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Linear Combinations and Span (Concept Question)

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Let [itex]A[/itex] be an [itex] m \hspace{1 mm} x \hspace{1 mm} n [/itex] matrix, and let [itex] \vec{b} [/itex] be a vector in [itex] \mathbb{R}^{m} [/itex]. Suppose that [itex] \vec{b} [/itex] is a linear combination of the columns of A. Then the columns of A span [itex] \mathbb{R}^{m} [/itex]
    2. Relevant equations


    3. The attempt at a solution
    I said that this statement was true using the following theorem from my textbook:
    Let [itex]A[/itex] be an [itex]m \hspace{1 mm}x \hspace{1 mm}n[/itex] matrix. Then the following statements are logically equivalent.
    a) For each [itex]\vec{b}[/itex] in [itex]\mathbb{R}^{m}[/itex], the equation [itex]A \vec{x} = \vec{b} [/itex] has a solution
    b) Each [itex]\vec{b} [/itex] in [itex] \mathbb{R}^{m} [/itex] is a linear combination of the columns of A
    c) The columns of A span [itex]\mathbb{R}^{m}[/itex]
    d) A has a pivot position in every row
    However, my book says that this statement is false and I am not sure why. I think I am probably missing something obvious, but I'm not sure what.
     
    Last edited: Feb 28, 2015
  2. jcsd
  3. Feb 28, 2015 #2
    In the statement, [itex]\vec{b} [/itex] is a particular vector, and not any arbitrary vector in [itex] \mathbb{R}^{m} [/itex]; that is to say, it is not necessarily true that any vector in [itex] \mathbb{R}^{m} [/itex] can be expressed as a linear combination of the columns of A.
     
  4. Feb 28, 2015 #3
    ok, so I was wondering if this counter example would be a good way to verify that it is false? I picked an arbitrary matrix and an arbitrary vector:
    $$
    A =
    \begin{bmatrix}
    0 & 3\\
    1& 5\\
    2 &8
    \end{bmatrix}
    $$
    $$ \vec{b} =
    \begin{bmatrix}
    1\\
    2
    \\5

    \end{bmatrix} $$
    I reduced A and found that there is not a pivot in every row, so I said that the columns of A do not span [itex] \mathbb{R}^{m} [/itex]. Is this a sufficient counter example?
     
  5. Mar 1, 2015 #4

    Mark44

    Staff: Mentor

    I'm not sure it meets the conditions of the original problem, which states that ##\vec{b}## is a linear combination of the columns of A. In any case, the condition for ##\vec{b}## seems to me to be something of a red herring. Your 3 x 2 matrix clearly (I hope) can't span R3, since there are only two columns.
     
  6. Mar 1, 2015 #5
    You're right that my example doesn't meet the given conditions. However, if I were to find an example that does meet the required conditions using a 3x2 matrix, could I then use that counter example to prove that the statement is false? Basically I'm still a little confused about where to go here...
     
  7. Mar 1, 2015 #6

    Mark44

    Staff: Mentor

    Yes, I believe so.
     
  8. Mar 1, 2015 #7
    Awesome. I was thinking in this case it may be easier to use a matrix made up of stars and squares(like the kind used to determine echelon forms) rather than actually come up with a linear combination.
     
  9. Mar 1, 2015 #8

    Mark44

    Staff: Mentor

    No, I would use a specific matrix.
     
  10. Mar 1, 2015 #9
    Ok, thanks for the advice
     
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