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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.

Short Answer

Expert verified

The system of the equation has a solution for every b in \({\mathbb{R}^{18}}\).

Step by step solution

01

Make assumptions for the given problem

Let the system \(A{\bf{x}} = {\bf{b}}\) has 18 equations and 20 variables. Then, by the rank theorem, you get

\({\rm{rank}}\,\,A + \dim \,\,{\rm{Nul}}\,\,A = n\).

02

Find the solution set using rank theorem

Since the number of columns in A is 20, the column space of A will be

\(\begin{array}{c}\dim \,{\rm{Col}}\,\,A \ge 20 - 2\\ \ge 18\end{array}\).

Therefore, the column space of A is \({\mathbb{R}^{18}}\).

Hence, the system of the equation has a solution for every b in \({\mathbb{R}^{18}}\).

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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 \(m \times n\).

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).

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show the coordinate mapping is one to one. (Hint: Suppose \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\) for some u and w in V, and show that \({\bf{u}} = {\bf{w}}\)).

In Exercise 4, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that a subset \(\left\{ {{{\bf{u}}_1},...,{{\bf{u}}_p}} \right\}\) in V is linearly independent if and only if the set of coordinate vectors \(\left\{ {{{\left( {{{\bf{u}}_{\bf{1}}}} \right)}_B},.....,{{\left( {{{\bf{u}}_p}} \right)}_B}} \right\}\) is linearly independent in \({\mathbb{R}^n}\)(Hint: Since the coordinate mapping is one-to-one, the following equations have the same solutions, \({c_{\bf{1}}}\),….,\({c_p}\))

\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) The zero vector V

\({\left( {{c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p}} \right)_B} = {\left( {\bf{0}} \right)_B}\) The zero vector in \({\mathbb{R}^n}\)a

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

15. Let\(A\)be an\(m \times n\)matrix, and let\(B\)be a\(n \times p\)matrix such that\(AB = 0\). Show that\({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\). (Hint: One of the four subspaces\({\mathop{\rm Nul}\nolimits} A\),\({\mathop{\rm Col}\nolimits} A,\,{\mathop{\rm Nul}\nolimits} B\), and\({\mathop{\rm Col}\nolimits} B\)is contained in one of the other three subspaces.)

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