/*! 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} Q2E Which of the sets W in Exercises... [FREE SOLUTION] | 91影视

91影视

Which of the sets W in Exercises 1 through 3 are subspaces of 3?

2. W={xyz:xyz}

Short Answer

Expert verified

W=xyz:xyz is not a subset of role="math" localid="1660732804183" 3.

Step by step solution

01

Consider the set.

A subset of the vector space nis called a (linear) subspace of if it has the following three properties:

a. W contains the zero vector in n.

b. W is closed under addition: If w1and w2are both in W , then so is w1+w2.

c. W is closed under scalar multiplication: If wis in W and k is an arbitrary scalar, then kwis in W .

The set W should satisfy all the above conditions.

W=xyz:xyz

02

Check for first condition.

The first condition is,

W=xyz:xyz

000:000

The first condition holds good.

03

Check for second condition.

The second condition is,

w1=xyz:xyz,w2=abc:abcw1+w2=x+ay+bz+c:x+ay+bz+c

The second condition holds good.

04

Check for third condition

The third condition is,

If k>0

kw1=kxkykz:kxkykz

If k<0

kw1=kxkykz:kxkykz

The third condition does not hold good.

05

Final answer.

W=xyz:xyz is not a subset of 3 .

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

In Exercises 21 through 25, find the reduced row-echelon form of the given matrix A. Then find a basis of the image of A and a basis of the kernel of A.

22.[248451793]

Let A and B be two matrices of the same size, with AB, both in reduced row-echelon form. Show thatKer(A)ker(B). Hint: Focus on the first column in which the two matrices differ, say, the kth columnsakandbkof A and B, respectively. Explain why at least one of the columnsakandbkfails to contain a leading 1. Thus, reversing the roles of matrices A and B if necessary, we can assume thatakdoes not contain a leading 1. We can writeak as a linear combination of preceding columns and use this representation to construct a vector in the kernel of A. Show that this vector fails to be in the kernel of B. Use Exercises 86 and 87 as a guide.

In Problem 46 through 55, Find all the cubics through the given points. You may use the results from Exercises 44 and 45 throughout. If there is a unique cubic, make a rough sketch of it. If there are infinitely many cubics, sketch two of them.

55.(0,0)(1,0)(2,0)(0,1)(1,1)(2,1)(0,2)(1,2)(2,2),(3,3).

In Problem 46 through 55, Find all the cubics through the given points. You may use the results from Exercises 44 and 45 throughout. If there is a unique cubic, make a rough sketch of it. If there are infinitely many cubics, sketch two of them.

46. (0,0),(1,0),(2,0),(3,0),(0,1),(0,2),(0,3),(1,1).

Explain why fitting a cubic through the mpoints P1(x1,y1),......,Pm(xm,ym)amounts to finding the kernel of an mx10matrix A. Give the entries of theof row A.

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.