derivative of inverse matrix by itself
$begingroup$
Let $A$ be a matrix, supposedly $ktimes k$ matrix.
I know that
$$frac{partial A^{-1}}{partial A} = -A^{-2} $$
I do not know how I am supposed to obtain the following results using this fact. I want to know the step of
$$frac{partial a^top A^{-1} b}{partial A} = -(A^top)^{-1}ab^top (A^top)^{-1} $$
Also, I want to know the solution to
$$frac{partial (A^top)^{-1}ab^top (A^top)^{-1} }{partial A} = ? $$
matrix-calculus
asked Dec 3 '18 at 5:42
user1292919user1292919
755512
$endgroup$
add a comment |
$begingroup$
Let $A$ be a matrix, supposedly $ktimes k$ matrix.
I know that
$$frac{partial A^{-1}}{partial A} = -A^{-2} $$
I do not know how I am supposed to obtain the following results using this fact. I want to know the step of
$$frac{partial a^top A^{-1} b}{partial A} = -(A^top)^{-1}ab^top (A^top)^{-1} $$
Also, I want to know the solution to
$$frac{partial (A^top)^{-1}ab^top (A^top)^{-1} }{partial A} = ? $$
matrix-calculus
asked Dec 3 '18 at 5:42
user1292919user1292919
755512
$endgroup$
2
$begingroup$
First formula; is not it $-A^ {-2}$ ?
$endgroup$
– Damien
Dec 3 '18 at 6:36
$begingroup$
@Damien Yes, i editted
$endgroup$
– user1292919
Dec 3 '18 at 15:26
$begingroup$
Do you just know the first identity you stated or also how to obtain it? Because if so, you should be able to figure out the other identities too. However, please let us know what your attempts are.
$endgroup$
– James
Dec 3 '18 at 15:41
add a comment |
$begingroup$
Let $A$ be a matrix, supposedly $ktimes k$ matrix.
I know that
$$frac{partial A^{-1}}{partial A} = -A^{-2} $$
I do not know how I am supposed to obtain the following results using this fact. I want to know the step of
$$frac{partial a^top A^{-1} b}{partial A} = -(A^top)^{-1}ab^top (A^top)^{-1} $$
Also, I want to know the solution to
$$frac{partial (A^top)^{-1}ab^top (A^top)^{-1} }{partial A} = ? $$
matrix-calculus
asked Dec 3 '18 at 5:42
user1292919user1292919
755512
$endgroup$
Let $A$ be a matrix, supposedly $ktimes k$ matrix.
I know that
$$frac{partial A^{-1}}{partial A} = -A^{-2} $$
I do not know how I am supposed to obtain the following results using this fact. I want to know the step of
$$frac{partial a^top A^{-1} b}{partial A} = -(A^top)^{-1}ab^top (A^top)^{-1} $$
Also, I want to know the solution to
$$frac{partial (A^top)^{-1}ab^top (A^top)^{-1} }{partial A} = ? $$
matrix-calculus
matrix-calculus
asked Dec 3 '18 at 5:42
user1292919user1292919
755512
asked Dec 3 '18 at 5:42
user1292919user1292919
755512
edited Dec 3 '18 at 15:26
user1292919
asked Dec 3 '18 at 5:42
user1292919user1292919
755512
asked Dec 3 '18 at 5:42
user1292919user1292919
755512
asked Dec 3 '18 at 5:42
user1292919user1292919
755512
755512
2
$begingroup$
First formula; is not it $-A^ {-2}$ ?
$endgroup$
– Damien
Dec 3 '18 at 6:36
$begingroup$
@Damien Yes, i editted
$endgroup$
– user1292919
Dec 3 '18 at 15:26
$begingroup$
Do you just know the first identity you stated or also how to obtain it? Because if so, you should be able to figure out the other identities too. However, please let us know what your attempts are.
$endgroup$
– James
Dec 3 '18 at 15:41
add a comment |
2
$begingroup$
First formula; is not it $-A^ {-2}$ ?
$endgroup$
– Damien
Dec 3 '18 at 6:36
$begingroup$
@Damien Yes, i editted
$endgroup$
– user1292919
Dec 3 '18 at 15:26
$begingroup$
Do you just know the first identity you stated or also how to obtain it? Because if so, you should be able to figure out the other identities too. However, please let us know what your attempts are.
$endgroup$
– James
Dec 3 '18 at 15:41
2
2
$begingroup$
First formula; is not it $-A^ {-2}$ ?
$endgroup$
– Damien
Dec 3 '18 at 6:36
$begingroup$
First formula; is not it $-A^ {-2}$ ?
$endgroup$
– Damien
Dec 3 '18 at 6:36
$begingroup$
@Damien Yes, i editted
$endgroup$
– user1292919
Dec 3 '18 at 15:26
$begingroup$
@Damien Yes, i editted
$endgroup$
– user1292919
Dec 3 '18 at 15:26
$begingroup$
Do you just know the first identity you stated or also how to obtain it? Because if so, you should be able to figure out the other identities too. However, please let us know what your attempts are.
$endgroup$
– James
Dec 3 '18 at 15:41
$begingroup$
Do you just know the first identity you stated or also how to obtain it? Because if so, you should be able to figure out the other identities too. However, please let us know what your attempts are.
$endgroup$
– James
Dec 3 '18 at 15:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Start with the defining equation for the matrix inverse and find its differential.
$$eqalign{
I &= A^{-1}A cr
0 &= dA^{-1},A + A^{-1},dA cr
0 &= dA^{-1},A + A^{-1},dA,A^{-1} cr
dA^{-1} &= -A^{-1},dA,A^{-1} cr
}$$
Next note the gradient of a matrix with respect to itself.
$$
{mathcal H}_{ijkl} = frac{partial A_{ij}}{partial A_{kl}} = delta_{ik}delta_{jl}
$$
Note that ${mathcal H}$ is a 4th order tensor with some interesting symmetry properties (isotropic). It is also the identity element for the Frobenius product, i.e. for any matrix $B$
$${mathcal H}:B=B:{mathcal H}=B$$
Now we can answer your first question. The function of interest is scalar-valued. Let's find its differential and gradient
$$eqalign{
phi &= a^TA^{-1}b cr &= ab^T:A^{-1} cr
dphi &= ab^T:dA^{-1} cr &= -ab^T:A^{-1},dA,A^{-1} cr
&= -A^{-T}ab^TA^{-T}:dA cr
frac{partialphi}{partial A} &= -A^{-T}ab^TA^{-T} cr
}$$
Now let's try the second question. This time the function of interest is matrix-valued.
$$eqalign{
F &= A^{-1}ab^TA^{-1} cr
dF &= dA^{-1}ab^TA^{-1} + A^{-1}ab^TdA^{-1} cr
&= -A^{-1},dA,A^{-1}ab^TA^{-1} - A^{-1}ab^TA^{-1},dA,A^{-1} cr
&= -A^{-1},dA,F - F,dA,A^{-1} cr
&= -Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA cr
frac{partial F}{partial A} &= -Big(A^{-1}{mathcal H}F^T+F{mathcal H}A^{-T}Big) cr
}$$
This gradient is a 4th order tensor.
If you prefer, you can vectorize the matrices to flatten the result.
$$eqalign{
{rm vec}(dF) &= -{rm vec}(-A^{-1},dA,F + F,dA,A^{-1}) cr
&= -(F^Totimes A^{-1} + A^{-T}otimes F),{rm vec}(dA) cr
df &= -(F^Totimes A^{-1} + A^{-T}otimes F),da cr
frac{partial f}{partial a} &= -Big(F^Totimes A^{-1} + A^{-T}otimes FBig) crcr
}$$
In some step above, a colon was used to denote the Frobenius (double-contraction) product
$$eqalign{
A &= {mathcal H}:B &implies &A_{ij} &= sum_{kl}{mathcal H}_{ijkl} B_{kl} cr alpha &= H:B &implies &alpha &= sum_{ij}H_{ij} B_{ij} = {rm Tr}(H^TB) cr
}$$
answered Dec 3 '18 at 15:35
greggreg
7,7701821
$endgroup$
1
$begingroup$
it might be obvious to you and others, but can you explain how to reach from this step $-A^{-1},dA,F - F,dA,A^{-1}$ to this step $-Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA$?
$endgroup$
– user550103
Dec 4 '18 at 6:04
1
$begingroup$
It's not obvious, it's just one of the properties of ${mathcal H}$. Work out a simple example like $(A{mathcal H}B:X)$ in index notation and, utilizing the properties of those Kronecker deltas, you'll find that it equals $(AXB^T)$
$endgroup$
– greg
Dec 4 '18 at 13:14
1
$begingroup$
It's also closely related to the equally non-obvious formula $${rm vec}(AXB^T) = (Botimes A),{rm vec}(X)$$ which uses a Kronecker product instead of Kronecker deltas.
$endgroup$
– greg
Dec 4 '18 at 13:22
$begingroup$
Thank you for the explanation, greg!
$endgroup$
– user550103
Dec 5 '18 at 9:58
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Start with the defining equation for the matrix inverse and find its differential.
$$eqalign{
I &= A^{-1}A cr
0 &= dA^{-1},A + A^{-1},dA cr
0 &= dA^{-1},A + A^{-1},dA,A^{-1} cr
dA^{-1} &= -A^{-1},dA,A^{-1} cr
}$$
Next note the gradient of a matrix with respect to itself.
$$
{mathcal H}_{ijkl} = frac{partial A_{ij}}{partial A_{kl}} = delta_{ik}delta_{jl}
$$
Note that ${mathcal H}$ is a 4th order tensor with some interesting symmetry properties (isotropic). It is also the identity element for the Frobenius product, i.e. for any matrix $B$
$${mathcal H}:B=B:{mathcal H}=B$$
Now we can answer your first question. The function of interest is scalar-valued. Let's find its differential and gradient
$$eqalign{
phi &= a^TA^{-1}b cr &= ab^T:A^{-1} cr
dphi &= ab^T:dA^{-1} cr &= -ab^T:A^{-1},dA,A^{-1} cr
&= -A^{-T}ab^TA^{-T}:dA cr
frac{partialphi}{partial A} &= -A^{-T}ab^TA^{-T} cr
}$$
Now let's try the second question. This time the function of interest is matrix-valued.
$$eqalign{
F &= A^{-1}ab^TA^{-1} cr
dF &= dA^{-1}ab^TA^{-1} + A^{-1}ab^TdA^{-1} cr
&= -A^{-1},dA,A^{-1}ab^TA^{-1} - A^{-1}ab^TA^{-1},dA,A^{-1} cr
&= -A^{-1},dA,F - F,dA,A^{-1} cr
&= -Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA cr
frac{partial F}{partial A} &= -Big(A^{-1}{mathcal H}F^T+F{mathcal H}A^{-T}Big) cr
}$$
This gradient is a 4th order tensor.
If you prefer, you can vectorize the matrices to flatten the result.
$$eqalign{
{rm vec}(dF) &= -{rm vec}(-A^{-1},dA,F + F,dA,A^{-1}) cr
&= -(F^Totimes A^{-1} + A^{-T}otimes F),{rm vec}(dA) cr
df &= -(F^Totimes A^{-1} + A^{-T}otimes F),da cr
frac{partial f}{partial a} &= -Big(F^Totimes A^{-1} + A^{-T}otimes FBig) crcr
}$$
In some step above, a colon was used to denote the Frobenius (double-contraction) product
$$eqalign{
A &= {mathcal H}:B &implies &A_{ij} &= sum_{kl}{mathcal H}_{ijkl} B_{kl} cr alpha &= H:B &implies &alpha &= sum_{ij}H_{ij} B_{ij} = {rm Tr}(H^TB) cr
}$$
answered Dec 3 '18 at 15:35
greggreg
7,7701821
$endgroup$
1
$begingroup$
it might be obvious to you and others, but can you explain how to reach from this step $-A^{-1},dA,F - F,dA,A^{-1}$ to this step $-Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA$?
$endgroup$
– user550103
Dec 4 '18 at 6:04
1
$begingroup$
It's not obvious, it's just one of the properties of ${mathcal H}$. Work out a simple example like $(A{mathcal H}B:X)$ in index notation and, utilizing the properties of those Kronecker deltas, you'll find that it equals $(AXB^T)$
$endgroup$
– greg
Dec 4 '18 at 13:14
1
$begingroup$
It's also closely related to the equally non-obvious formula $${rm vec}(AXB^T) = (Botimes A),{rm vec}(X)$$ which uses a Kronecker product instead of Kronecker deltas.
$endgroup$
– greg
Dec 4 '18 at 13:22
$begingroup$
Thank you for the explanation, greg!
$endgroup$
– user550103
Dec 5 '18 at 9:58
add a comment |
$begingroup$
Start with the defining equation for the matrix inverse and find its differential.
$$eqalign{
I &= A^{-1}A cr
0 &= dA^{-1},A + A^{-1},dA cr
0 &= dA^{-1},A + A^{-1},dA,A^{-1} cr
dA^{-1} &= -A^{-1},dA,A^{-1} cr
}$$
Next note the gradient of a matrix with respect to itself.
$$
{mathcal H}_{ijkl} = frac{partial A_{ij}}{partial A_{kl}} = delta_{ik}delta_{jl}
$$
Note that ${mathcal H}$ is a 4th order tensor with some interesting symmetry properties (isotropic). It is also the identity element for the Frobenius product, i.e. for any matrix $B$
$${mathcal H}:B=B:{mathcal H}=B$$
Now we can answer your first question. The function of interest is scalar-valued. Let's find its differential and gradient
$$eqalign{
phi &= a^TA^{-1}b cr &= ab^T:A^{-1} cr
dphi &= ab^T:dA^{-1} cr &= -ab^T:A^{-1},dA,A^{-1} cr
&= -A^{-T}ab^TA^{-T}:dA cr
frac{partialphi}{partial A} &= -A^{-T}ab^TA^{-T} cr
}$$
Now let's try the second question. This time the function of interest is matrix-valued.
$$eqalign{
F &= A^{-1}ab^TA^{-1} cr
dF &= dA^{-1}ab^TA^{-1} + A^{-1}ab^TdA^{-1} cr
&= -A^{-1},dA,A^{-1}ab^TA^{-1} - A^{-1}ab^TA^{-1},dA,A^{-1} cr
&= -A^{-1},dA,F - F,dA,A^{-1} cr
&= -Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA cr
frac{partial F}{partial A} &= -Big(A^{-1}{mathcal H}F^T+F{mathcal H}A^{-T}Big) cr
}$$
This gradient is a 4th order tensor.
If you prefer, you can vectorize the matrices to flatten the result.
$$eqalign{
{rm vec}(dF) &= -{rm vec}(-A^{-1},dA,F + F,dA,A^{-1}) cr
&= -(F^Totimes A^{-1} + A^{-T}otimes F),{rm vec}(dA) cr
df &= -(F^Totimes A^{-1} + A^{-T}otimes F),da cr
frac{partial f}{partial a} &= -Big(F^Totimes A^{-1} + A^{-T}otimes FBig) crcr
}$$
In some step above, a colon was used to denote the Frobenius (double-contraction) product
$$eqalign{
A &= {mathcal H}:B &implies &A_{ij} &= sum_{kl}{mathcal H}_{ijkl} B_{kl} cr alpha &= H:B &implies &alpha &= sum_{ij}H_{ij} B_{ij} = {rm Tr}(H^TB) cr
}$$
answered Dec 3 '18 at 15:35
greggreg
7,7701821
$endgroup$
1
$begingroup$
it might be obvious to you and others, but can you explain how to reach from this step $-A^{-1},dA,F - F,dA,A^{-1}$ to this step $-Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA$?
$endgroup$
– user550103
Dec 4 '18 at 6:04
1
$begingroup$
It's not obvious, it's just one of the properties of ${mathcal H}$. Work out a simple example like $(A{mathcal H}B:X)$ in index notation and, utilizing the properties of those Kronecker deltas, you'll find that it equals $(AXB^T)$
$endgroup$
– greg
Dec 4 '18 at 13:14
1
$begingroup$
It's also closely related to the equally non-obvious formula $${rm vec}(AXB^T) = (Botimes A),{rm vec}(X)$$ which uses a Kronecker product instead of Kronecker deltas.
$endgroup$
– greg
Dec 4 '18 at 13:22
$begingroup$
Thank you for the explanation, greg!
$endgroup$
– user550103
Dec 5 '18 at 9:58
add a comment |
$begingroup$
Start with the defining equation for the matrix inverse and find its differential.
$$eqalign{
I &= A^{-1}A cr
0 &= dA^{-1},A + A^{-1},dA cr
0 &= dA^{-1},A + A^{-1},dA,A^{-1} cr
dA^{-1} &= -A^{-1},dA,A^{-1} cr
}$$
Next note the gradient of a matrix with respect to itself.
$$
{mathcal H}_{ijkl} = frac{partial A_{ij}}{partial A_{kl}} = delta_{ik}delta_{jl}
$$
Note that ${mathcal H}$ is a 4th order tensor with some interesting symmetry properties (isotropic). It is also the identity element for the Frobenius product, i.e. for any matrix $B$
$${mathcal H}:B=B:{mathcal H}=B$$
Now we can answer your first question. The function of interest is scalar-valued. Let's find its differential and gradient
$$eqalign{
phi &= a^TA^{-1}b cr &= ab^T:A^{-1} cr
dphi &= ab^T:dA^{-1} cr &= -ab^T:A^{-1},dA,A^{-1} cr
&= -A^{-T}ab^TA^{-T}:dA cr
frac{partialphi}{partial A} &= -A^{-T}ab^TA^{-T} cr
}$$
Now let's try the second question. This time the function of interest is matrix-valued.
$$eqalign{
F &= A^{-1}ab^TA^{-1} cr
dF &= dA^{-1}ab^TA^{-1} + A^{-1}ab^TdA^{-1} cr
&= -A^{-1},dA,A^{-1}ab^TA^{-1} - A^{-1}ab^TA^{-1},dA,A^{-1} cr
&= -A^{-1},dA,F - F,dA,A^{-1} cr
&= -Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA cr
frac{partial F}{partial A} &= -Big(A^{-1}{mathcal H}F^T+F{mathcal H}A^{-T}Big) cr
}$$
This gradient is a 4th order tensor.
If you prefer, you can vectorize the matrices to flatten the result.
$$eqalign{
{rm vec}(dF) &= -{rm vec}(-A^{-1},dA,F + F,dA,A^{-1}) cr
&= -(F^Totimes A^{-1} + A^{-T}otimes F),{rm vec}(dA) cr
df &= -(F^Totimes A^{-1} + A^{-T}otimes F),da cr
frac{partial f}{partial a} &= -Big(F^Totimes A^{-1} + A^{-T}otimes FBig) crcr
}$$
In some step above, a colon was used to denote the Frobenius (double-contraction) product
$$eqalign{
A &= {mathcal H}:B &implies &A_{ij} &= sum_{kl}{mathcal H}_{ijkl} B_{kl} cr alpha &= H:B &implies &alpha &= sum_{ij}H_{ij} B_{ij} = {rm Tr}(H^TB) cr
}$$
answered Dec 3 '18 at 15:35
greggreg
7,7701821
$endgroup$
Start with the defining equation for the matrix inverse and find its differential.
$$eqalign{
I &= A^{-1}A cr
0 &= dA^{-1},A + A^{-1},dA cr
0 &= dA^{-1},A + A^{-1},dA,A^{-1} cr
dA^{-1} &= -A^{-1},dA,A^{-1} cr
}$$
Next note the gradient of a matrix with respect to itself.
$$
{mathcal H}_{ijkl} = frac{partial A_{ij}}{partial A_{kl}} = delta_{ik}delta_{jl}
$$
Note that ${mathcal H}$ is a 4th order tensor with some interesting symmetry properties (isotropic). It is also the identity element for the Frobenius product, i.e. for any matrix $B$
$${mathcal H}:B=B:{mathcal H}=B$$
Now we can answer your first question. The function of interest is scalar-valued. Let's find its differential and gradient
$$eqalign{
phi &= a^TA^{-1}b cr &= ab^T:A^{-1} cr
dphi &= ab^T:dA^{-1} cr &= -ab^T:A^{-1},dA,A^{-1} cr
&= -A^{-T}ab^TA^{-T}:dA cr
frac{partialphi}{partial A} &= -A^{-T}ab^TA^{-T} cr
}$$
Now let's try the second question. This time the function of interest is matrix-valued.
$$eqalign{
F &= A^{-1}ab^TA^{-1} cr
dF &= dA^{-1}ab^TA^{-1} + A^{-1}ab^TdA^{-1} cr
&= -A^{-1},dA,A^{-1}ab^TA^{-1} - A^{-1}ab^TA^{-1},dA,A^{-1} cr
&= -A^{-1},dA,F - F,dA,A^{-1} cr
&= -Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA cr
frac{partial F}{partial A} &= -Big(A^{-1}{mathcal H}F^T+F{mathcal H}A^{-T}Big) cr
}$$
This gradient is a 4th order tensor.
If you prefer, you can vectorize the matrices to flatten the result.
$$eqalign{
{rm vec}(dF) &= -{rm vec}(-A^{-1},dA,F + F,dA,A^{-1}) cr
&= -(F^Totimes A^{-1} + A^{-T}otimes F),{rm vec}(dA) cr
df &= -(F^Totimes A^{-1} + A^{-T}otimes F),da cr
frac{partial f}{partial a} &= -Big(F^Totimes A^{-1} + A^{-T}otimes FBig) crcr
}$$
In some step above, a colon was used to denote the Frobenius (double-contraction) product
$$eqalign{
A &= {mathcal H}:B &implies &A_{ij} &= sum_{kl}{mathcal H}_{ijkl} B_{kl} cr alpha &= H:B &implies &alpha &= sum_{ij}H_{ij} B_{ij} = {rm Tr}(H^TB) cr
}$$
answered Dec 3 '18 at 15:35
greggreg
7,7701821
edited Dec 3 '18 at 16:07
answered Dec 3 '18 at 15:35
greggreg
7,7701821
answered Dec 3 '18 at 15:35
greggreg
7,7701821
answered Dec 3 '18 at 15:35
greggreg
7,7701821
7,7701821
1
$begingroup$
it might be obvious to you and others, but can you explain how to reach from this step $-A^{-1},dA,F - F,dA,A^{-1}$ to this step $-Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA$?
$endgroup$
– user550103
Dec 4 '18 at 6:04
1
$begingroup$
It's not obvious, it's just one of the properties of ${mathcal H}$. Work out a simple example like $(A{mathcal H}B:X)$ in index notation and, utilizing the properties of those Kronecker deltas, you'll find that it equals $(AXB^T)$
$endgroup$
– greg
Dec 4 '18 at 13:14
1
$begingroup$
It's also closely related to the equally non-obvious formula $${rm vec}(AXB^T) = (Botimes A),{rm vec}(X)$$ which uses a Kronecker product instead of Kronecker deltas.
$endgroup$
– greg
Dec 4 '18 at 13:22
$begingroup$
Thank you for the explanation, greg!
$endgroup$
– user550103
Dec 5 '18 at 9:58
add a comment |
1
$begingroup$
it might be obvious to you and others, but can you explain how to reach from this step $-A^{-1},dA,F - F,dA,A^{-1}$ to this step $-Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA$?
$endgroup$
– user550103
Dec 4 '18 at 6:04
1
$begingroup$
It's not obvious, it's just one of the properties of ${mathcal H}$. Work out a simple example like $(A{mathcal H}B:X)$ in index notation and, utilizing the properties of those Kronecker deltas, you'll find that it equals $(AXB^T)$
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– greg
Dec 4 '18 at 13:14
1
$begingroup$
It's also closely related to the equally non-obvious formula $${rm vec}(AXB^T) = (Botimes A),{rm vec}(X)$$ which uses a Kronecker product instead of Kronecker deltas.
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– greg
Dec 4 '18 at 13:22
$begingroup$
Thank you for the explanation, greg!
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– user550103
Dec 5 '18 at 9:58
1
1
$begingroup$
it might be obvious to you and others, but can you explain how to reach from this step $-A^{-1},dA,F - F,dA,A^{-1}$ to this step $-Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA$?
$endgroup$
– user550103
Dec 4 '18 at 6:04
$begingroup$
it might be obvious to you and others, but can you explain how to reach from this step $-A^{-1},dA,F - F,dA,A^{-1}$ to this step $-Big(A^{-1}{mathcal H}F^T + F{mathcal H}A^{-T}Big):dA$?
$endgroup$
– user550103
Dec 4 '18 at 6:04
1
1
$begingroup$
It's not obvious, it's just one of the properties of ${mathcal H}$. Work out a simple example like $(A{mathcal H}B:X)$ in index notation and, utilizing the properties of those Kronecker deltas, you'll find that it equals $(AXB^T)$
$endgroup$
– greg
Dec 4 '18 at 13:14
$begingroup$
It's not obvious, it's just one of the properties of ${mathcal H}$. Work out a simple example like $(A{mathcal H}B:X)$ in index notation and, utilizing the properties of those Kronecker deltas, you'll find that it equals $(AXB^T)$
$endgroup$
– greg
Dec 4 '18 at 13:14
1
1
$begingroup$
It's also closely related to the equally non-obvious formula $${rm vec}(AXB^T) = (Botimes A),{rm vec}(X)$$ which uses a Kronecker product instead of Kronecker deltas.
$endgroup$
– greg
Dec 4 '18 at 13:22
$begingroup$
It's also closely related to the equally non-obvious formula $${rm vec}(AXB^T) = (Botimes A),{rm vec}(X)$$ which uses a Kronecker product instead of Kronecker deltas.
$endgroup$
– greg
Dec 4 '18 at 13:22
$begingroup$
Thank you for the explanation, greg!
$endgroup$
– user550103
Dec 5 '18 at 9:58
$begingroup$
Thank you for the explanation, greg!
$endgroup$
– user550103
Dec 5 '18 at 9:58
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2
$begingroup$
First formula; is not it $-A^ {-2}$ ?
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– Damien
Dec 3 '18 at 6:36
$begingroup$
@Damien Yes, i editted
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– user1292919
Dec 3 '18 at 15:26
$begingroup$
Do you just know the first identity you stated or also how to obtain it? Because if so, you should be able to figure out the other identities too. However, please let us know what your attempts are.
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– James
Dec 3 '18 at 15:41