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

If the null space of A \({\bf{7}} \times {\bf{6}}\) matrix A is 4-dimensional, what is the dimension of the column space of A?

Short Answer

Expert verified

The dimension of the column space of A is 1.

Step by step solution

01

Find the rank of the matrix

Using the nullity theorem, you get:

\(\begin{aligned} {\rm{rank}}\,A &= n - \dim \,{\rm{Nul}}\,A\\ &= 6 - 5\\ &= 1\end{aligned}\)

02

Find the dimension of the column space

The dimension of the column space of A is equal to its rank.

So, the dimension of the column space of 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

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.

Let \(B = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{9}}\end{array}} \right)\,} \right\}\). Since the coordinate mapping determined by B is a linear transformation from \({\mathbb{R}^{\bf{2}}}\) into \({\mathbb{R}^{\bf{2}}}\), this mapping must be implemented by some \({\bf{2}} \times {\bf{2}}\) matrix A. Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)).

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?

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).

Justify the following equalities:

a.\({\rm{dim Row }}A{\rm{ + dim Nul }}A = n{\rm{ }}\)

b.\({\rm{dim Col }}A{\rm{ + dim Nul }}{A^T} = m\)

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