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Expert-verified Found in: Page 191 ### Linear Algebra and its Applications

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).

1. The submatrix $${A_1}$$ is a $$m \times r$$ matrix with rank$$r$$.
2. The submatrix $${A_2}$$ is invertible.
See the step by step solution

## Step 1: Explain that an $$m \times n$$ matrix A has an $$m \times r$$ submatrix $${A_1}$$

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$$r$$.

## Step 2: Explain that $${A_1}$$ has an invertible $$r \times r$$ submatrix $${A_2}$$

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. ### Want to see more solutions like these? 