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Chapter 4: Q2E (page 191)

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 a vector space \(V\), and suppose \({{\mathop{\rm b}\nolimits} _1} = - {{\mathop{\rm c}\nolimits} _1} + 4{{\mathop{\rm c}\nolimits} _2}\) and \({{\mathop{\rm b}\nolimits} _2} = 5{{\mathop{\rm c}\nolimits} _1} - 3{{\mathop{\rm c}\nolimits} _2}\).

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

b. Find \({\left[ {\mathop{\rm x}\nolimits} \right]_c}\) for \({\mathop{\rm x}\nolimits} = 5{{\mathop{\rm b}\nolimits} _1} + 3{{\mathop{\rm b}\nolimits} _2}\). Use part (a).

Short Answer

Expert verified
  1. The change-of-coordinates matrix is \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}{ - 1}&5\\4&{ - 3}\end{array}} \right]\).
  2. The \(C - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \left[ {\begin{array}{*{20}{c}}{10}\\{11}\end{array}} \right]\).

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 \({{\mathop{\rm b}\nolimits} _1} = - {{\mathop{\rm c}\nolimits} _1} + 4{{\mathop{\rm c}\nolimits} _2}\) and \({{\mathop{\rm b}\nolimits} _2} = 5{{\mathop{\rm c}\nolimits} _1} - 3{{\mathop{\rm c}\nolimits} _2}\). Therefore, \({\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]_C} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right],{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]_C} = \left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]\).

\(\begin{aligned} \mathop P\limits_{C \leftarrow B} &= \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\mathop{\rm b}\nolimits} _1}} \right]}_C}}&{{{\left[ {{{\mathop{\rm b}\nolimits} _2}} \right]}_C}}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}&5\\4&{ - 3}\end{array}} \right]\end{aligned}\)

Thus, the change-of-coordinates matrix from \(B\) to \(C\) is \(\mathop P\limits_{C \leftarrow B} = \left[ {\begin{array}{*{20}{c}}{ - 1}&5\\4&{ - 3}\end{array}} \right]\).

03

Determine \({\left[ {\mathop{\rm x}\nolimits}  \right]_c}\) for \({\mathop{\rm x}\nolimits}  = 5{{\mathop{\rm b}\nolimits} _1} + 3{{\mathop{\rm b}\nolimits} _2}\)

b)

Suppose \({\mathop{\rm x}\nolimits} = 5{{\mathop{\rm b}\nolimits} _1} + 3{{\mathop{\rm b}\nolimits} _2}\), then \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}5\\3\end{array}} \right]\).

\(\begin{aligned} {\left[ {\mathop{\rm x}\nolimits} \right]_C} &= \mathop P\limits_{C \leftarrow B} {\left[ {\mathop{\rm x}\nolimits} \right]_B}\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}&5\\4&{ - 3}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\3\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 5 + 15}\\{20 - 9}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{10}\\{11}\end{array}} \right]\end{aligned}\)

Therefore, the \(C - \)coordinate vector is \({\left[ {\mathop{\rm x}\nolimits} \right]_C} = \left[ {\begin{array}{*{20}{c}}{10}\\{11}\end{array}} \right]\).

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Most popular questions from this chapter

(calculus required) Define \(T:C\left( {0,1} \right) \to C\left( {0,1} \right)\) as follows: For f in \(C\left( {0,1} \right)\), let \(T\left( t \right)\) be the antiderivative \({\mathop{\rm F}\nolimits} \) of \({\mathop{\rm f}\nolimits} \) such that \({\mathop{\rm F}\nolimits} \left( 0 \right) = 0\). Show that \(T\) is a linear transformation, and describe the kernel of \(T\). (See the notation in Exercise 20 of Section 4.1.)

The null space of a \({\bf{5}} \times {\bf{6}}\) matrix A is 4-dimensional, what is the dimension of the column space of A.

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

21. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 2}&{ - 4.2}&{ - 4.8}&{ - 3.6}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

(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 \({c_j}'{\mathop{\rm s}\nolimits} \).)

Question 11: Let\(S\)be a finite minimal spanning set of a vector space\(V\). That is,\(S\)has the property that if a vector is removed from\(S\), then the new set will no longer span\(V\). Prove that\(S\)must be a basis for\(V\).
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