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Chapter 4: Q16E (page 191)

Qis any matrix such that

\({\left( {\bf{v}} \right)_C} = Q{\left( {\bf{v}} \right)_B}\)for each v in V (9)

Set \({\bf{v}} = {{\bf{b}}_{\bf{1}}}\) in (9). Then (9) shows that \({\left( {{{\bf{b}}_{\bf{1}}}} \right)_C}\) is the first column of Q because (a) _____. Similarly, for \(k = {\bf{2}}\),鈥�..n the kth column of Q is (b) _____ because (c) _____. This shows the matrix \(\mathop P\limits_{C \leftarrow B} \) defined by (5) in Theorem 15 is the only matrix that satisfies condition (4).

Short Answer

Expert verified

(a) \({\left( {{{\bf{b}}_1}} \right)_C} = Q{{\bf{e}}_1}\)

(b) \({\left( {{{\bf{b}}_k}} \right)_C}\)

(c) \({\left( {{{\bf{b}}_k}} \right)_C} = Q{{\bf{e}}_k}\)

Step by step solution

01

Check for blank (a)

The columns of Q are C coordinate vectors of the vectors in basis B,that is,

\(Q = \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}{{\left( {{{\bf{b}}_2}} \right)}_C}.....{{\left( {{{\bf{b}}_k}} \right)}_C}} \right)\).

If you set \({\bf{v}} = {{\bf{b}}_1}\) in (1), then \({\left( {{{\bf{b}}_1}} \right)_C}\) represents the first column of Q because

\(\begin{aligned} {\left( {{{\bf{b}}_1}} \right)_C} &= Q{\left( {{{\bf{b}}_1}} \right)_B}\\ &= \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}\,{{\left( {{{\bf{b}}_2}} \right)}_C}\,\,....\,\,{{\left( {{{\bf{b}}_k}} \right)}_C}} \right)\left( {\begin{array}{*{20}{c}}1\\0\\ \vdots \\0\end{array}} \right)\\ &= Q{{\bf{e}}_1}.\end{aligned}\)

02

Check for blank (b)

For \(k = 2\),鈥�n, the kthcolumn of Q is

\(\begin{aligned} {\left( {{{\bf{b}}_k}} \right)_C} &= Q{\left( {{{\bf{b}}_1}} \right)_B}\\ &= \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}\,\,{{\left( {{{\bf{b}}_2}} \right)}_C}\,\,...\,\,{{\left( {{{\bf{b}}_k}} \right)}_C}} \right)\left( {\begin{array}{*{20}{c}}0\\0\\ \vdots \\1\end{array}} \right)\\ &= Q{e_k}.\end{aligned}\)

So, the kth column of Q is \({\left( {{{\bf{b}}_k}} \right)_C}\).

03

Check for blank (c)

\(\mathop P\limits_{C + B} = \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}\,\,{{\left( {{{\bf{b}}_2}} \right)}_C}\,\,....\,\,{{\left( {{{\bf{b}}_n}} \right)}_C}} \right)\)is the only matrix that satisfies the condition

\({\left( {\bf{x}} \right)_c} = \mathop P\limits_{C = S} {\left( {\bf{x}} \right)_S}\).

It means,

\(\begin{aligned}{c}{\left( {{{\bf{b}}_k}} \right)_C} &= Q{\left( {{{\bf{b}}_1}} \right)_B}\\ &= \left( {{{\left( {{{\bf{b}}_1}} \right)}_C}\,{{\left( {{{\bf{b}}_2}} \right)}_C}\,...\,\,{{\left( {{{\bf{b}}_k}} \right)}_C}} \right)\left( {\begin{array}{*{20}{c}}0\\0\\ \vdots \\1\end{array}} \right)\\ &= Q{{\bf{e}}_k}.\end{aligned}\)

So, \({\left( {{{\bf{b}}_k}} \right)_C} = Q{{\bf{e}}_k}\).

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Most popular questions from this chapter

Let \(A\) be an \(m \times n\) matrix of rank \(r > 0\) and let \(U\) be an echelon form of \(A\). Explain why there exists an invertible matrix \(E\) such that \(A = EU\), and use this factorization to write \(A\) as the sum of \(r\) rank 1 matrices. [Hint: See Theorem 10 in Section 2.4.]

What would you have to know about the solution set of a homogenous system of 18 linear equations 20 variables in order to understand that every associated nonhomogenous equation has a solution? Discuss.

Explain what is wrong with the following discussion: Let \({\bf{f}}\left( t \right) = {\bf{3}} + t\) and \({\bf{g}}\left( t \right) = {\bf{3}}t + {t^{\bf{2}}}\), and note that \({\bf{g}}\left( t \right) = t{\bf{f}}\left( t \right)\). Then, \(\left\{ {{\bf{f}},{\bf{g}}} \right\}\) is linearly dependent because g is a multiple of f.

In Exercise 18, Ais an \(m \times n\) matrix. Mark each statement True or False. Justify each answer.

18. a. If B is any echelon form of A, then the pivot columns of B form a basis for the column space of A.

b. Row operations preserve the linear dependence relations among the rows of A.

c. The dimension of the null space of A is the number of columns of A that are not pivot columns.

d. The row space of \({A^T}\) is the same as the column space of A.

e. If A and B are row equivalent, then their row spaces are the same.

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

19. \(A = \left( {\begin{array}{*{20}{c}}{.9}&1&0\\0&{ - .9}&0\\0&0&{.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}0\\1\\1\end{array}} \right)\).
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