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Chapter 4: Q4.5-33E (page 191)

[M] According to Theorem 11, a linearly independent set \(\left\{ {{{\bf{v}}_1},....{{\bf{v}}_k}} \right\}\) in \({\mathbb{R}^n}\) can be expanded to a basis for \({\mathbb{R}^n}\). One way to this is to create \(A = \left[ {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}& \cdots &{{{\bf{v}}_k}\,\;\,\begin{array}{*{20}{c}}{{{\bf{e}}_1}}& \cdots &{{{\bf{e}}_n}}\end{array}}\end{array}} \right]\), with \({{\bf{e}}_1}\),….,\({{\bf{e}}_n}\) the columns of identity matrix; the pivot columns of A form a basis for \({\mathbb{R}^n}\).

a. Use the method described to extend the following vectors to a basis for \({\mathbb{R}^5}\).

\({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{ - {\bf{9}}}\\{ - {\bf{7}}}\\{\bf{8}}\\{ - {\bf{5}}}\\{\bf{7}}\end{array}} \right]\), \({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}{\bf{9}}\\{\bf{4}}\\{\bf{1}}\\{\bf{6}}\\{ - {\bf{7}}}\end{array}} \right]\), \({{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{7}}\\{ - {\bf{8}}}\\{\bf{5}}\\{ - {\bf{7}}}\end{array}} \right]\)

b. Explain why the method works in general: Why are the original vectors \({{\bf{v}}_1}\),….,\({{\bf{v}}_k}\) included in the basis found for Col A? Why is \({\bf{Col}}\,A = {\mathbb{R}^n}\)?

Short Answer

Expert verified

a. \(\left\{ {{{\bf{v}}_1},\,\,{{\bf{v}}_2},\;\;{{\bf{v}}_3},\;\;{{\bf{e}}_2},\;{{\bf{e}}_3}} \right\}\)

b. \({\rm{Col}}\left( A \right) = {\mathbb{R}^n}\)

Step by step solution

01

Form the matrix using vectors

The matrix using vectors can be written as:

\(A = \left[ {\begin{array}{*{20}{c}}{ - 9}&9&6\\{ - 7}&4&7\\8&1&{ - 8}\\{ - 5}&6&5\\7&{ - 7}&{ - 7}\end{array}} \right]\)

02

Find the matrix \(\left[ {\begin{array}{*{20}{c}}A&{{I_{\bf{5}}}}\end{array}} \right]\)

The matrix \(\left[ {\begin{array}{*{20}{c}}A&{{I_5}}\end{array}} \right]\) can be expressed as:

\(\left[ {\begin{array}{*{20}{c}}A&{{I_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 9}&9&6\\{ - 7}&4&7\\8&1&{ - 8}\\{ - 5}&6&5\\7&{ - 7}&{ - 7}\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{array}} \right]\)

03

Find the row reduced echelon form of \(\left[ {\begin{array}{*{20}{c}}A&{{I_{\bf{5}}}}\end{array}} \right]\)

Let

\(B = \left[ {\begin{array}{*{20}{c}}{ - 9}&9&6\\{ - 7}&4&7\\8&1&{ - 8}\\{ - 5}&6&5\\7&{ - 7}&{ - 7}\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{array}} \right]\).

Use the following MATLAB code to find the row reduce echelon form:

\(\begin{array}{l} > > B = \left[ { - 9\,\,9\,\,6\,\,1\,\,0\,\,0\,\,0\,\,0;\,\, - 7\,\,4\,\,7\,\,0\,\,1\,\,0\,\,0\,\,0\,;\,\,8\,\,1\,\, - 8\,\,0\,\,0\,\,1\,\,0\,\,0;\,\, - 5\,\,6\,\,5\,\,0\,\,0\,\,0\,\,1\,\,0;\,\,7\,\, - 7\,\, - 7\,\,0\,\,0\,\,0\,\,0\,\,1} \right];\\ > > U = {\rm{rref}}\left( A \right)\end{array}\)\(\left[ {\begin{array}{*{20}{c}}{ - 9}&9&6\\{ - 7}&4&7\\8&1&{ - 8}\\{ - 5}&6&5\\7&{ - 7}&{ - 7}\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\\0&0&0\\0&0&0\end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}}{ - \frac{1}{3}}&0&0&1&{\frac{3}{7}}\\0&0&0&1&{\frac{5}{7}}\\{ - \frac{1}{3}}&0&0&0&{ - \frac{3}{7}}\\0&1&0&3&{\frac{{22}}{7}}\\0&0&1&0&{ - \frac{{53}}{7}}\end{array}} \right]\)

As the pivot columns are 1, 2, 3, 5, and 6, the extended basis is:

\(\left\{ {{{\bf{v}}_1},\,\,{{\bf{v}}_2},\;\;{{\bf{v}}_3},\;\;{{\bf{e}}_2},\;{{\bf{e}}_3}} \right\}\)

04

Check why Col A\( = {\mathbb{R}^n}\)

If the set \(\left\{ {{{\bf{v}}_1},\;{{\bf{v}}_2},\;....,\;{{\bf{v}}_k}} \right\}\) is linearly independent, there is a pivot element in each column. Therefore,

\({\rm{Col}}\left( A \right) = {\mathbb{R}^n}\)

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Most popular questions from this chapter

Let V be a vector space that contains a linearly independent set \(\left\{ {{u_{\bf{1}}},{u_{\bf{2}}},{u_{\bf{3}}},{u_{\bf{4}}}} \right\}\). Describe how to construct a set of vectors \(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\) in V such that \(\left\{ {{v_{\bf{1}}},{v_{\bf{3}}}} \right\}\) is a basis for Span\(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\).

In Exercise 2, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

2. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{5}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{7}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{5}}}\end{array}} \right)\)

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

17. A submatrix of a matrix A is any matrix that results from deleting some (or no) rows and/or columns of A. It can be shown that A has rank \(r\) if and only if A contains an invertible \(r \times r\) submatrix and no longer square submatrix is invertible. Demonstrate part of this statement by explaining (a) why an \(m \times n\) matrix A of rank \(r\) has an \(m \times r\) submatrix \({A_1}\) of rank \(r\), and (b) why \({A_1}\) has an invertible \(r \times r\) submatrix \({A_2}\).

The concept of rank plays an important role in the design of engineering control systems, such as the space shuttle system mentioned in this chapter’s introductory example. A state-space model of a control system includes a difference equation of the form

\({{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}\)for \(k = 0,1,....\) (1)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(\left\{ {{{\mathop{\rm x}\nolimits} _k}} \right\}\) is a sequence of “state vectors” in \({\mathbb{R}^n}\) that describe the state of the system at discrete times, and \(\left\{ {{{\mathop{\rm u}\nolimits} _k}} \right\}\) is a control, or input, sequence. The pair \(\left( {A,B} \right)\) is said to be controllable if

\({\mathop{\rm rank}\nolimits} \left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right) = n\) (2)

The matrix that appears in (2) is called the controllability matrix for the system. If \(\left( {A,B} \right)\) is controllable, then the system can be controlled, or driven from the state 0 to any specified state \({\mathop{\rm v}\nolimits} \) (in \({\mathbb{R}^n}\)) in at most \(n\) steps, simply by choosing an appropriate control sequence in \({\mathbb{R}^m}\). This fact is illustrated in Exercise 18 for \(n = 4\) and \(m = 2\). For a further discussion of controllability, see this text’s website (Case study for Chapter 4).

(M) Show that \(\left\{ {t,sin\,t,cos\,{\bf{2}}t,sin\,t\,cos\,t} \right\}\) is a linearly independent set of functions defined on \(\mathbb{R}\). Start by assuming that

\({c_{\bf{1}}} \cdot t + {c_{\bf{2}}} \cdot sin\,t + {c_{\bf{3}}} \cdot cos\,{\bf{2}}t + {c_{\bf{4}}} \cdot sin\,t\,cos\,t = {\bf{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{5}} \right)\)

Equation (5) must hold for all real t, so choose several specific values of t (say, \(t = {\bf{0}},\,.{\bf{1}},\,.{\bf{2}}\)) until you get a system of enough equations to determine that the \({c_j}\) must be zero.

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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