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Question about hat matrix X(X'X)^(-1)X'

  1. Oct 14, 2009 #1
    Hi everyone,

    I am working on the following problem.

    Suppose the set of vectors X1,..,Xk is a basis for linear space V1.
    Suppose the set of vectors Y1,..,Yk is also a basis for linear space
    V1.
    Clearly the linear space spanned by the Xs equals the linear space
    spanned by the Ys.

    Construct an algebraic argument to show that
    X(X'X)^(-1)X'=Y(Y'Y)^(-1)Y'

    I am very confused, I am not sure what is meant by algebraic argument
    in this instance, and I would welcome your ideas on how to tackle this
    question.

    Thanks in advance.
     
  2. jcsd
  3. Oct 14, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I am confused as to what is meant by "X" and "Y" here since you only mention, [itex]\{X_k\}[/itex] and [itex]\{Y_k\}[/itex] previously. But the reference to X', X-1, Y', and Y-1 indicate that X and Y are matrices or linear operators, not vectors. What is the relationship between X and [itex]\{X_k\}[/itex], between Y and [itex]\{Y_k\}[/itex]?

    If X and Y are matrices or linear operators then:
    It looks to me that once you use (AB)-1= B-1A-1 it becomes very easy to show that the two are the same!
     
  4. Oct 14, 2009 #3
    Sorry I meant to write

    X=[X1: X2 :...: Xk]
    Y=[Y1: Y2 :...: Yk]

    You are right, X and Y are matrices. Thanks a lot for the help.
     
  5. Oct 14, 2009 #4
    I have tried (AB)-1= B-1A-1, but I am not getting what I need. Would you be kind enough to give me another hint.
     
  6. Oct 15, 2009 #5

    HallsofIvy

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    Science Advisor

    You want to show that [itex]X(X'X)^{-1}X'=Y(Y'Y)^{-1}Y'[/itex].

    Actually, you can show much more than that.

    As I said, "[itex](AB)^{-1}= B^{-1}A^{-1}[/itex] so that [itex](X'X)^{-1}= X^{-1}X'^{-1}[/itex] and then [itex]X(X'X)^{-1}X'= X(X^{-1}X'^{-1})X'[/itex]
    Can you do that?
     
  7. Oct 15, 2009 #6
    I understand the lines you wrote. But I do not know where to take it from here.

    What should I do with
    LaTeX Code: X(X^{-1}Xsingle-quote^{-1})Xsingle-quote
    ?
    Thanks a lot for the help.
     
    Last edited: Oct 15, 2009
  8. Oct 15, 2009 #7
    Do I need to write
    X=PYP^{-1}
    where P changes the basis from Y to X?
    And plug in?
     
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