Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Q27E (page 191)

Which of the subspaces \({\rm{Row }}A\) , \({\rm{Col }}A\), \({\rm{Nul }}A\), \({\rm{Row}}\,{A^T}\) , \({\rm{Col}}\,{A^T}\) , and \({\rm{Nul}}\,{A^T}\) are in \({\mathbb{R}^m}\) and which are in \({\mathbb{R}^n}\) ? How many distinct subspaces are in this list?.

Short Answer

Expert verified

There are four distinct subspaces in the list.

Step by step solution

01

Assume an arbitrary system and relate the given statement with it

Assume that the system has matrix A of the size \(m \times n\). So, The null space in \(A\) must be the subspace of \({\mathbb{R}^n}\); the rows in \(A\) must be the subspace of \({\mathbb{R}^n}\); thecolumns in \(A\) must be the subspace of \({\mathbb{R}^m}\).

As the matrix \({A^T}\) has the size \(n \times m\), the null space in \({A^T}\) is a subspace of \({\mathbb{R}^m}\); the rows in \({A^T}\) are the subspace of \({\mathbb{R}^m}\); the columns in \({A^T}\) are the subspace of \({\mathbb{R}^n}\).

02

Draw a conclusion

As \({\rm{Row}}\,{A^T} = {\rm{Col}}\,A\) and \({\rm{Row}}\,A = {\rm{Col}}\,{A^T}\) , there are four possible distinct subspaces— \({\rm{Row}}\,A\),\({\rm{Col}}\,A\), \({\rm{Nul}}\,A\), \({\rm{Nul}}\,{A^T}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercise 2, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

2. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{5}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{7}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{5}}}\end{array}} \right)\)

In Exercise 6, find the coordinate vector of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

6. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{6}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{0}}\end{array}} \right)\)

[M] Repeat Exercise 35 for a random integer-valued matrixwhose rank is at most 4. One way to makeis 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.)

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 \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

In Exercise 7, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

7. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{\bf{9}}\end{array}} \right),{b_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{2}}}\\{\bf{4}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{9}}}\\{\bf{6}}\end{array}} \right)\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free