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Q17SE

Expert-verifiedFound in: Page 191

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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).**

- The submatrix \({A_1}\) is a \(m \times r\) matrix with rank\(r\).
- The submatrix \({A_2}\) is invertible.

a)

Suppose \(A\) is an \(m \times n\) matrix with rank\(r\).

Assume that \({A_1}\) consists of \(r\) pivot columns of \(A\). The columns of \({A_1}\) are linearly independent.

Thus, the submatrix \({A_1}\) is an \(m \times r\) matrix with rank\(r\).

b)

**The rank theorem **states that the dimensions of the column space and the row space of an \(m \times n\) matrix \(A\) are equal. This common dimension, the rank of \(A\), also equals the number of pivot positions in \(A\) and satisfies the equation \({\mathop{\rm rank}\nolimits} A + \dim {\mathop{\rm Nul}\nolimits} A = n\).

Using the rank theorem on \({A_1}\), you get the dimension of Row\({A_1}\) as \(r\). Also, \({A_1}\) has \(r\) linearly independent rows. Suppose \({A_2}\) consists of \(r\) linearly independent rows of \({A_1}\). So, \({A_2}\) is a \(r \times r\) matrix with linearly independent rows. According to the invertible matrix theorem, \({A_2}\) is invertible.

Thus, the submatrix \({A_2}\) is invertible.

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