understanding linear algebra of a forget gate












2












$begingroup$


This blog covers the basics of LSTMs.



A forget gate is defined as :



$$f_t = sigma(W_f cdot [h_{t-1}, x_t]+ b_f)$$



At this point the linear algebra confuses me more than it should. The syntax of $Wcdot [h,x]$ is confusing in this context. I think a vector should go into the activation function since the output $f$ is a vector, but the syntax of the forget gate above implies that the input has $2$ columns because $[h,x]$ will be an $ntimes 2$ matrix



For the sake of example lets say ...



begin{align} W &= begin{bmatrix} 0 & 1 \
2 &3 end{bmatrix}\
h &= begin{bmatrix} -1 \
2 end{bmatrix}\
x &= begin{bmatrix} 3 \
0 end{bmatrix}\
b &= begin{bmatrix} 1 \
-2 end{bmatrix}end{align}



Can anyone give the final vector that goes into the sigmoid function ?



I think the math is



$$ begin{bmatrix} 0 & 1 \ 2 & 3 end{bmatrix}begin{bmatrix} -3 & 3 \ 2 & 0 end{bmatrix} + begin{bmatrix} 1 \ -2end{bmatrix} = begin{bmatrix} 2 & 0 \ 4 & 6end{bmatrix}+ begin{bmatrix} 1 \ -2end{bmatrix} = text{ Something wrong}$$






neural-network lstm rnn







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New contributor




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$endgroup$

















    2












    $begingroup$


    This blog covers the basics of LSTMs.



    A forget gate is defined as :



    $$f_t = sigma(W_f cdot [h_{t-1}, x_t]+ b_f)$$



    At this point the linear algebra confuses me more than it should. The syntax of $Wcdot [h,x]$ is confusing in this context. I think a vector should go into the activation function since the output $f$ is a vector, but the syntax of the forget gate above implies that the input has $2$ columns because $[h,x]$ will be an $ntimes 2$ matrix



    For the sake of example lets say ...



    begin{align} W &= begin{bmatrix} 0 & 1 \
    2 &3 end{bmatrix}\
    h &= begin{bmatrix} -1 \
    2 end{bmatrix}\
    x &= begin{bmatrix} 3 \
    0 end{bmatrix}\
    b &= begin{bmatrix} 1 \
    -2 end{bmatrix}end{align}



    Can anyone give the final vector that goes into the sigmoid function ?



    I think the math is



    $$ begin{bmatrix} 0 & 1 \ 2 & 3 end{bmatrix}begin{bmatrix} -3 & 3 \ 2 & 0 end{bmatrix} + begin{bmatrix} 1 \ -2end{bmatrix} = begin{bmatrix} 2 & 0 \ 4 & 6end{bmatrix}+ begin{bmatrix} 1 \ -2end{bmatrix} = text{ Something wrong}$$






    neural-network lstm rnn







    share|improve this question









    New contributor




    sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2





      $begingroup$


      This blog covers the basics of LSTMs.



      A forget gate is defined as :



      $$f_t = sigma(W_f cdot [h_{t-1}, x_t]+ b_f)$$



      At this point the linear algebra confuses me more than it should. The syntax of $Wcdot [h,x]$ is confusing in this context. I think a vector should go into the activation function since the output $f$ is a vector, but the syntax of the forget gate above implies that the input has $2$ columns because $[h,x]$ will be an $ntimes 2$ matrix



      For the sake of example lets say ...



      begin{align} W &= begin{bmatrix} 0 & 1 \
      2 &3 end{bmatrix}\
      h &= begin{bmatrix} -1 \
      2 end{bmatrix}\
      x &= begin{bmatrix} 3 \
      0 end{bmatrix}\
      b &= begin{bmatrix} 1 \
      -2 end{bmatrix}end{align}



      Can anyone give the final vector that goes into the sigmoid function ?



      I think the math is



      $$ begin{bmatrix} 0 & 1 \ 2 & 3 end{bmatrix}begin{bmatrix} -3 & 3 \ 2 & 0 end{bmatrix} + begin{bmatrix} 1 \ -2end{bmatrix} = begin{bmatrix} 2 & 0 \ 4 & 6end{bmatrix}+ begin{bmatrix} 1 \ -2end{bmatrix} = text{ Something wrong}$$






      neural-network lstm rnn







      share|improve this question









      New contributor




      sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      This blog covers the basics of LSTMs.



      A forget gate is defined as :



      $$f_t = sigma(W_f cdot [h_{t-1}, x_t]+ b_f)$$



      At this point the linear algebra confuses me more than it should. The syntax of $Wcdot [h,x]$ is confusing in this context. I think a vector should go into the activation function since the output $f$ is a vector, but the syntax of the forget gate above implies that the input has $2$ columns because $[h,x]$ will be an $ntimes 2$ matrix



      For the sake of example lets say ...



      begin{align} W &= begin{bmatrix} 0 & 1 \
      2 &3 end{bmatrix}\
      h &= begin{bmatrix} -1 \
      2 end{bmatrix}\
      x &= begin{bmatrix} 3 \
      0 end{bmatrix}\
      b &= begin{bmatrix} 1 \
      -2 end{bmatrix}end{align}



      Can anyone give the final vector that goes into the sigmoid function ?



      I think the math is



      $$ begin{bmatrix} 0 & 1 \ 2 & 3 end{bmatrix}begin{bmatrix} -3 & 3 \ 2 & 0 end{bmatrix} + begin{bmatrix} 1 \ -2end{bmatrix} = begin{bmatrix} 2 & 0 \ 4 & 6end{bmatrix}+ begin{bmatrix} 1 \ -2end{bmatrix} = text{ Something wrong}$$






      neural-network lstm rnn




      neural-network lstm rnn






      share|improve this question









      New contributor




      sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 14 hours ago









      Siong Thye Goh

      1,197418




      1,197418






      New contributor




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      asked 17 hours ago









      samsam

      112




      112




      New contributor




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      New contributor





      sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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          2 Answers
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          $begingroup$

          I interpret it as



          $$f_t = sigma left(W_fcdot begin{bmatrix} h_{t-1} \ x_tend{bmatrix} + b_fright)$$



          That is $W_f$ has as many columns as the entries of $h_{t-1}$ and $x_t$. $W_f$ also has as many rows as $b_f$. This would make the dimension matches and prodcues a vector output.






          share|improve this answer











          $endgroup$





















            1












            $begingroup$

            Note that $$[h_{t-1}, x_t]$$
            is the concatenation of two vectors.
            In your example, it would be: $$[h_{t-1}, x_t] = [-1, 2 , 3, 0]$$
            and then the dimensions of $W_f$ would be $2 times 4$, where $2$ is the dimension of the output of the LSTM cell, i.e. the activation $h_t$, that you defined to be of dimension $2$.



            Hence, $$W_f cdot [h_{t-1}, x_t] $$ is a multiplication of a matrix of dimension $2times4$ by a vector of $4$, which will return a vector of dimesion $2$. And then the sigmoid function will be applied point wise on each of the two elements of the result.



            Hope it makes sense.






            share|improve this answer











            $endgroup$













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              2 Answers
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              2 Answers
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              1












              $begingroup$

              I interpret it as



              $$f_t = sigma left(W_fcdot begin{bmatrix} h_{t-1} \ x_tend{bmatrix} + b_fright)$$



              That is $W_f$ has as many columns as the entries of $h_{t-1}$ and $x_t$. $W_f$ also has as many rows as $b_f$. This would make the dimension matches and prodcues a vector output.






              share|improve this answer











              $endgroup$


















                1












                $begingroup$

                I interpret it as



                $$f_t = sigma left(W_fcdot begin{bmatrix} h_{t-1} \ x_tend{bmatrix} + b_fright)$$



                That is $W_f$ has as many columns as the entries of $h_{t-1}$ and $x_t$. $W_f$ also has as many rows as $b_f$. This would make the dimension matches and prodcues a vector output.






                share|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I interpret it as



                  $$f_t = sigma left(W_fcdot begin{bmatrix} h_{t-1} \ x_tend{bmatrix} + b_fright)$$



                  That is $W_f$ has as many columns as the entries of $h_{t-1}$ and $x_t$. $W_f$ also has as many rows as $b_f$. This would make the dimension matches and prodcues a vector output.






                  share|improve this answer











                  $endgroup$



                  I interpret it as



                  $$f_t = sigma left(W_fcdot begin{bmatrix} h_{t-1} \ x_tend{bmatrix} + b_fright)$$



                  That is $W_f$ has as many columns as the entries of $h_{t-1}$ and $x_t$. $W_f$ also has as many rows as $b_f$. This would make the dimension matches and prodcues a vector output.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 14 hours ago

























                  answered 14 hours ago









                  Siong Thye GohSiong Thye Goh

                  1,197418




                  1,197418























                      1












                      $begingroup$

                      Note that $$[h_{t-1}, x_t]$$
                      is the concatenation of two vectors.
                      In your example, it would be: $$[h_{t-1}, x_t] = [-1, 2 , 3, 0]$$
                      and then the dimensions of $W_f$ would be $2 times 4$, where $2$ is the dimension of the output of the LSTM cell, i.e. the activation $h_t$, that you defined to be of dimension $2$.



                      Hence, $$W_f cdot [h_{t-1}, x_t] $$ is a multiplication of a matrix of dimension $2times4$ by a vector of $4$, which will return a vector of dimesion $2$. And then the sigmoid function will be applied point wise on each of the two elements of the result.



                      Hope it makes sense.






                      share|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        Note that $$[h_{t-1}, x_t]$$
                        is the concatenation of two vectors.
                        In your example, it would be: $$[h_{t-1}, x_t] = [-1, 2 , 3, 0]$$
                        and then the dimensions of $W_f$ would be $2 times 4$, where $2$ is the dimension of the output of the LSTM cell, i.e. the activation $h_t$, that you defined to be of dimension $2$.



                        Hence, $$W_f cdot [h_{t-1}, x_t] $$ is a multiplication of a matrix of dimension $2times4$ by a vector of $4$, which will return a vector of dimesion $2$. And then the sigmoid function will be applied point wise on each of the two elements of the result.



                        Hope it makes sense.






                        share|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Note that $$[h_{t-1}, x_t]$$
                          is the concatenation of two vectors.
                          In your example, it would be: $$[h_{t-1}, x_t] = [-1, 2 , 3, 0]$$
                          and then the dimensions of $W_f$ would be $2 times 4$, where $2$ is the dimension of the output of the LSTM cell, i.e. the activation $h_t$, that you defined to be of dimension $2$.



                          Hence, $$W_f cdot [h_{t-1}, x_t] $$ is a multiplication of a matrix of dimension $2times4$ by a vector of $4$, which will return a vector of dimesion $2$. And then the sigmoid function will be applied point wise on each of the two elements of the result.



                          Hope it makes sense.






                          share|improve this answer











                          $endgroup$



                          Note that $$[h_{t-1}, x_t]$$
                          is the concatenation of two vectors.
                          In your example, it would be: $$[h_{t-1}, x_t] = [-1, 2 , 3, 0]$$
                          and then the dimensions of $W_f$ would be $2 times 4$, where $2$ is the dimension of the output of the LSTM cell, i.e. the activation $h_t$, that you defined to be of dimension $2$.



                          Hence, $$W_f cdot [h_{t-1}, x_t] $$ is a multiplication of a matrix of dimension $2times4$ by a vector of $4$, which will return a vector of dimesion $2$. And then the sigmoid function will be applied point wise on each of the two elements of the result.



                          Hope it makes sense.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited 3 hours ago

























                          answered 13 hours ago









                          EscachatorEscachator

                          309111




                          309111






















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