/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q36E Let \(V\) be a vector space, and... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Q36E (page 191)

Let \(V\) be a vector space, and let \(T:V \to V\)be a linear transformation. Given \(z\) in \(V\) , suppose \({x_p}\) in \(V\) satisfies \(T\left( {{x_p}} \right) = z\)and let \(u\) be any vector in the kernel of \(T\) . Show that \(u + {x_p}\)satisfies the nonhomogeneous equation \(T\left( x \right) = z\).

Short Answer

Expert verified

It is shown that \(u + {x_p}\) satisfies the nonhomogeneous equation \(T\left( x \right) = z\).

Step by step solution

01

Describe the given statement

It is given that \(z\) is in \(V\), and \({x_p}\) is in \(V\), which satisfies \(T\left( {{x_p}} \right) = z\). Also, \({\rm{u}}\) is in the kernel of \(T\). This implies that \(T\left( u \right) = 0\).

It is given that T is a linear transformation. It means that T satisfies the distributive property.

02

Use the linearity property and the relation\(T\left( {{x_p}} \right) = z\)

Let \(x\)be \(u + {x_p}\), then according to the linearity property, \(T\left( {u + {x_p}} \right) = T\left( u \right) + T\left( {{x_p}} \right)\).

\(\begin{aligned} T\left( {u + {x_p}} \right) &= T\left( u \right) + T\left( {{x_p}} \right)\\ &= 0 + z\\ &= z\end{aligned}\)

03

Draw a conclusion

If \(T\left( x \right) = z\), then \(u + {x_p}\) satisfies the nonhomogeneous equation \(T\left( x \right) = z\).

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 91影视!

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

Suppose a \({\bf{5}} \times {\bf{6}}\) matrix A has four pivot columns. What is dim Nul A? Is \({\bf{Col}}\,A = {\mathbb{R}^{\bf{3}}}\)? Why or why not?

If a \({\bf{3}} \times {\bf{8}}\) matrix A has a rank 3, find dim Nul A, dim Row A, and rank \({A^T}\).

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that a subset \(\left\{ {{{\bf{u}}_1},...,{{\bf{u}}_p}} \right\}\) in V is linearly independent if and only if the set of coordinate vectors \(\left\{ {{{\left( {{{\bf{u}}_{\bf{1}}}} \right)}_B},.....,{{\left( {{{\bf{u}}_p}} \right)}_B}} \right\}\) is linearly independent in \({\mathbb{R}^n}\)(Hint: Since the coordinate mapping is one-to-one, the following equations have the same solutions, \({c_{\bf{1}}}\),鈥�.,\({c_p}\))

\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) The zero vector V

\({\left( {{c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p}} \right)_B} = {\left( {\bf{0}} \right)_B}\) The zero vector in \({\mathbb{R}^n}\)a

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.)

Use coordinate vector to test whether the following sets of poynomial span \({{\bf{P}}_{\bf{2}}}\). Justify your conclusions.

a. \({\bf{1}} - {\bf{3}}t + {\bf{5}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{5}}t - {\bf{7}}{t^{\bf{2}}}\), \( - {\bf{4}} + {\bf{5}}t - {\bf{6}}{t^{\bf{2}}}\), \({\bf{1}} - {t^{\bf{2}}}\)

b. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

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.