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

Let \({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}1\\0\\1\end{array}} \right]\), \({{\bf{v}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\1\end{array}} \right]\), \({{\bf{v}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{0}}\end{array}} \right]\), and let H be the set of vectors in \({\mathbb{R}^{\bf{3}}}\) whose second and third entries are equal. Then every vector in H has a unique expansion as a linear combination of \({{\bf{v}}_{\bf{1}}}\), \({{\bf{v}}_{\bf{2}}}\), \({{\bf{v}}_{\bf{3}}}\), because

\(\left[ {\begin{array}{*{20}{c}}s\\t\\t\end{array}} \right] = s\left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{1}}\end{array}} \right] + \left( {t - s} \right)\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right] + s\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right]\)for any \(s\), and \(t\). Is \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) a basis for H? Why or why not?

Short Answer

Expert verified

The set\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)is not a basis for H.

Step by step solution

01

State the invertible matrix theorem

Recall the invertible matrix theorem, which states that if the square matrix is invertible, then the columns are linearly independent, and they form a basis for \({\mathbb{R}^n}\).

02

State the set of vectors in Span

The vectors\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)must span H in order to provide a basis for H. So,\(H = {\rm{Span}}\left\{ {{v_1},{v_2},{v_3}} \right\}\).

The vectors in set H have identical second and third entries, and H is a subset of\(\left\{ {{v_1},{v_2},{v_3}} \right\}\).But there are some vectors that are not in H.

So,\(H\)is not in\({\rm{Span}}\left\{ {{v_1},{v_2},{v_3}} \right\}\).

Therefore,\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)is not a basis for H.

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

Verify that rank \({{\mathop{\rm uv}\nolimits} ^T} \le 1\) if \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 3}\\5\end{array}} \right]\) and \({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right]\).

If the null space of a\({\bf{5}} \times {\bf{6}}\)matrix A is 4-dimensional, what is the dimension of the row space of A?

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 \({\left( {\bf{x}} \right)_B}\)).

(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 \({c_j}'{\mathop{\rm s}\nolimits} \).)

Let be a basis of\({\mathbb{R}^n}\). .Produce a description of an \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\)matrix A that implements the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\). Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)). (See Exercise 21.)

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