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Matrices theorem help

  1. Apr 5, 2006 #1
    Let A be a mxn matrix with columns C1,...Cn. If rank A = n, show taht
    [itex] {A^T C_{1},...,A^T C_{N}} [/itex]is a basis of Rn


    since Rank A = n, then the columns are linearly independant

    so does that automatically mean that any multiple,, like A transpose for example, will keep the independance of the Columns?

    A theorem also tells us that if the Rank A = n, then the column span Rn. So the columns span Rn in this case

    is this adequate for a proof?
     
  2. jcsd
  3. Apr 5, 2006 #2

    0rthodontist

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    No, it is not. Multiplication by A is not just multiplication by a scalar. It could also rotate the vectors.

    You need to show that the set {A^T C1, ... A^T Cn} is linearly independent, which implies it is a basis in Rn since it has n elements. Assume otherwise. Then you have a1A^T C1 + ... + anA^T Cn = 0 for a1 ... an not all zero. Can you manipulate this to get a contradiction?
     
  4. Apr 5, 2006 #3

    AKG

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    The answer is below. Highlight the first white line if you need a hint. Every time you need a hint, highlight the next line. For your own good, highlight as few lines as possible, i.e. try to do most of it yourself.

    {ATC1, ..., ATCn} is a basis of Rn

    iff {ATC1, ..., ATCn} is linearly independent

    iff the columns of the matrix (ATC1 ... ATCn) are linearly independent

    iff det[(ATC1 ... ATCn)] is non-zero

    iff det[AT(C1 ... Cn)] is non-zero

    iff det[ATA] is non-zero

    iff det[AT]det[A] is non-zero

    iff det[A]det[A] is non-zero

    iff det[A] is non-zero


    iff rank[A] = n
     
    Last edited: Apr 5, 2006
  5. Apr 5, 2006 #4
    This would be true if A were a nxn matrix, but the question specifies that A is a mxn matrix.
     
    Last edited: Apr 5, 2006
  6. Apr 5, 2006 #5
    ok supose that
    [tex] a_{1}A^T C_{1} + ... + a_{n}A^T C_{n} = 0 [/tex]
    where not all ai are zero

    we know that A is not a zero matrix because the rank A = n <= m
    not A transpose is not zero
    then the Columns must be all zero
    But again A is not zero. Thus by contradiction we get that all the ai must be zero

    IS that good?
     
  7. Apr 5, 2006 #6

    AKG

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    Oops, I overlooked that.
     
  8. Apr 5, 2006 #7

    AKG

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    No. Why is the underlined stuff true? Try this:

    {ATC1, ..., ATCn} is a basis
    iff {ATC1, ..., ATCn} is linearly independent
    iff c1ATC1 + ... + cnATCn = 0 implies c1 = ... = cn = 0
    iff AT(c1C1 + ... + cnCn) = 0 implies c1 = ... = cn = 0
    iff AT(c1C1 + ... + cnCn) = 0 implies c1C1 + ... + cnCn = 0 [since the Ci are linearly independent since Rank[A] = n, so [itex]\sum c_iC_i = 0 \Leftrightarrow \forall i,\, c_i=0[/itex]]
    iff AT(X) = 0 implies X = 0

    But rank[AT] = rank[A] = n. Is there some theorem which says that, given this, that italicized line must be true?
     
  9. Apr 5, 2006 #8

    0rthodontist

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    A^T has all independent columns since it has rank n. Therefore, A^T(X) = 0 can only be true if X is 0. Otherwise you would have a linear combination of the columns of A^T, with not all nonzero coefficients, yielding 0, which we know cannot happen.
     
  10. Apr 6, 2006 #9

    Hurkyl

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    You mean rows.


    Anyways, I think everyone's overlooking something important!

    [tex] a_{1}A^T C_{1} + ... + a_{n}A^T C_{n} = 0 [/tex]

    This means that the vector [itex]a_1 C_1 + \cdots + a_n C_n[/itex] is an element of the null space of [itex]A^T[/itex]. People seem to be arguing that the null space is trivial, and therefore the vector must be zero.

    But that's wrong!

    [itex]A^T[/itex] is a rank n nxm matrix. Since it's operating on an m-dimensional space, it must have a null space of dimension (m-n).


    You have to use the fact that these m-long column vectors are special -- each of the [itex]C_n[/itex] is the transpose of one of the rows of [itex]A^T[/itex].


    In fact, if we place the [itex]A^T C_i[/itex] into a matrix, we get:

    [tex]
    [ A^T C_1 \mid A^T C_2 \mid \cdots \mid A^T C_n ] = A^T A
    [/tex]

    There's actually a 1-line proof that this (square) matrix is nonsingular, but I think the approach you guys are using ought to work, as long as you start using the fact that the [itex]C_i[/itex] are special m-long vectors, and not arbitrary m-long vectors.
     
    Last edited: Apr 6, 2006
  11. Apr 6, 2006 #10

    0rthodontist

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    :redface: Oh yeah... I was picturing it wrong.
     
  12. Apr 6, 2006 #11

    AKG

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    When m=n, this is easy, and that's what I did in my first post because I assumed we were dealing with square matrices. If m is not n, what is the 1-line proof?
     
  13. Apr 7, 2006 #12

    Hurkyl

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    [tex]
    A^TA \vec{v} = 0 \implies \vec{v}^T A^T A \vec{v} = 0
    \implies (A\vec{v})^T (A \vec{v}) = 0 \implies
    A \vec{v} = 0 \implies \vec{v} = 0[/tex]
    Where the last one follows because A has rank n and is operating on R^n.
     
  14. Apr 7, 2006 #13

    AKG

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    Very nice, thanks Hurkyl.
     
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