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Another L.A. problem, am i doing this correcto?

  1. Apr 21, 2006 #1
    Suppose that A is a 3 x 3 matrix whose nullspace is a line through the origin in 3-space. Can the row or column space of A also be a line through the origin?
    Well if we have a matrix whose nullspace is a line through the origin then we have..
    [X;Y;Z] = [a;b;c] t
    And we know that the dimensions of our nullspace is 1..
    Also we know that nullspace(a)+rank(a) = n
    where n is the columns.
    so..
    1+rank(a)=3
    this implies the rank = 2.
    Which has a dimension of two, which means it is a plane through the origin. If the dimension was one, then we would have a line through the origin.
     
  2. jcsd
  3. Apr 21, 2006 #2

    AKG

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    It's right more or less, but poorly worded. First, I don't think:

    Well if we have a matrix whose nullspace is a line through the origin then we have..
    [X;Y;Z] = [a;b;c] t


    is necessary. Also:

    Also we know that nullspace(a)+rank(a) = n

    is not technically correct; you mean dim(nullspace(A)) + rank(A) = n. "nullspace(A)" is not a number, it is a subspace. Sometimes, the dimension of the nullspace is called the nullity of A, and denoted N(A) so this notation may save you from having to write "dim(nullspace(A))", but don't just write "nullspace(A)".

    this implies the rank = 2.
    Which has a dimension of two, which means it is a plane through the origin.


    should be:

    this implies the rank = 2.
    Which implies that the row and columns spaces each have a dimension of two, which means each of them is a plane through the origin.


    Notice in your solution, you never actually mention the row or column spaces, so it's not perfectly clear what you're talking about. Also, you never technically say that the row and column spaces are not lines. You say that they are planes, but why say that when the question asks whether or not they are lines. Don't say that they are planes, say that they are not lines. Note that:

    If the dimension was one, then we would have a line through the origin.

    does not say that they are not lines. You say that if the dimension were 1, then they'd be lines, but what you need to say is that since the dimensions are not 1, they are not lines. A -> B and ~A -> ~B are not logically equivalent, e.g.
     
  4. Apr 21, 2006 #3
    Ahh..thanks.
     
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