Use Exercise 28 to explain why the equation\(Ax = b\)has a solution for all\({\rm{b}}\)in\({\mathbb{R}^m}\)if and only if the equation\({A^T}x = 0\)has only the trivial solution.

Short Answer

Expert verified

For the condition given, the number of non-pivot columns in \({A^T}\) is 0, which implies that \({A^T}x = 0\) has a trivial solution.

Step by step solution

01

Assume an arbitrary system and relate the given statement with it

Consider the nonhomogeneous system \(Ax = b\) with matrix \(A\) of the size \(m \times n\). If the system has a pivot position in each row, it will have solution for all values of \(b\) in the subspace of \({\mathbb{R}^m}\) .

02

Use the result of Exercise 28b

If the system \(Ax = b\) has a pivot position in each row, then the number of columns in \(A\) is \(m\), that is, \({\rm{dimCol}}\,\,A = m\). By the result of Exercise 28b,\({\rm{dimCol}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,{A^T} = m\). Put the values in \({\rm{dimCol}}\,\,A = m\) to get

\(\begin{aligned} {\rm{dim}}\,{\rm{Col}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,{A^T} &= m\\m + {\rm{dim}}\,{\rm{Nul}}\,\,{A^T} &= m\\{\rm{dim}}\,{\rm{Nul}}\,\,{A^T} &= 0.\end{aligned}\)

03

Draw a conclusion

As \({\rm{dim}}\,{\rm{Nul}}\,\,{A^T} = 0\), the number of non-pivot columns in \({A^T}\) is 0, which implies that \({A^T}x = 0\) has a trivial solution.

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Most popular questions from this chapter

In Exercise 17, Ais an \(m \times n\] matrix. Mark each statement True or False. Justify each answer.

17. a. The row space of A is the same as the column space of \({A^T}\].

b. If B is any echelon form of A, and if B has three nonzero rows, then the first three rows of A form a basis for Row A.

c. The dimensions of the row space and the column space of A are the same, even if Ais not square.

d. The sum of the dimensions of the row space and the null space of A equals the number of rows in A.

e. On a computer, row operations can change the apparent rank of a matrix.

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: 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).

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

13. Show that if \(P\) is an invertible \(m \times m\) matrix, then rank\(PA\)=rank\(A\).(Hint: Apply Exercise12 to \(PA\) and \({P^{ - 1}}\left( {PA} \right)\).)

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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