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!
matrix-calculus
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up vote
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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!
matrix-calculus
add a comment |
up vote
0
down vote
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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!
matrix-calculus
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
matrix-calculus
asked Nov 15 at 23:35
Sphery
628
628
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1 Answer
1
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1
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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.
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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