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Chapter 4: Q9E (page 191)

In Exercises 7-10, let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},{{\mathop{\rm b}\nolimits} _2}} \right\}\) and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},{{\mathop{\rm c}\nolimits} _2}} \right\}\) be bases for \({\mathbb{R}^2}\). In each exercise, find the change-of-coordinates matrix from \(B\) to \(C\) and the change-of-coordinates matrix from \(C\) to \(B\).

9. \({{\mathop{\rm b}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}{ - 6}\\{ - 1}\end{array}} \right),{{\mathop{\rm b}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}2\\0\end{array}} \right),{{\mathop{\rm c}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right),{{\mathop{\rm c}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}6\\{ - 2}\end{array}} \right)\).

Short Answer

Expert verified

The change-of-coordinates matrix from \(B\) to \(C\) is \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}9&{ - 2}\\{ - 4}&1\end{array}} \right)\). The change-of-coordinates matrix from \(C\) to \(B\) is \(\mathop P\limits_{B \leftarrow C} = \left( {\begin{array}{*{20}{c}}1&2\\4&9\end{array}} \right)\).

Step by step solution

01

Change-of-coordinate matrix

Let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\) and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},...,{{\mathop{\rm c}\nolimits} _n}} \right\}\) be bases of a vector space \(V\). Then, according toTheorem 15,there is a unique \(n \times n\) matrix \(\mathop P\limits_{C \leftarrow B} \) such that

\({\left( {\mathop{\rm x}\nolimits} \right)_C} = \mathop P\limits_{C \leftarrow B} {\left( {\mathop{\rm x}\nolimits} \right)_B}\). 鈥�(1)

The columns of \(\mathop P\limits_{C \leftarrow B} \) are the \(C - \)coordinate vectors of the vectors in the ba\(\frac{1}{2}\)sis \(B\). That is, \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}{{{\left( {{{\mathop{\rm b}\nolimits} _1}} \right)}_C}}&{{{\left( {{{\mathop{\rm b}\nolimits} _2}} \right)}_C}}& \cdots &{{{\left( {{{\mathop{\rm b}\nolimits} _n}} \right)}_C}}\end{array}} \right)\).

02

Determine the change-of-coordinates from \(B\) to \(C\)

Write the augmented matrix as shown below:

Perform an elementary row operation to produce a row-reduced echelon form of the matrix.

At row 1, multiply row 1 by .

\( \sim \left( {\begin{array}{*{20}{c}}1&3&{ - 3}&1\\0&{ - 3}&{12}&{ - 9}\end{array}} \right)\)

At row 2, add rows 2 and 1.

\( \sim \left( {\begin{array}{*{20}{c}}1&3&{ - 3}&1\\0&1&{ - 4}&1\end{array}} \right)\)

At row 1, multiply row 2 by 3 and subtract it from row 1.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&9&{ - 2}\\0&1&{ - 4}&1\end{array}} \right)\)

Therefore, \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}9&{ - 2}\\{ - 4}&1\end{array}} \right)\).

Thus, the change-of-coordinates matrix from \(B\) to \(C\) is \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}9&{ - 2}\\{ - 4}&1\end{array}} \right)\).

03

Determine the change-of-coordinates from \(C\) to \(B\)

It is known that \({\left( {\mathop P\limits_{C \leftarrow B} } \right)^{ - 1}}\) is the matrix that converts \(C - \)coordinates into \(B - \)coordinates. That is, \({\left( {\mathop P\limits_{C \leftarrow B} } \right)^{ - 1}} = \mathop P\limits_{B \leftarrow C} \).

\(\begin{aligned} \mathop P\limits_{B \leftarrow C} &= {\left( {\mathop P\limits_{C \leftarrow B} } \right)^{ - 1}}\\ &= {\left( {\begin{array}{*{20}{c}}9&{ - 2}\\{ - 4}&1\end{array}} \right)^{ - 1}}\\ &= \frac{1}{1}\left( {\begin{array}{*{20}{c}}1&2\\4&9\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&2\\4&9\end{array}} \right)\end{aligned}\)

Thus, the change-of-coordinates matrix from \(C\) to \(B\) is \(\mathop P\limits_{B \leftarrow C} = \left( {\begin{array}{*{20}{c}}1&2\\4&9\end{array}} \right)\).

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Most popular questions from this chapter

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

If the null space of A \({\bf{7}} \times {\bf{6}}\) matrix A is 4-dimensional, what is the dimension of the column space of A?

Consider the polynomials \({{\bf{p}}_{\bf{1}}}\left( t \right) = {\bf{1}} + t\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = {\bf{1}} - t\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{4}}\), \({{\bf{p}}_{\bf{4}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), and \({{\bf{p}}_{\bf{5}}}\left( t \right) = {\bf{1}} + {\bf{2}}t + {t^{\bf{2}}}\), and let H be the subspace of \({P_{\bf{5}}}\) spanned by the set \(S = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}},\,{{\bf{p}}_{\bf{4}}},\,{{\bf{p}}_{\bf{5}}}} \right\}\). Use the method described in the proof of the Spanning Set Theorem (Section 4.3) to produce a basis for H. (Explain how to select appropriate members of S.)

Consider the polynomials , and \({p_{\bf{3}}}\left( t \right) = {\bf{2}}\) \({p_{\bf{1}}}\left( t \right) = {\bf{1}} + t,{p_{\bf{2}}}\left( t \right) = {\bf{1}} - t\)(for all t). By inspection, write a linear dependence relation among \({p_{\bf{1}}},{p_{\bf{2}}},\) and \({p_{\bf{3}}}\). Then find a basis for Span\(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}},{p_{\bf{3}}}} \right\}\).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)
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