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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

Find a basis for the set of vectors in \({\mathbb{R}^{\bf{2}}}\) on the line \(y = {\bf{5}}x\).

A basis for \(Nul A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}\).

See the step by step solution

Step by Step Solution

Step 1: Define the null space of a matrix

The set of all homogeneous equation solutions, \(A{\bf{x}} = 0\), is \(Nul A\). It is called null space of A.

Step 2: State the basis for the set of vectors

The line equation \(y = 5x\) can also be written as \(y - 5x = 0\).

In the matrix form, it is \(A = \left( {\begin{array}{*{20}{c}}5&{ - 1}\end{array}} \right)\).

From the above equation, \(y\) corresponds to the pivot position. So, \(y\) is the basic variable, and \(x\) is the free variable.

Let \(x = a\).

Substitute the value \(x = a\) in the equation \(y = 5x\) to obtain the general solution

\(y = 5a\).

Obtain the vector in the parametric form using \(x = a\) and \(y = 5a\).

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right) = \left( {\begin{array}{*{20}{c}}a\\{5a}\end{array}} \right)\\ = a\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)\end{array}\)

It can also be written as shown below:

\(\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right) = x\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)\)

The basis of vector is shown below:

\( \Rightarrow \left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}\)

Thus, \({\bf{x}} = x\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)\), and a basis for \(Nul A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}\).

Most popular questions for Math Textbooks

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).


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