What does the following matrix expression equal to after differentiating it wrt. $dot{mathrm{x}}$ and then...











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Suppose I have the following expression:



$$dot{mathbf{x}}^intercalleft(mathbf{A}-mathbf{B}mathbf{D}^{-1}mathbf{B}^intercalright)dot{mathbf{x}}$$



where $mathbf{x}(t)in mathbb{R}^{m}$ is a vector and $mathbf{A}in mathbb{R}^{mtimes m}, mathbf{B}in mathbb{R}^{mtimes (m-n)}$ and $mathbf{D}in mathbb{R}^{(m-n)times(m-n)}$ are matrices that are dependent on $mathbf{x}(t)$.



Derivation of the expression with respect to $dot{mathbf{x}}$ yields



$$2dot{mathbf{x}}^intercalleft(mathbf{A}-mathbf{B}mathbf{D}^{-1}mathbf{B}^intercalright)$$



according to the rules discussed in this Wikipedia article about matrix calculus.



What happens if I would like to further derive this expression w.r.t. time?



So my question in short: What does this expression equal to?



$$dfrac{text{d}}{text{d}t}left(dfrac{partial}{partial dot{mathbf{x}}}left(dot{mathbf{x}}^intercalleft(mathbf{A}(mathbf{x}(t))-mathbf{B}(mathbf{x}(t))mathbf{D}^{-1}(mathbf{x}(t))mathbf{B}^intercal(mathbf{x}(t))right)dot{mathbf{x}}right)right)$$



Thank you in advance!










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    Suppose I have the following expression:



    $$dot{mathbf{x}}^intercalleft(mathbf{A}-mathbf{B}mathbf{D}^{-1}mathbf{B}^intercalright)dot{mathbf{x}}$$



    where $mathbf{x}(t)in mathbb{R}^{m}$ is a vector and $mathbf{A}in mathbb{R}^{mtimes m}, mathbf{B}in mathbb{R}^{mtimes (m-n)}$ and $mathbf{D}in mathbb{R}^{(m-n)times(m-n)}$ are matrices that are dependent on $mathbf{x}(t)$.



    Derivation of the expression with respect to $dot{mathbf{x}}$ yields



    $$2dot{mathbf{x}}^intercalleft(mathbf{A}-mathbf{B}mathbf{D}^{-1}mathbf{B}^intercalright)$$



    according to the rules discussed in this Wikipedia article about matrix calculus.



    What happens if I would like to further derive this expression w.r.t. time?



    So my question in short: What does this expression equal to?



    $$dfrac{text{d}}{text{d}t}left(dfrac{partial}{partial dot{mathbf{x}}}left(dot{mathbf{x}}^intercalleft(mathbf{A}(mathbf{x}(t))-mathbf{B}(mathbf{x}(t))mathbf{D}^{-1}(mathbf{x}(t))mathbf{B}^intercal(mathbf{x}(t))right)dot{mathbf{x}}right)right)$$



    Thank you in advance!










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Suppose I have the following expression:



      $$dot{mathbf{x}}^intercalleft(mathbf{A}-mathbf{B}mathbf{D}^{-1}mathbf{B}^intercalright)dot{mathbf{x}}$$



      where $mathbf{x}(t)in mathbb{R}^{m}$ is a vector and $mathbf{A}in mathbb{R}^{mtimes m}, mathbf{B}in mathbb{R}^{mtimes (m-n)}$ and $mathbf{D}in mathbb{R}^{(m-n)times(m-n)}$ are matrices that are dependent on $mathbf{x}(t)$.



      Derivation of the expression with respect to $dot{mathbf{x}}$ yields



      $$2dot{mathbf{x}}^intercalleft(mathbf{A}-mathbf{B}mathbf{D}^{-1}mathbf{B}^intercalright)$$



      according to the rules discussed in this Wikipedia article about matrix calculus.



      What happens if I would like to further derive this expression w.r.t. time?



      So my question in short: What does this expression equal to?



      $$dfrac{text{d}}{text{d}t}left(dfrac{partial}{partial dot{mathbf{x}}}left(dot{mathbf{x}}^intercalleft(mathbf{A}(mathbf{x}(t))-mathbf{B}(mathbf{x}(t))mathbf{D}^{-1}(mathbf{x}(t))mathbf{B}^intercal(mathbf{x}(t))right)dot{mathbf{x}}right)right)$$



      Thank you in advance!










      share|cite|improve this question













      Suppose I have the following expression:



      $$dot{mathbf{x}}^intercalleft(mathbf{A}-mathbf{B}mathbf{D}^{-1}mathbf{B}^intercalright)dot{mathbf{x}}$$



      where $mathbf{x}(t)in mathbb{R}^{m}$ is a vector and $mathbf{A}in mathbb{R}^{mtimes m}, mathbf{B}in mathbb{R}^{mtimes (m-n)}$ and $mathbf{D}in mathbb{R}^{(m-n)times(m-n)}$ are matrices that are dependent on $mathbf{x}(t)$.



      Derivation of the expression with respect to $dot{mathbf{x}}$ yields



      $$2dot{mathbf{x}}^intercalleft(mathbf{A}-mathbf{B}mathbf{D}^{-1}mathbf{B}^intercalright)$$



      according to the rules discussed in this Wikipedia article about matrix calculus.



      What happens if I would like to further derive this expression w.r.t. time?



      So my question in short: What does this expression equal to?



      $$dfrac{text{d}}{text{d}t}left(dfrac{partial}{partial dot{mathbf{x}}}left(dot{mathbf{x}}^intercalleft(mathbf{A}(mathbf{x}(t))-mathbf{B}(mathbf{x}(t))mathbf{D}^{-1}(mathbf{x}(t))mathbf{B}^intercal(mathbf{x}(t))right)dot{mathbf{x}}right)right)$$



      Thank you in advance!







      matrix-calculus






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      asked Nov 15 at 23:35









      Sphery

      628




      628






















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          Define the variables
          $$eqalign{
          v &= {dot x},,,,,,M&=A-BD^{-1}B^T cr
          {mathcal E} &= v^TMv &= M:vv^T cr
          }$$

          where colon is a convenient product notation for the trace, i.e. $,,A:B={rm Tr}(A^TB)$



          Now calculate the differential and gradient of the scalar function
          $$eqalign{
          d{mathcal E}
          &= M:(dv,v^T+v,dv^T) cr
          &= (M+M^T):(dv,v^T) cr
          &= (M+M^T)v:dv cr
          p=frac{partial{mathcal E}}{partial v} &= (M+M^T)v cr
          }$$

          NB:   If $M=M^T$ we can simplify this to $,,p = 2Mv$.



          Now take the time derivative of the $p$ vector
          $$eqalign{
          {dot p} &= 2{dot M}v + 2M{dot v} cr
          }$$

          Since you were vague on how your matrices depend on the $x$-vector, that's about as far as I can take it.






          share|cite|improve this answer





















          • I would have been more intrigued by what is $dot{mathbf{M}}$ in terms of $mathbf{A}$, $mathbf{B}$ and $mathbf{D}$. I have since discovered, that $dfrac{text{d}mathbf{A}(mathbf{x}(t))}{text{d}t} = dfrac{partialmathbf{A}}{partial mathbf{x}}left(mathbf{I}_motimesmathbf{dot{mathbf{x}}}right)$ if $mathbf{x}(t)in mathbb{R}^{m}$.
            – Sphery
            Nov 17 at 18:29











          Your Answer





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          1 Answer
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          1 Answer
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          active

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          active

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          up vote
          1
          down vote













          Define the variables
          $$eqalign{
          v &= {dot x},,,,,,M&=A-BD^{-1}B^T cr
          {mathcal E} &= v^TMv &= M:vv^T cr
          }$$

          where colon is a convenient product notation for the trace, i.e. $,,A:B={rm Tr}(A^TB)$



          Now calculate the differential and gradient of the scalar function
          $$eqalign{
          d{mathcal E}
          &= M:(dv,v^T+v,dv^T) cr
          &= (M+M^T):(dv,v^T) cr
          &= (M+M^T)v:dv cr
          p=frac{partial{mathcal E}}{partial v} &= (M+M^T)v cr
          }$$

          NB:   If $M=M^T$ we can simplify this to $,,p = 2Mv$.



          Now take the time derivative of the $p$ vector
          $$eqalign{
          {dot p} &= 2{dot M}v + 2M{dot v} cr
          }$$

          Since you were vague on how your matrices depend on the $x$-vector, that's about as far as I can take it.






          share|cite|improve this answer





















          • I would have been more intrigued by what is $dot{mathbf{M}}$ in terms of $mathbf{A}$, $mathbf{B}$ and $mathbf{D}$. I have since discovered, that $dfrac{text{d}mathbf{A}(mathbf{x}(t))}{text{d}t} = dfrac{partialmathbf{A}}{partial mathbf{x}}left(mathbf{I}_motimesmathbf{dot{mathbf{x}}}right)$ if $mathbf{x}(t)in mathbb{R}^{m}$.
            – Sphery
            Nov 17 at 18:29















          up vote
          1
          down vote













          Define the variables
          $$eqalign{
          v &= {dot x},,,,,,M&=A-BD^{-1}B^T cr
          {mathcal E} &= v^TMv &= M:vv^T cr
          }$$

          where colon is a convenient product notation for the trace, i.e. $,,A:B={rm Tr}(A^TB)$



          Now calculate the differential and gradient of the scalar function
          $$eqalign{
          d{mathcal E}
          &= M:(dv,v^T+v,dv^T) cr
          &= (M+M^T):(dv,v^T) cr
          &= (M+M^T)v:dv cr
          p=frac{partial{mathcal E}}{partial v} &= (M+M^T)v cr
          }$$

          NB:   If $M=M^T$ we can simplify this to $,,p = 2Mv$.



          Now take the time derivative of the $p$ vector
          $$eqalign{
          {dot p} &= 2{dot M}v + 2M{dot v} cr
          }$$

          Since you were vague on how your matrices depend on the $x$-vector, that's about as far as I can take it.






          share|cite|improve this answer





















          • I would have been more intrigued by what is $dot{mathbf{M}}$ in terms of $mathbf{A}$, $mathbf{B}$ and $mathbf{D}$. I have since discovered, that $dfrac{text{d}mathbf{A}(mathbf{x}(t))}{text{d}t} = dfrac{partialmathbf{A}}{partial mathbf{x}}left(mathbf{I}_motimesmathbf{dot{mathbf{x}}}right)$ if $mathbf{x}(t)in mathbb{R}^{m}$.
            – Sphery
            Nov 17 at 18:29













          up vote
          1
          down vote










          up vote
          1
          down vote









          Define the variables
          $$eqalign{
          v &= {dot x},,,,,,M&=A-BD^{-1}B^T cr
          {mathcal E} &= v^TMv &= M:vv^T cr
          }$$

          where colon is a convenient product notation for the trace, i.e. $,,A:B={rm Tr}(A^TB)$



          Now calculate the differential and gradient of the scalar function
          $$eqalign{
          d{mathcal E}
          &= M:(dv,v^T+v,dv^T) cr
          &= (M+M^T):(dv,v^T) cr
          &= (M+M^T)v:dv cr
          p=frac{partial{mathcal E}}{partial v} &= (M+M^T)v cr
          }$$

          NB:   If $M=M^T$ we can simplify this to $,,p = 2Mv$.



          Now take the time derivative of the $p$ vector
          $$eqalign{
          {dot p} &= 2{dot M}v + 2M{dot v} cr
          }$$

          Since you were vague on how your matrices depend on the $x$-vector, that's about as far as I can take it.






          share|cite|improve this answer












          Define the variables
          $$eqalign{
          v &= {dot x},,,,,,M&=A-BD^{-1}B^T cr
          {mathcal E} &= v^TMv &= M:vv^T cr
          }$$

          where colon is a convenient product notation for the trace, i.e. $,,A:B={rm Tr}(A^TB)$



          Now calculate the differential and gradient of the scalar function
          $$eqalign{
          d{mathcal E}
          &= M:(dv,v^T+v,dv^T) cr
          &= (M+M^T):(dv,v^T) cr
          &= (M+M^T)v:dv cr
          p=frac{partial{mathcal E}}{partial v} &= (M+M^T)v cr
          }$$

          NB:   If $M=M^T$ we can simplify this to $,,p = 2Mv$.



          Now take the time derivative of the $p$ vector
          $$eqalign{
          {dot p} &= 2{dot M}v + 2M{dot v} cr
          }$$

          Since you were vague on how your matrices depend on the $x$-vector, that's about as far as I can take it.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 16 at 17:16









          john316

          111




          111












          • I would have been more intrigued by what is $dot{mathbf{M}}$ in terms of $mathbf{A}$, $mathbf{B}$ and $mathbf{D}$. I have since discovered, that $dfrac{text{d}mathbf{A}(mathbf{x}(t))}{text{d}t} = dfrac{partialmathbf{A}}{partial mathbf{x}}left(mathbf{I}_motimesmathbf{dot{mathbf{x}}}right)$ if $mathbf{x}(t)in mathbb{R}^{m}$.
            – Sphery
            Nov 17 at 18:29


















          • I would have been more intrigued by what is $dot{mathbf{M}}$ in terms of $mathbf{A}$, $mathbf{B}$ and $mathbf{D}$. I have since discovered, that $dfrac{text{d}mathbf{A}(mathbf{x}(t))}{text{d}t} = dfrac{partialmathbf{A}}{partial mathbf{x}}left(mathbf{I}_motimesmathbf{dot{mathbf{x}}}right)$ if $mathbf{x}(t)in mathbb{R}^{m}$.
            – Sphery
            Nov 17 at 18:29
















          I would have been more intrigued by what is $dot{mathbf{M}}$ in terms of $mathbf{A}$, $mathbf{B}$ and $mathbf{D}$. I have since discovered, that $dfrac{text{d}mathbf{A}(mathbf{x}(t))}{text{d}t} = dfrac{partialmathbf{A}}{partial mathbf{x}}left(mathbf{I}_motimesmathbf{dot{mathbf{x}}}right)$ if $mathbf{x}(t)in mathbb{R}^{m}$.
          – Sphery
          Nov 17 at 18:29




          I would have been more intrigued by what is $dot{mathbf{M}}$ in terms of $mathbf{A}$, $mathbf{B}$ and $mathbf{D}$. I have since discovered, that $dfrac{text{d}mathbf{A}(mathbf{x}(t))}{text{d}t} = dfrac{partialmathbf{A}}{partial mathbf{x}}left(mathbf{I}_motimesmathbf{dot{mathbf{x}}}right)$ if $mathbf{x}(t)in mathbb{R}^{m}$.
          – Sphery
          Nov 17 at 18:29


















           

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