Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Determine the dimensions of and \({\mathop{\rm Col}\nolimits} A\) for the matrices shown in Exercise 13-18

18. \(A = \left( {\begin{array}{*{20}{c}}1&4&{ - 1}\\0&7&0\\0&0&0\end{array}} \right)\)

Short Answer

Expert verified

The dimension of \({\mathop{\rm Col}\nolimits} A\) is 2, and the dimension of \({\mathop{\rm Nul}\nolimits} A\) is 1.

Step by step solution

01

State the condition for the dimensions of \({\mathop{\rm Nul}\nolimits} A\) and \({\mathop{\rm Col}\nolimits} A\)

Thedimension of \({\mathop{\rm Nul}\nolimits} A\) is thenumber of free variablesin the equation \(A{\mathop{\rm x}\nolimits} = 0\), and the dimension of \({\mathop{\rm Col}\nolimits} A\)is thenumber of pivot columnsin \(A\).

02

Determine the dimensions of \({\mathop{\rm Nul}\nolimits} A\) and \({\mathop{\rm Col}\nolimits} A\)

The given matrix A is in echelon form. As it has two pivot columns, the dimension of \({\mathop{\rm Col}\nolimits} A\) is 2. The equation \(A{\mathop{\rm x}\nolimits} = 0\) has one free variable since the matrix has one column without a pivot. Therefore, the dimension of \({\mathop{\rm Nul}\nolimits} A\) is 1.

Thus, the dimension of \({\mathop{\rm Col}\nolimits} A\) is 2 and of \({\mathop{\rm Nul}\nolimits} A\) is 1.

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.

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

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

Let \(A\) be an \(m \times n\) matrix of rank \(r > 0\) and let \(U\) be an echelon form of \(A\). Explain why there exists an invertible matrix \(E\) such that \(A = EU\), and use this factorization to write \(A\) as the sum of \(r\) rank 1 matrices. [Hint: See Theorem 10 in Section 2.4.]

Question 11: Let\(S\)be a finite minimal spanning set of a vector space\(V\). That is,\(S\)has the property that if a vector is removed from\(S\), then the new set will no longer span\(V\). Prove that\(S\)must be a basis for\(V\).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

Use Exercise 28 to explain why the equation\(Ax = b\)has a solution for all\({\rm{b}}\)in\({\mathbb{R}^m}\)if and only if the equation\({A^T}x = 0\)has only the trivial solution.

See all solutions

Recommended explanations on Math Textbooks

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