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Vector spaces, Spans and Matrix Determinants

  1. Dec 13, 2007 #1
    I think I have something mixed up so if someone can please point out my error.

    1. the set of all linear combinations is called a span.

    2. If a family of vectors is linearly independent none of them can be written as a linear combination of finitely many other vectors in the collection.

    3. If the determinant of a matrix is not equal to zero the vectors are linearly independent.

    Therefore, if the determinant of the matrix does not equal zero the vectors can not be written as a linear combination, hence there is no span.
     
    Last edited: Dec 13, 2007
  2. jcsd
  3. Dec 13, 2007 #2
    The only problem I see here is that a "span" is something that applies to a set of vectors, or a space. Then all of a sudden you're talking about matrices. It does not make sense in any way I can think of to talk about the "span" of a matrix-- the words are just not meaningful.

    Now maybe the idea is, you want to take the basis set of vectors for some space, and create a matrix whose columns consist of those basis vectors. If this is the case then yes, the determinant would be nonzero (this is actually one way of testing whether a set of vectors form a basis), and yes, it will not be possible to represent any of the basis vectors as a linear combination of the others (since this is part of what a basis means). The span of the basis vectors meanwhile will be the entire space.
     
  4. Dec 13, 2007 #3

    radou

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    As Coin pointed out, the term "span" applies to a set of vectors, so you'll have to be a bit more precise. So, you could say that the set of all linear combinations of a set of vectors forms the span of that very set (or linear shell, as I have been tought).
     
  5. Dec 13, 2007 #4
    Ok, to clear things up a little

    Given vectors X1, X2..., X3 in R^n, a vector in the form X = t1 X1 + t2 X2 + ... tk Xk

    1. the set of such linear combinations is called a span of the Xi and denoted by span{X1, X2, ...Xk}

    2. If a family of vectors is linearly independent none of them can be written as a linear combination of finitely many other vectors in the collection.

    3. If the determinant of a matrix is not equal to zero the vectors are linearly independent.

    Therefore, if the determinant of the matrix does not equal zero the vectors can not be written as a linear combination, hence there is no span.

    Now my text gives me a solution where a matrix whose columns consist of basis vectors has a determinant of -42. It also states that the vectors span R^4.

    Now, if the non zero determinant means that no linear combination can be written, and with no linear combination there is no span as it is a combination of the linear combinations. So, either the vectors span R^4 or the determinant is non-zero. I can't see how it can be both.

    The vectors are: {[1 3 -1 0]T, [-2 1 0 0]T, [0 2 1 -1]T, [3 6 -3 -2]T}
     
  6. Dec 13, 2007 #5

    HallsofIvy

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    No, "non zero determinant" means that none of the vectors in the set can be written as a linear combination of the others. It surely does not mean "no linear combination can be written". A linear combination of a set of vectors is just a sum of numbers times those vectors. That can always be done.
     
  7. Dec 13, 2007 #6
    Yeah, I am starting to see that.

    I also found another theorem which counters what I posted before.

    An nxn matrix is invertible if and only if
    1) the rows are linearly independent (otherwise det = 0)*
    2) the columns are linearly independent (otherwise det = 0)*
    3) rows of A span R^n

    So in order for the rows of A to span R^n the det must be non-zero, not the other way around.

    *a matrix is not invertible if the determinant = 0*
     
    Last edited: Dec 13, 2007
  8. Dec 13, 2007 #7
    It works both ways: if det(A) is non-zero, then the rows (or columns) of A span R^n. And, conversely, if the rows (or columns) of A span R^n, then det(A) is non-zero.

    Right, and that one goes both ways as well: if det(A) = 0, A is not invertible.

    All of which is to say that invertibility, the span covering the entire space, and non-zero determinant are all basically different ways of saying the same thing. Most people would use the expression "matrix A has full rank" to denote this property.
     
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