1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Error on p228 Of D.Lay's Linear Algebra

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant 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."


    3. 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?
     
  2. jcsd
  3. Apr 20, 2010 #2

    Mark44

    Staff: Mentor

    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.
     
  4. Apr 20, 2010 #3
    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.
     
  5. Apr 20, 2010 #4

    Mark44

    Staff: Mentor

    I agree that the "thus" clause doesn't follow from the hypothesis.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook