dimensional vector spaces, and let \(T \in L(V,W)\text{.}\) Additionally, let \(\mcb_1\) and \(\mcb_2\) be bases for \(V\text{,}\) and let \(\mcc_1\) and \(\mcc_2\) be bases for \(W\text{.}\) Then

\begin{equation*} [T]_{\mcb_2,\mcc_2} = [I]_{\mcc_1,\mcc_2}[T]_{\mcb_1,\mcc_1}[I]_{\mcb_2,\mcb_1}\text{.} \end{equation*}

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Proof.

We will use Theorem 5.5.15:

\begin{equation*} [T]_{\mcb_2,\mcc_2} = [ITI]_{\mcb_2,\mcc_2} = [I]_{\mcc_1,\mcc_2}[T]_{\mcb_1,\mcc_1}[I]_{\mcb_2,\mcb_1}\text{.} \end{equation*}

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The most important (and most common) use of this theorem happens when \(V = W\text{,}\) \(\mcb_1 = \mcc_1\text{,}\) and \(\mcb_2 = \mcc_2\text{.}\)

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Corollary 5.6.9.

Let \(\mcb\) and \(\mcc\) be bases for a finite-dimensional vector space \(V\text{,}\) and let \(T \in L(V)\text{.}\) Then

\begin{equation*} [T]_{\mcc} = (P_{\mcc,\mcb})^{-1}[T]_{\mcb}P_{\mcc,\mcb}\text{.} \end{equation*}

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Proof.

This result is due to Theorem 5.6.8 and Lemma 5.6.6.

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We will end this section with an example that takes advantage of Corollary 5.6.9.

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Example 5.6.10.

We consider the linear transformation \(T:\rr^2 \to \rr^2\) which is reflection across the line \(y=\tfrac{1}{2}x\text{.}\) While the action of \(T\) is not impossible to write down in the usual coordinate system, it is even easier using the alternate basis \(\mcb = \{\bfv_1, \bfv_2 \}\text{,}\) where

\begin{equation*} \bfv_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \hspace{12pt} \bfv_2 = \begin{bmatrix} -1 \\ 2 \end{bmatrix}\text{.} \end{equation*}

To see why this linear transformation is easier to describe in the \(\mcb\)-coordinates, we recall how easy reflection across the \(y\)-axis is to describe relative to the standard basis—simply negate the first coordinate! The \(\mcb\)-basis vectors in this case lie on the axis of reflection and along the line perpendicular to that axis.

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We note that \(T(\bfv_1) = \bfv_1\) and that \(T(\bfv_2) = -\bfv_2\text{.}\) This shows that the coordinate matrix of \(T\) with respect to \(\mcb\) is

\begin{equation*} [T]_{\mcb} = \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix}\text{.} \end{equation*}

(Writing the action of \(T\) this way makes it especially easy to see that performing this transformation twice puts us back where we started.) We will use Corollary 5.6.9 to calculate the matrix for \(T\) relative to the standard basis. That is, we wish to calculate \([T]_{\mce}\text{.}\)

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We first note that the matrix \(P_{\mcb,\mce}\) is, according to Lemma 5.6.5,

\begin{equation*} P_{\mcb, \mce} = \begin{bmatrix} 2 \amp -1 \\ 1 \amp 2 \end{bmatrix} \text{.} \end{equation*}

Then, with the help of Lemma 5.6.6, we have

\begin{equation*} P_{\mce, \mcb} = (P_{\mcb,\mce})^{-1} = \begin{bmatrix} 2/5 \amp 1/5 \\ -1/5 \amp 2/5 \end{bmatrix}\text{.} \end{equation*}

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We can put these together to find \([T]_{\mce}\text{:}\)

\begin{equation*} [T]_{\mce} = P_{\mcb,\mce}[T]_{\mcb}P_{\mce,\mcb} = \begin{bmatrix} 2 \amp -1 \\ 1 \amp 2 \end{bmatrix} \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix} \begin{bmatrix} 2/5 \amp 1/5 \\ -1/5 \amp 2/5 \end{bmatrix} = \begin{bmatrix} 3/5 \amp 4/5 \\ 4/5 \amp -3/5 \end{bmatrix}\text{.} \end{equation*}

The action of the transformation, as written in the final line here, is perhaps better understood in words rather than symbols. To reflect across the line \(y=\frac{1}{2}x\text{,}\) first shift from the standard coordinates to the alternate \(\mcb\)-coordinates. (This is accomplished by \(P_{\mce, \mcb}\text{.}\)) In this new coordinate system, the action of \(T\) is easily described. (Thus, \([T]_{\mcb}\text{.}\)) After that action is carried out, then we shift back to the standard coordinate system. (That is the work of \(P_{\mcb, \mce}\text{.}\)) From start to finish, this gives us a matrix which carries out the action of \(T\) relative to \(\mce\text{.}\)

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Reading Questions 5.6.2 Reading Questions

1.

Let \(\bfv_1 = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\text{,}\) \(\bfv_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\text{,}\) \(\bfw_1 = \begin{bmatrix} 2 \\ -1 \end{bmatrix}\text{,}\) and \(\bfw_2 = \begin{bmatrix} 4 \\ 3 \end{bmatrix}\text{.}\) Consider the bases \(\mcb = \{\bfv_1, \bfv_2 \}\) and \(\mcc = \{ \bfw_1, \bfw_2 \}\) of \(\rr^2\text{.}\)

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Find the change-of-basis matrix \(P_{\mcb, \mcc}\text{.}\)

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2.

Using the definitions of the vectors and bases from the previous reading question, find \(P_{\mcc, \mcb}\text{.}\)

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Exercises 5.6.3 Exercises

1.

Let \(\mcb = \{ \bfv_1, \bfv_2 \}\) and \(\mcc = \{ \bfw_1, \bfw_2 \}\) be bases of \(\rr^2\text{,}\) where

\begin{equation*} \bfv_1 = \begin{bmatrix} 5 \\ 2 \end{bmatrix}, \hspace{6pt} \bfv_2 = \begin{bmatrix} -1 \\ -1 \end{bmatrix}, \hspace{6pt} \bfw_1 = \begin{bmatrix} 3 \\ -3 \end{bmatrix}, \hspace{6pt} \bfw_2 = \begin{bmatrix} -5 \\ -3 \end{bmatrix}\text{.} \end{equation*}

Find the change-of-basis matrices \(P_{\mcb, \mcc}\) and \(P_{\mcc, \mcb}\text{.}\)

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Answer.

Here are the change-of-basis matrices:

\begin{equation*} P_{\mcb,\mcc} = \begin{bmatrix} 5/24 \amp 1/12 \\ -7/8 \amp 1/4 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} P_{\mcc,\mcb} = \begin{bmatrix} 2 \amp -2/3 \\ 7 \amp 5/3 \end{bmatrix}\text{.} \end{equation*}

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2.

Let \(\mcb = \{ \bfv_1, \bfv_2, \bfv_3 \}\) and \(\mcc = \{ \bfw_1, \bfw_2, \bfw_3 \}\) be bases of \(\ff_5^3\text{,}\) where

\begin{align*} \bfv_1 \amp = \begin{bmatrix} 4 \\ 3 \\ 3 \end{bmatrix}, \hspace{6pt} \bfv_2 = \begin{bmatrix} 2 \\ 2 \\ 0 \end{bmatrix}, \hspace{6pt} \bfv_3 = \begin{bmatrix} 3 \\ 1 \\ 3 \end{bmatrix}\\ \bfw_1 \amp = \begin{bmatrix} 4 \\ 2 \\ 4 \end{bmatrix}, \hspace{6pt} \bfw_2 = \begin{bmatrix} 4 \\ 1 \\ 4 \end{bmatrix}, \hspace{6pt} \bfw_3 = \begin{bmatrix} 4 \\ 2 \\ 2 \end{bmatrix} \text{.} \end{align*}

Find the change-of-basis matrices \(P_{\mcb, \mcc}\) and \(P_{\mcc, \mcb}\text{.}\)

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3.

Consider the basis \(\mcb = \{\bfv_1, \bfv_2 \}\) of \(\ff_3^2\) where

\begin{equation*} \bfv_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \hspace{6pt} \bfv_2 = \begin{bmatrix} 0 \\ 2 \end{bmatrix}\text{.} \end{equation*}

If \(\mce\) is the standard basis for \(\ff_3^2\text{,}\) find \(P_{\mcb,\mce}\) and \(P_{\mce, \mcb}\text{.}\)
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Use your work in part (a) to find \([\bfv]_{\mcb}\) if \(\bfv = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\text{.}\)
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Answer.

Here are the change-of-basis matrices:

\begin{equation*} P_{\mcb,\mce} = \begin{bmatrix} 1 \amp 0 \\ 2 \amp 2 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} P_{\mce,\mcb} = \begin{bmatrix} 1 \amp 0 \\ 2 \amp 2 \end{bmatrix}\text{.} \end{equation*}

(Yes, they are the same!)

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We find that

\begin{equation*} [\bfv]_{\mcb} = P_{\mce,\mcb}[\bfv]_{\mce} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\text{.} \end{equation*}

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4.

Let \(\mcb = \{ p_1, p_2, p_3 \}\) be a basis for \(P_2\text{,}\) where

\begin{equation*} p_1 = 3t-5t^2, \hspace{6pt} p_2 = 3-3t+5t^2, \hspace{6pt} p_3 = -1-t-t^2\text{.} \end{equation*}

If \(\mce\) is the standard basis for \(P_2\text{,}\) find \(P_{\mcb,\mce}\) and \(P_{\mce, \mcb}\text{.}\)
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Use your work in part (a) to find \([p]_{\mcb}\) if \(p = -3 + \tfrac{1}{2}t + \tfrac{3}{2}t^2\text{.}\)
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If \(T:P_2 \to P_2\) is the linear transformation that takes the derivative, find \([T]_{\mcb}\text{.}\)
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5.

Let \(\mcb = \{ p_1, p_2, p_3 \}\) be a basis for \(P_2\text{,}\) where

\begin{equation*} p_1 = 1+5t-5t^2, \hspace{6pt} p_2 = -2+t+t^2, \hspace{6pt} p_3 = 5-t-4t^2\text{.} \end{equation*}

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If \(\mce\) is the standard basis for \(P_2\text{,}\) find \(P_{\mcb,\mce}\) and \(P_{\mce, \mcb}\text{.}\)
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If \(T:P_2 \to P_2\) is the linear transformation

\begin{equation*} T(p) = p(2)t + p(1)t^2\text{,} \end{equation*}

find \([T]_{\mcb}\text{.}\)

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6.

Let \(\mcb = \{A_1, A_2, A_3, A_4 \}\) be a basis for \(M_2(\ff_5)\text{,}\) where

\begin{equation*} A_1 = \begin{bmatrix} 4 \amp 0 \\ 4 \amp 4 \end{bmatrix}, \hspace{3pt} A_2 = \begin{bmatrix} 1 \amp 1 \\ 1 \amp 3 \end{bmatrix}, \hspace{3pt} A_3 = \begin{bmatrix} 0 \amp 2 \\ 0 \amp 1 \end{bmatrix}, \hspace{3pt} A_4 = \begin{bmatrix} 3 \amp 1 \\ 1 \amp 1 \end{bmatrix}\text{.} \end{equation*}

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If \(\mce\) is the standard basis for \(M_2(\ff_5)\text{,}\) find \(P_{\mcb,\mce}\) and \(P_{\mce, \mcb}\text{.}\)
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If \(T:M_2(\ff_5) \to M_2(\ff_5)\) is the linear transformation

\begin{equation*} T \left( \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \right) = \begin{bmatrix} 2a \amp 4b+c \\ a+d \amp 3b+d \end{bmatrix}\text{,} \end{equation*}

find \([T]_{\mcb}\text{.}\)

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7.

Consider the linear transformation \(T:\rr^2 \to \rr^2\) which is projection onto the line \(y=-x\text{.}\)

Propose a basis \(\mcb\) for \(\rr^2\) where \([T]_{\mcb}\) will be easy to determine.
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Find \([T]_{\mcb}\text{.}\)
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If \(\mce\) is the standard basis for \(\rr^2\text{,}\) find \(P_{\mcb, \mce}\) and \(P_{\mce, \mcb}\text{.}\)
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Using your work in previous parts of this problem, find \([T]_{\mce}\text{.}\)
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8.

Consider the linear transformation \(T:\rr^2 \to \rr^2\) which is reflection across the line \(y=-\tfrac{1}{4}x\text{.}\)

Propose a basis \(\mcb\) for \(\rr^2\) where \([T]_{\mcb}\) will be easy to determine.
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Find \([T]_{\mcb}\text{.}\)
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If \(\mce\) is the standard basis for \(\rr^2\text{,}\) find \(P_{\mcb, \mce}\) and \(P_{\mce, \mcb}\text{.}\)
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Using your work in previous parts of this problem, find \([T]_{\mce}\text{.}\)
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Answer.

We will use the basis \(\mcb = \{\bfv_1,\bfv_2\}\text{,}\) where

\begin{equation*} \bfv_1 = \begin{bmatrix} 4 \\ -1 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} \bfv_2 = \begin{bmatrix} 1 \\ 4 \end{bmatrix}\text{.} \end{equation*}

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\(\displaystyle [T]_{\mcb} = \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix}\)
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We find that

\begin{equation*} P_{\mcb,\mce} = \begin{bmatrix} 4 \amp 1 \\ -1 \amp 4 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} P_{\mce,\mcb} = \begin{bmatrix} 4/17 \amp -1/17 \\ 1/17 \amp 4/17 \end{bmatrix}\text{.} \end{equation*}

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\(\displaystyle [T]_{\mce} = \begin{bmatrix} 15/17 \amp -8/17 \\ -8/17 \amp -15/17 \end{bmatrix}\)
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Writing Exercises

9.

Let \(V\) be an \(n\)-dimensional vector space over \(\ff\text{,}\) and let \(\mcb\) and \(\mcc\) be two bases for \(V\text{.}\) Prove that the columns of the matrix \(P_{\mcb,\mcc}\) are linearly independent.

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10.

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Solution.

The \(i\)th column of \(P_{\mcb,\mcc}\) is \([\bfv_i]_{\mcc}\text{.}\) Or, stated differently, if \(C_{\mcc}:V \to \ff^n\) is the coordinate mapping, then the \(i\)th column of \(P_{\mcb,\mcc}\) is \(C_{\mcc}(\bfv_i)\text{.}\)

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We know (by Theorem 5.5.3) that the coordinate mapping is an isomorphism. Since \(\mcb\) is a basis for \(V\text{,}\) \(\mcb\) is a spanning set for \(V\text{.}\) But then the set containing the columns of \(P_{\mcb,\mcc}\) is \(C_{\mcc}(\mcb)\text{,}\) and since \(\mcb\) spans \(V\) we know that \(C_{\mcc}(\mcb)\) will span \(\ff^n\text{.}\)

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11.

Let \(A \in M_n(\ff)\) be invertible. Prove that there exist bases \(\mcb\) and \(\mcc\) for \(\ff^n\) such that \(A = P_{\mcb,\mcc}\text{.}\)

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