Syntax Explanation - Tranpose of Dot Product

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On page 29 of https://arxiv.org/pdf/1802.01528.pdf there is syntax that I find confusing:



excerpt from paper
What does the syntax [wT, b]T mean? Is this a 2 dimensional matrix formed from a horizontal row w (transposed from column) and a column of b, all transposed? What is the final shape? Same question for x hat.



Last, how does w.x+b then become w hat . x hat?






linear-algebra vectors matrix-calculus







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    0












    $begingroup$


    On page 29 of https://arxiv.org/pdf/1802.01528.pdf there is syntax that I find confusing:



    excerpt from paper
    What does the syntax [wT, b]T mean? Is this a 2 dimensional matrix formed from a horizontal row w (transposed from column) and a column of b, all transposed? What is the final shape? Same question for x hat.



    Last, how does w.x+b then become w hat . x hat?






    linear-algebra vectors matrix-calculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      On page 29 of https://arxiv.org/pdf/1802.01528.pdf there is syntax that I find confusing:



      excerpt from paper
      What does the syntax [wT, b]T mean? Is this a 2 dimensional matrix formed from a horizontal row w (transposed from column) and a column of b, all transposed? What is the final shape? Same question for x hat.



      Last, how does w.x+b then become w hat . x hat?






      linear-algebra vectors matrix-calculus







      share|cite|improve this question









      $endgroup$




      On page 29 of https://arxiv.org/pdf/1802.01528.pdf there is syntax that I find confusing:



      excerpt from paper
      What does the syntax [wT, b]T mean? Is this a 2 dimensional matrix formed from a horizontal row w (transposed from column) and a column of b, all transposed? What is the final shape? Same question for x hat.



      Last, how does w.x+b then become w hat . x hat?






      linear-algebra vectors matrix-calculus




      linear-algebra vectors matrix-calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 11 at 21:54









      Matthew ArthurMatthew Arthur

      183




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          2 Answers
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          $begingroup$

          If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
          $$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
          Similarly
          $$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$



          Finally,
          $$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How does the Transpose play into this?
            $endgroup$
            – Matthew Arthur
            Jan 11 at 22:02






          • 1




            $begingroup$
            @MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
            $endgroup$
            – angryavian
            Jan 11 at 22:05





















          0












          $begingroup$

          If we have
          $$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
          then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
          $$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
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            $begingroup$

            If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
            $$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
            Similarly
            $$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$



            Finally,
            $$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How does the Transpose play into this?
              $endgroup$
              – Matthew Arthur
              Jan 11 at 22:02






            • 1




              $begingroup$
              @MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
              $endgroup$
              – angryavian
              Jan 11 at 22:05


















            3












            $begingroup$

            If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
            $$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
            Similarly
            $$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$



            Finally,
            $$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How does the Transpose play into this?
              $endgroup$
              – Matthew Arthur
              Jan 11 at 22:02






            • 1




              $begingroup$
              @MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
              $endgroup$
              – angryavian
              Jan 11 at 22:05
















            3












            3








            3





            $begingroup$

            If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
            $$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
            Similarly
            $$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$



            Finally,
            $$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$






            share|cite|improve this answer









            $endgroup$



            If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
            $$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
            Similarly
            $$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$



            Finally,
            $$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 11 at 22:01









            angryavianangryavian

            40.5k23280




            40.5k23280












            • $begingroup$
              How does the Transpose play into this?
              $endgroup$
              – Matthew Arthur
              Jan 11 at 22:02






            • 1




              $begingroup$
              @MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
              $endgroup$
              – angryavian
              Jan 11 at 22:05




















            • $begingroup$
              How does the Transpose play into this?
              $endgroup$
              – Matthew Arthur
              Jan 11 at 22:02






            • 1




              $begingroup$
              @MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
              $endgroup$
              – angryavian
              Jan 11 at 22:05


















            $begingroup$
            How does the Transpose play into this?
            $endgroup$
            – Matthew Arthur
            Jan 11 at 22:02




            $begingroup$
            How does the Transpose play into this?
            $endgroup$
            – Matthew Arthur
            Jan 11 at 22:02




            1




            1




            $begingroup$
            @MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
            $endgroup$
            – angryavian
            Jan 11 at 22:05






            $begingroup$
            @MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
            $endgroup$
            – angryavian
            Jan 11 at 22:05













            0












            $begingroup$

            If we have
            $$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
            then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
            $$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              If we have
              $$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
              then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
              $$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                If we have
                $$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
                then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
                $$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$






                share|cite|improve this answer









                $endgroup$



                If we have
                $$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
                then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
                $$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 11 at 22:06









                Morgan RodgersMorgan Rodgers

                9,66721439




                9,66721439






























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