Linear Algebra and Its Applications, 5th Edition

Linear Algebra and Its Applications, 5th Edition

Authors: David C. Lay, Steven R. Lay, Judi J. McDonald

ISBN-13: 978-0321982384

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See our solution for Question 21E from Chapter 4.3 from Lay's Linear Algebra and Its Applications, 5th Edition.

Problem 21E

Chapter:
Problem:
In Exercises 21 and 22, mark each statement True or False. Justify each answer. a. A single vector by itself is linearly dependent. b. c. The columns of an invertible n × n matrix form

Step-by-Step Solution

Given Information
We are given with some statements, we have to find whether they are True or False

Step-1: (a) .
Statement: A single vector by itself is linearly dependent

The statement is False. To check the dependency write the linear equation for a nonzero vector v and non zero scaler c

\[cv = 0\]As, v and c are non zero, the equation can not be satisfied. The equation can only be satisifed for nonzero scaler, when the single vector is zero

False

Step-2: (b) .
Statement: If $H = Span\left\{ {{b_1},...{b_p}} \right\}$, then $\left\{ {{b_1},...{b_p}} \right\}$ is a basis for H

The statement is False. The statement is only true if the vectors $\left\{ {{b_1},...{b_p}} \right\}$ are lienarly independent. However, if the vectors are linearly dependent, they can not form a basis for H. False

Step-3: (c) .
Statement: The columns of an invertible $n \times $ matrix form a basis for $R^n$

The statement is True from the invertible matrix theorem. If the matrix is invertible, then the columns are linearly independent. So, they form a basis for $R^n$

True

Step-4: (d) .
Statement: A basis is a spanning set that is as large as possible

The statement is False. For $R^n$, there can be more number of spanning sets than the basis. Not every spanning set can form a basis.

False

Step-5: (e) .
Statement: In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix.

The statement is False. When we reduce a matrix A to a matrix B by elementary row operations, the columns of B are often totally different from those of A. However, the solution of both the equations (Ax=0 and Bx=0) gives the same solution set.

False