Error on p228 Of D.Lay's Linear Algebra

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SUMMARY

The discussion identifies a critical error in the third edition of David C. Lay's "Linear Algebra and its Applications," specifically on page 228. The statement "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" is deemed incorrect. The correct interpretation is that while every element of Nul A can be expressed as a linear combination of u, v, and w, the use of "thus" in the original statement misrepresents the relationship between linear combinations and spanning sets. The participants agree that additional clarification regarding the dimensionality and independence of the vectors is necessary for accurate understanding.

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  • Understanding of linear combinations in vector spaces
  • Familiarity with the concept of spanning sets in linear algebra
  • Knowledge of null spaces and their properties
  • Basic comprehension of linear independence
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  • Study the properties of null spaces in linear algebra
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Students of linear algebra, educators teaching the subject, and anyone seeking to clarify concepts related to vector spaces and null spaces will benefit from this discussion.

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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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