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Q 18SE

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Linear Algebra and its Applications
Found in: Page 191
Linear Algebra and its Applications

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

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Short Answer

Question 18: Suppose A is a \(4 \times 4\) matrix and B is a \(4 \times 2\) matrix, and let \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) represent a sequence of input vectors in \({\mathbb{R}^2}\).

  1. Set \({{\mathop{\rm x}\nolimits} _0} = 0\), compute \({{\mathop{\rm x}\nolimits} _1},...,{{\mathop{\rm x}\nolimits} _4}\) from equation (1), and write a formula for \({{\mathop{\rm x}\nolimits} _4}\) involving the controllability matrix \(M\) appearing in equation (2). (Note: The matrix \(M\) is constructed as a partitioned matrix. Its overall size here is \(4 \times 8\).)
  2. Suppose \(\left( {A,B} \right)\) is controllable and v is any vector in \({\mathbb{R}^4}\). Explain why there exists a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) in \({\mathbb{R}^2}\) such that \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

a. The formula for \({{\mathop{\rm x}\nolimits} _4}\) is \({{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits} \).

b. The control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) makes \({{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits} \).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: State the difference equation in exercise 17

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)

The pair \(\left( {A,B} \right)\) is said to be controllable if rank\(\left[ {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right] = n\).

Step 2: Write the formula for \({{\mathop{\rm x}\nolimits} _4}\) involving the controllability matrix

a)

Consider \(A\) as a \(m \times n\) matrix with rank\(r\) and \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) as a sequence of input vectors in \({\mathbb{R}^2}\).

Use the equation \({{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}\) for \(k = 0,1,...\) with \({{\mathop{\rm x}\nolimits} _0} = 0\) as shown below:

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = A{{\mathop{\rm x}\nolimits} _0} + B{{\mathop{\rm u}\nolimits} _0}\\ = B{{\mathop{\rm u}\nolimits} _0}\\{{\mathop{\rm x}\nolimits} _2} = A{{\mathop{\rm x}\nolimits} _1} + B{{\mathop{\rm u}\nolimits} _1}\\ = AB{{\mathop{\rm u}\nolimits} _0} + B{{\mathop{\rm u}\nolimits} _1}\end{array}\)

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _3} = A{{\mathop{\rm x}\nolimits} _2} + B{{\mathop{\rm u}\nolimits} _2}\\ = A\left( {AB{{\mathop{\rm u}\nolimits} _0} + B{{\mathop{\rm u}\nolimits} _1}} \right) + B{{\mathop{\rm u}\nolimits} _2}\\ = {A^2}B{{\mathop{\rm u}\nolimits} _0} + AB{{\mathop{\rm u}\nolimits} _1} + B{{\mathop{\rm u}\nolimits} _2}\end{array}\)

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _4} = A{{\mathop{\rm x}\nolimits} _3} + B{{\mathop{\rm u}\nolimits} _3}\\ = A\left( {{A^2}B{{\mathop{\rm u}\nolimits} _0} + AB{{\mathop{\rm u}\nolimits} _1} + B{{\mathop{\rm u}\nolimits} _2}} \right) + B{{\mathop{\rm u}\nolimits} _3}\\ = {A^3}B{{\mathop{\rm u}\nolimits} _0} + {A^2}B{{\mathop{\rm u}\nolimits} _1} + AB{{\mathop{\rm u}\nolimits} _2} + B{{\mathop{\rm u}\nolimits} _3}\\ = \left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}&{{A^3}B}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _3}}\\{{{\mathop{\rm u}\nolimits} _2}}\\{{{\mathop{\rm u}\nolimits} _1}}\\{{{\mathop{\rm u}\nolimits} _0}}\end{array}} \right)\\ = M{\mathop{\rm u}\nolimits} \end{array}\)

Matrix \(M\) has four rows since \(B\) does, and \(M\) has eight columns since \(B\) and each of the matrices \({A^k}B\) has two columns. Vector \({\mathop{\rm u}\nolimits} \) in the final equation is in \({\mathbb{R}^8}\) since each \({{\mathop{\rm u}\nolimits} _k}\) is in \({\mathbb{R}^2}\).

Step 3: Explain why there exists a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) in \({\mathbb{R}^2}\) 

b)

If the matrix pair \(\left( {A,B} \right)\) is controllable, then the controllability matrix has rank 4, with a pivot in every row, and the columns of \(M\) span \({\mathbb{R}^4}\). Therefore, for any vector \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^4}\), there exists a vector u in \({\mathbb{R}^8}\) such that \({\mathop{\rm v}\nolimits} = M{\mathop{\rm u}\nolimits} \).

According to part (a), \({{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits} \) when \({\mathop{\rm u}\nolimits} \) is partitioned into a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\).

Thus, the control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) makes \({{\mathop{\rm x}\nolimits} _4} = M{\mathop{\rm u}\nolimits} \).

Most popular questions for Math Textbooks

(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.

Let \(A\) be any \(2 \times 3\) matrix such that \({\mathop{\rm rank}\nolimits} A = 1\), let u be the first column of \(A\), and suppose \({\mathop{\rm u}\nolimits} \ne 0\). Explain why there is a vector v in \({\mathbb{R}^3}\) such that \(A = {{\mathop{\rm uv}\nolimits} ^T}\). How could this construction be modified if the first column of \(A\) were zero?

If the null space of an \({\bf{8}} \times {\bf{5}}\) matrix A is 2-dimensional, what is the dimension of the row space of A?

Let \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Find \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) such that \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right] = {{\mathop{\rm uv}\nolimits} ^T}\) .

[M] Repeat Exercise 35 for a random integer-valued matrix whose rank is at most 4. One way to make is to create a random integ\(6 \times 7\)er-valued \(6 \times 4\) matrix \(J\) and a random integer-valued \(4 \times 7\) matrix \(K\), and set \(A = JK\). (See supplementary Exercise 12 at the end of the chapter; and see the study guide for the matrix-generating program.)
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