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Math Help - how to multiply out "transpose"?

  1. #1
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    how to multiply out "transpose"?

    Hi

    I am really confused with the "transpose" related stuff.

    Can anyone work me through it?

    and the paritial differentiation with respect to it

    ===========================
    why is

    (Y-X\beta)^T(Y-X\beta)=Y^TY-2\beta^TX^TY^T+\beta^TX^TX\beta

    instead of

    (Y-X\beta)^T(Y-X\beta)<br />
=(Y^T-\beta^TX^T)(Y-X\beta)<br />
=Y^TY-Y^TX\beta-\beta^TX^TY+\beta^TX^TX\beta

    and differentiate with respect to beta is

    =-2X^TY+X^TX\beta
    Attached Thumbnails Attached Thumbnails how to multiply out &quot;transpose&quot;?-1.png   how to multiply out &quot;transpose&quot;?-2.png  
    Last edited by casperyc; April 24th 2010 at 03:51 PM.
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  2. #2
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    Quote Originally Posted by casperyc View Post
    Hi

    I am really confused with the "transpose" related stuff.

    Can anyone work me through it?

    and the paritial differentiation with respect to it
    If we are working with matrices, how do we know beta and beta hat commute? Or are we working with vector? What is beta?

    (\hat{\beta}-\beta)(\hat{\beta}^T-\beta^T)=\hat{\beta}\hat{\beta}^T-\hat{\beta}\beta^T-\beta\hat{\beta}^T+\beta\beta^T
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    If we are working with matrices, how do we know beta and beta hat commute? Or are we working with vector? What is beta?

    (\hat{\beta}-\beta)(\hat{\beta}^T-\beta^T)=\hat{\beta}\hat{\beta}^T-\hat{\beta}\beta^T-\beta\hat{\beta}^T+\beta\beta^T
    Hi there,

    I think what you have writtern is pretty obvious going to give me my answer.

    So, my actual problem is dont understand the order of the multiplication,

    say  (A-B)(A^T-B^T)

    what I would do is

    (A-B)(A^T-B^T)=A A^T - AB^T - B A^T + BB^T

    which give me a different 'middle' term...

    \beta is just a vector


    and if you could see the second pic (which is just next to it)
    it confuses me more....
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  4. #4
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    Quote Originally Posted by casperyc View Post
    Hi there,

    I think what you have writtern is pretty obvious going to give me my answer.

    So, my actual problem is dont understand the order of the multiplication,

    say  (A-B)(A^T-B^T)

    what I would do is

    (A-B)(A^T-B^T)=A A^T - AB^T - B A^T + BB^T

    which give me a different 'middle' term...

    \beta is just a vector
    Your multiplication is correct. You will receive two different middle terms. If we can show the terms commute, we could combine them.
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  5. #5
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    If beta is a vector, \beta\hat{\beta}^T is inner product.
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    If beta is a vector, \beta\hat{\beta}^T is inner product.

    Thanks.
    I will think it through

    Then what about the differentiation thing?
    (which I have typed it out in the post)
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  7. #7
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    Are the Xs and Ys matrices?
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  8. #8
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    Quote Originally Posted by dwsmith View Post
    Are the Xs and Ys matrices?
    just suppose

    y=n \times 1

    X=n \times p

    \beta=p \times 1
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  9. #9
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    Quote Originally Posted by dwsmith View Post
    If we are working with matrices, how do we know beta and beta hat commute? Or are we working with vector? What is beta?

    (\hat{\beta}-\beta)(\hat{\beta}^T-\beta^T)=\hat{\beta}\hat{\beta}^T-\hat{\beta}\beta^T-\beta\hat{\beta}^T+\beta\beta^T
    Since these are vectors and we are dealing with inner product, (\hat{\beta}-\beta)(\hat{\beta}^T-\beta^T)=\hat{\beta}\hat{\beta}^T-\hat{\beta}\beta^T-\beta\hat{\beta}^T+\beta\beta^T=\hat{\beta}\hat{\b  eta}^T-2\hat{\beta}\beta^T+\beta\beta^T
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  10. #10
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    If in the second problem if beta is a scalar and X and Y are vectors, the quantity will simplify down to Y^TY-2Y^TX\beta+X^TX\beta
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