Error on p228 Of D.Lay's Linear Algebra

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Homework Help Overview

The discussion revolves around a perceived error in the third edition of David C. Lay's Linear Algebra and its Applications, specifically regarding a statement about the spanning set for the null space of a matrix A. Participants are examining the implications of the statement made on page 228 concerning linear combinations of vectors u, v, and w.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of the statement that every linear combination of u, v, and w is an element of Nul A, and whether this implies that {u, v, w} is a spanning set for Nul A. Some are exploring the conditions under which this statement holds true, such as the dimensionality and linear independence of the vectors involved.

Discussion Status

The discussion is ongoing, with differing opinions on the accuracy of the original statement and the implications of the use of "thus." Some participants are providing additional context that may clarify the original statement, while others are emphasizing the need for more precise language to avoid misinterpretation.

Contextual Notes

There are assumptions being made about the dimensionality of the null space and the properties of the vectors u, v, and w, which are not explicitly stated in the original text. Participants are also noting the lack of errata for the book and are curious about other potential errors.

Wesleytf
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Homework Statement


Not so much a problem, just an error I noticed... It's in the third edition (and update of the third edition) of David C. Lay's Linear Algebra and its Applications.


Homework Equations


p.228, last sentence:
"Every linear combination of u, v, and w is an element of Nul A. Thus, {u, v, w} is a spanning set for Nul A."


The Attempt at a Solution


The "thus" is incorrect--I think what he's trying to say is that every element of Nul A is a linear combination of u, v and w. Thus, {u, v, w} is a spanning set for Nul A.

As stated, I think it is false.
For instance, take this case:

Given: {u, v, w} is a basis for Nul A

Superimpose his statement:
Every linear combination of u and v is an element of Nul A.
Thus, {u, v} is a spanning set for Nul A. (--><--) (contradiction).


I found this while doing the reading for class several weeks back. I looked online and I couldn't find any errata for this book--does anyone know of any other errors in the book? Anyone disagree that this is incorrect, or at least unclear?
 
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I don't see anything wrong with the statement in the book, although it could be fleshed out by saying
1. dim(null A) = 3.
2. u, v, and w are in null A.
3. u, v, and w are linearly independent.

If we understand these to be true, then what you said does not necessarily follow.
You have u and v in null A.
u and v are linearly independent.
But {u, v} is not a basis for null (A) and hence doesn't span null A, since there aren't enough vectors to form a basis.
 
My objection is not to the truth of the statement that those vectors span the Nul space--in the book, that's exactly what's presented. It's the use of "thus" that is inaccurate. The statement I presented afterwards was just meant to show how the argument that the equivalent statement "because the L.C.s of the vectors are elements, they span" is incorrect. That is what the "thus" implies, and I think this is where the mistake is. If you look at the statement independently, it seems that is has to be incorrect.
 
I agree that the "thus" clause doesn't follow from the hypothesis.
 

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