Mixing
Derivative of matrix inverse w.r.t. vector
$begingroup$
I need to differentiate the inverse of the $Ktimes K$ symmetric matrix $A$ w.r.t some vector (that $A$ depends on). Is there a rule for this? In case I do the derivative w.r.t. to some scalar there's this rule that
$DA^{-1}= -A^{-1}*DA*A^{-1}$.
Is this also valid in case of derivative w.r.t. vector or if not is there something equivalent?
Thanks!
linear-algebra matrices derivatives
edited Feb 19 '14 at 19:29
Davide Giraudo
128k17155268
asked Feb 17 '14 at 20:15
dummydummy
364
$endgroup$
add a comment |
$begingroup$
I need to differentiate the inverse of the $Ktimes K$ symmetric matrix $A$ w.r.t some vector (that $A$ depends on). Is there a rule for this? In case I do the derivative w.r.t. to some scalar there's this rule that
$DA^{-1}= -A^{-1}*DA*A^{-1}$.
Is this also valid in case of derivative w.r.t. vector or if not is there something equivalent?
Thanks!
linear-algebra matrices derivatives
edited Feb 19 '14 at 19:29
Davide Giraudo
128k17155268
asked Feb 17 '14 at 20:15
dummydummy
364
$endgroup$
$begingroup$
You can not derive a matrix with respect to a vector
$endgroup$
– Ahmad Bazzi
Sep 2 '16 at 11:37
$begingroup$
@AhmadBazzi The verb form of derivative is actually differentiate. When one derives something, he extrapolates it only from rudiments and shows it is true.
$endgroup$
– Chase Ryan Taylor
Nov 22 '17 at 6:37
add a comment |
$begingroup$
I need to differentiate the inverse of the $Ktimes K$ symmetric matrix $A$ w.r.t some vector (that $A$ depends on). Is there a rule for this? In case I do the derivative w.r.t. to some scalar there's this rule that
$DA^{-1}= -A^{-1}*DA*A^{-1}$.
Is this also valid in case of derivative w.r.t. vector or if not is there something equivalent?
Thanks!
linear-algebra matrices derivatives
edited Feb 19 '14 at 19:29
Davide Giraudo
128k17155268
asked Feb 17 '14 at 20:15
dummydummy
364
$endgroup$
I need to differentiate the inverse of the $Ktimes K$ symmetric matrix $A$ w.r.t some vector (that $A$ depends on). Is there a rule for this? In case I do the derivative w.r.t. to some scalar there's this rule that
$DA^{-1}= -A^{-1}*DA*A^{-1}$.
Is this also valid in case of derivative w.r.t. vector or if not is there something equivalent?
Thanks!
linear-algebra matrices derivatives
linear-algebra matrices derivatives
edited Feb 19 '14 at 19:29
Davide Giraudo
128k17155268
asked Feb 17 '14 at 20:15
dummydummy
364
edited Feb 19 '14 at 19:29
Davide Giraudo
128k17155268
asked Feb 17 '14 at 20:15
dummydummy
364
edited Feb 19 '14 at 19:29
Davide Giraudo
128k17155268
edited Feb 19 '14 at 19:29
Davide Giraudo
128k17155268
edited Feb 19 '14 at 19:29
Davide Giraudo
128k17155268
128k17155268
asked Feb 17 '14 at 20:15
dummydummy
364
asked Feb 17 '14 at 20:15
dummydummy
364
asked Feb 17 '14 at 20:15
dummydummy
364
364
$begingroup$
You can not derive a matrix with respect to a vector
$endgroup$
– Ahmad Bazzi
Sep 2 '16 at 11:37
$begingroup$
@AhmadBazzi The verb form of derivative is actually differentiate. When one derives something, he extrapolates it only from rudiments and shows it is true.
$endgroup$
– Chase Ryan Taylor
Nov 22 '17 at 6:37
add a comment |
$begingroup$
You can not derive a matrix with respect to a vector
$endgroup$
– Ahmad Bazzi
Sep 2 '16 at 11:37
$begingroup$
@AhmadBazzi The verb form of derivative is actually differentiate. When one derives something, he extrapolates it only from rudiments and shows it is true.
$endgroup$
– Chase Ryan Taylor
Nov 22 '17 at 6:37
$begingroup$
You can not derive a matrix with respect to a vector
$endgroup$
– Ahmad Bazzi
Sep 2 '16 at 11:37
$begingroup$
You can not derive a matrix with respect to a vector
$endgroup$
– Ahmad Bazzi
Sep 2 '16 at 11:37
$begingroup$
@AhmadBazzi The verb form of derivative is actually differentiate. When one derives something, he extrapolates it only from rudiments and shows it is true.
$endgroup$
– Chase Ryan Taylor
Nov 22 '17 at 6:37
$begingroup$
@AhmadBazzi The verb form of derivative is actually differentiate. When one derives something, he extrapolates it only from rudiments and shows it is true.
$endgroup$
– Chase Ryan Taylor
Nov 22 '17 at 6:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can find a list of rules for differentiating matrices by vectors or scalars by looking up "Matrix calculus".
A list of such rules is available on Wikipedia here in the "Identities" section.
answered Feb 19 '14 at 19:37
BatmanBatman
16.5k11735
$endgroup$
add a comment |
$begingroup$
Consider some function $A:u=[u_1,cdots,u_k]^Tin mathbb{R}^krightarrow A(u)=[a_{i,j}(u)]in GL_n(mathbb{R})$ and let
$$f:urightarrow (A(u))^{-1}.$$
Then $Df_u:h=[h_1,cdots,h_k]^Tinmathbb{R}^krightarrow -A^{-1}(u)DA_u(h)A^{-1}(u)in M_n(mathbb{R})$, where $DA_u(h)=[b_{i,j}(h)]in M_n(mathbb{R})$ (a linear application in $h$) is defined by
$b_{i,j}(h)=sum_{pleq k} dfrac{partial a_{i,j}(u)}{partial u_p}h_pinmathbb{R}$.
answered May 3 '17 at 12:32
loup blancloup blanc
24.1k21851
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
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votes
$begingroup$
You can find a list of rules for differentiating matrices by vectors or scalars by looking up "Matrix calculus".
A list of such rules is available on Wikipedia here in the "Identities" section.
answered Feb 19 '14 at 19:37
BatmanBatman
16.5k11735
$endgroup$
add a comment |
$begingroup$
You can find a list of rules for differentiating matrices by vectors or scalars by looking up "Matrix calculus".
A list of such rules is available on Wikipedia here in the "Identities" section.
answered Feb 19 '14 at 19:37
BatmanBatman
16.5k11735
$endgroup$
add a comment |
$begingroup$
You can find a list of rules for differentiating matrices by vectors or scalars by looking up "Matrix calculus".
A list of such rules is available on Wikipedia here in the "Identities" section.
answered Feb 19 '14 at 19:37
BatmanBatman
16.5k11735
$endgroup$
You can find a list of rules for differentiating matrices by vectors or scalars by looking up "Matrix calculus".
A list of such rules is available on Wikipedia here in the "Identities" section.
answered Feb 19 '14 at 19:37
BatmanBatman
16.5k11735
answered Feb 19 '14 at 19:37
BatmanBatman
16.5k11735
answered Feb 19 '14 at 19:37
BatmanBatman
16.5k11735
answered Feb 19 '14 at 19:37
BatmanBatman
16.5k11735
16.5k11735
add a comment |
add a comment |
$begingroup$
Consider some function $A:u=[u_1,cdots,u_k]^Tin mathbb{R}^krightarrow A(u)=[a_{i,j}(u)]in GL_n(mathbb{R})$ and let
$$f:urightarrow (A(u))^{-1}.$$
Then $Df_u:h=[h_1,cdots,h_k]^Tinmathbb{R}^krightarrow -A^{-1}(u)DA_u(h)A^{-1}(u)in M_n(mathbb{R})$, where $DA_u(h)=[b_{i,j}(h)]in M_n(mathbb{R})$ (a linear application in $h$) is defined by
$b_{i,j}(h)=sum_{pleq k} dfrac{partial a_{i,j}(u)}{partial u_p}h_pinmathbb{R}$.
answered May 3 '17 at 12:32
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
Consider some function $A:u=[u_1,cdots,u_k]^Tin mathbb{R}^krightarrow A(u)=[a_{i,j}(u)]in GL_n(mathbb{R})$ and let
$$f:urightarrow (A(u))^{-1}.$$
Then $Df_u:h=[h_1,cdots,h_k]^Tinmathbb{R}^krightarrow -A^{-1}(u)DA_u(h)A^{-1}(u)in M_n(mathbb{R})$, where $DA_u(h)=[b_{i,j}(h)]in M_n(mathbb{R})$ (a linear application in $h$) is defined by
$b_{i,j}(h)=sum_{pleq k} dfrac{partial a_{i,j}(u)}{partial u_p}h_pinmathbb{R}$.
answered May 3 '17 at 12:32
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
Consider some function $A:u=[u_1,cdots,u_k]^Tin mathbb{R}^krightarrow A(u)=[a_{i,j}(u)]in GL_n(mathbb{R})$ and let
$$f:urightarrow (A(u))^{-1}.$$
Then $Df_u:h=[h_1,cdots,h_k]^Tinmathbb{R}^krightarrow -A^{-1}(u)DA_u(h)A^{-1}(u)in M_n(mathbb{R})$, where $DA_u(h)=[b_{i,j}(h)]in M_n(mathbb{R})$ (a linear application in $h$) is defined by
$b_{i,j}(h)=sum_{pleq k} dfrac{partial a_{i,j}(u)}{partial u_p}h_pinmathbb{R}$.
answered May 3 '17 at 12:32
loup blancloup blanc
24.1k21851
$endgroup$
Consider some function $A:u=[u_1,cdots,u_k]^Tin mathbb{R}^krightarrow A(u)=[a_{i,j}(u)]in GL_n(mathbb{R})$ and let
$$f:urightarrow (A(u))^{-1}.$$
Then $Df_u:h=[h_1,cdots,h_k]^Tinmathbb{R}^krightarrow -A^{-1}(u)DA_u(h)A^{-1}(u)in M_n(mathbb{R})$, where $DA_u(h)=[b_{i,j}(h)]in M_n(mathbb{R})$ (a linear application in $h$) is defined by
$b_{i,j}(h)=sum_{pleq k} dfrac{partial a_{i,j}(u)}{partial u_p}h_pinmathbb{R}$.
answered May 3 '17 at 12:32
loup blancloup blanc
24.1k21851
answered May 3 '17 at 12:32
loup blancloup blanc
24.1k21851
answered May 3 '17 at 12:32
loup blancloup blanc
24.1k21851
answered May 3 '17 at 12:32
loup blancloup blanc
24.1k21851
24.1k21851
add a comment |
add a comment |
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$begingroup$
You can not derive a matrix with respect to a vector
$endgroup$
– Ahmad Bazzi
Sep 2 '16 at 11:37
$begingroup$
@AhmadBazzi The verb form of derivative is actually differentiate. When one derives something, he extrapolates it only from rudiments and shows it is true.
$endgroup$
– Chase Ryan Taylor
Nov 22 '17 at 6:37