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Chapter 4: Q1E (page 191)

\(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},{{\mathop{\rm b}\nolimits} _2Let }} \right\}\)and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},{{\mathop{\rm c}\nolimits} _2}} \right\}\) be bases for a vector space \(V\), and suppose \({{\mathop{\rm b}\nolimits} _1} = 6{{\mathop{\rm c}\nolimits} _1} - 2{{\mathop{\rm c}\nolimits} _2}\) and \({{\mathop{\rm b}\nolimits} _2} = 9{{\mathop{\rm c}\nolimits} _1} - 4{{\mathop{\rm c}\nolimits} _2}\).

a. Find the change-of-coordinates matrix from \(B\) to \(C\).

b. Find \({\left[ {\mathop{\rm x}\nolimits} \right]_c}\) for \({\mathop{\rm x}\nolimits} = - 3{{\mathop{\rm b}\nolimits} _1} + 2{{\mathop{\rm b}\nolimits} _2}\). Use part (a).

Short Answer

Expert verified
  1. The change-of-coordinates matrix is \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}6&9\\{ - 2}&{ - 4}\end{array}} \right]\).
  2. The \(C - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \left[ {\begin{array}{*{20}{c}}0\\{ - 2}\end{array}} \right]\).

Step by step solution

01

Change-of-coordinate matrix

Theorem 15states that 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 the bases of a vector space \(V\). Then, there is a unique \(n \times n\) matrix \(\mathop P\limits_{C \leftarrow B} \) such that \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \mathop P\limits_{C \leftarrow B} {\left[ {\mathop{\rm x}\nolimits} \right]_B}\).

The columns of \(\mathop P\limits_{C \leftarrow B} \) are the \(C - \)coordinate vectors of the vectors in the basis \(B\). Thus, \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]}_C}}&{{{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]}_C}}& \cdots &{{{\left[ {{{\mathop{\rm b}\nolimits} _n}} \right]}_C}}\end{array}} \right]\).

02

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

a)

Suppose \({{\mathop{\rm b}\nolimits} _1} = 6{{\mathop{\rm c}\nolimits} _1} - 2{{\mathop{\rm c}\nolimits} _2}\) and \({{\mathop{\rm b}\nolimits} _2} = 9{{\mathop{\rm c}\nolimits} _1} - 4{{\mathop{\rm c}\nolimits} _2}\). Therefore, \({\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]_C} = \left[ {\begin{array}{*{20}{c}}6\\{ - 2}\end{array}} \right],{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]_C} = \left[ {\begin{array}{*{20}{c}}9\\{ - 4}\end{array}} \right]\).

\(\begin{aligned} \mathop P\limits_{C \leftarrow B} &= \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]}_C}}&{{{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]}_C}}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}6&9\\{ - 2}&{ - 4}\end{array}} \right]\end{aligned}\)

Thus, the change-of-coordinates matrix from \(B\) to \(C\) is \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}6&9\\{ - 2}&{ - 4}\end{array}} \right]\).

03

Determine \({\left[ {\mathop{\rm x}\nolimits}  \right]_c}\) for \({\mathop{\rm x}\nolimits}  =  - 3{{\mathop{\rm b}\nolimits} _1} + 2{{\mathop{\rm b}\nolimits} _2}\)

b)

Suppose \({\mathop{\rm x}\nolimits} = - 3{{\mathop{\rm b}\nolimits} _1} + 2{{\mathop{\rm b}\nolimits} _2}\), then \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\2\end{array}} \right]\).

\(\begin{aligned} {\left[ {\mathop{\rm x}\nolimits} \right]_C} &= \mathop P\limits_{C \leftarrow B} {\left[ {\mathop{\rm x}\nolimits} \right]_B}\\ &= \left[ {\begin{array}{*{20}{c}}6&9\\{ - 2}&{ - 4}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 3}\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 18 + 18}\\{6 - 8}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}0\\{ - 2}\end{array}} \right]\end{aligned}\)

Therefore, the \(C - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \left[ {\begin{array}{*{20}{c}}0\\{ - 2}\end{array}} \right]\).

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Most popular questions from this chapter

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)

If A is a \({\bf{6}} \times {\bf{8}}\) matrix, what is the smallest possible dimension of Null A?

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\2\\1\\0\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}{ - 8}&5&{ - 2}&0\\{ - 5}&2&1&{ - 2}\\{10}&{ - 8}&6&{ - 3}\\3&{ - 2}&1&0\end{array}} \right)\)

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\(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}},{p_{\bf{3}}}} \right\}\).

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