Mixing
Question on the adjoint of a function
Clash Royale CLAN TAG#URR8PPP
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1) Let $E$ a vector space of finite dimension with scalar product $left<cdot,cdotright>$. We defined the adjoint of $f:Eto E$ the function $g$ defined as $$left<f(x),yright>=left<x,g(y)right>.$$
Let $S$ s.t. $left<x,yright>=x^TSy$. Let $F$ the matrix of $f$ and $G$ the matrix of $g$. If $S=I$, then
$$left<Fx,yright>=x^TF^Ty=left<x,F^Tyright>$$
and thus $G=F^T$. But if $Sneq I$, we have $$left<Fx,yright>=x^T F^T S y,$$
and how to get $G$ s.t. $left<x,Gyright>=left<Fx,yright>$ ?
2) We define in a more general way, the adjoint of $fin L(E,F)$ as $gin L(F^*,E^*)$ s.t. $g=fcirc u$. What is the link between the adjoint defined in 1) and the adjoint defined in 2) ?
linear-algebra adjoint-operators
edited Jul 25 at 14:16
Michael McGovern
5701314
asked Jul 25 at 13:19
user386627
714214
add a comment |Â
up vote
2
down vote
favorite
1) Let $E$ a vector space of finite dimension with scalar product $left<cdot,cdotright>$. We defined the adjoint of $f:Eto E$ the function $g$ defined as $$left<f(x),yright>=left<x,g(y)right>.$$
Let $S$ s.t. $left<x,yright>=x^TSy$. Let $F$ the matrix of $f$ and $G$ the matrix of $g$. If $S=I$, then
$$left<Fx,yright>=x^TF^Ty=left<x,F^Tyright>$$
and thus $G=F^T$. But if $Sneq I$, we have $$left<Fx,yright>=x^T F^T S y,$$
and how to get $G$ s.t. $left<x,Gyright>=left<Fx,yright>$ ?
2) We define in a more general way, the adjoint of $fin L(E,F)$ as $gin L(F^*,E^*)$ s.t. $g=fcirc u$. What is the link between the adjoint defined in 1) and the adjoint defined in 2) ?
linear-algebra adjoint-operators
edited Jul 25 at 14:16
Michael McGovern
5701314
asked Jul 25 at 13:19
user386627
714214
2
Instead of<and>it is better to uselangleandrangle.
– Jendrik Stelzner
Jul 25 at 13:35
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 25 at 13:39
1
Please post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta.
– Shaun
Jul 25 at 13:39
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
1) Let $E$ a vector space of finite dimension with scalar product $left<cdot,cdotright>$. We defined the adjoint of $f:Eto E$ the function $g$ defined as $$left<f(x),yright>=left<x,g(y)right>.$$
Let $S$ s.t. $left<x,yright>=x^TSy$. Let $F$ the matrix of $f$ and $G$ the matrix of $g$. If $S=I$, then
$$left<Fx,yright>=x^TF^Ty=left<x,F^Tyright>$$
and thus $G=F^T$. But if $Sneq I$, we have $$left<Fx,yright>=x^T F^T S y,$$
and how to get $G$ s.t. $left<x,Gyright>=left<Fx,yright>$ ?
2) We define in a more general way, the adjoint of $fin L(E,F)$ as $gin L(F^*,E^*)$ s.t. $g=fcirc u$. What is the link between the adjoint defined in 1) and the adjoint defined in 2) ?
linear-algebra adjoint-operators
edited Jul 25 at 14:16
Michael McGovern
5701314
asked Jul 25 at 13:19
user386627
714214
1) Let $E$ a vector space of finite dimension with scalar product $left<cdot,cdotright>$. We defined the adjoint of $f:Eto E$ the function $g$ defined as $$left<f(x),yright>=left<x,g(y)right>.$$
Let $S$ s.t. $left<x,yright>=x^TSy$. Let $F$ the matrix of $f$ and $G$ the matrix of $g$. If $S=I$, then
$$left<Fx,yright>=x^TF^Ty=left<x,F^Tyright>$$
and thus $G=F^T$. But if $Sneq I$, we have $$left<Fx,yright>=x^T F^T S y,$$
and how to get $G$ s.t. $left<x,Gyright>=left<Fx,yright>$ ?
2) We define in a more general way, the adjoint of $fin L(E,F)$ as $gin L(F^*,E^*)$ s.t. $g=fcirc u$. What is the link between the adjoint defined in 1) and the adjoint defined in 2) ?
linear-algebra adjoint-operators
edited Jul 25 at 14:16
Michael McGovern
5701314
asked Jul 25 at 13:19
user386627
714214
edited Jul 25 at 14:16
Michael McGovern
5701314
edited Jul 25 at 14:16
Michael McGovern
5701314
edited Jul 25 at 14:16
Michael McGovern
5701314
5701314
asked Jul 25 at 13:19
user386627
714214
asked Jul 25 at 13:19
user386627
714214
asked Jul 25 at 13:19
user386627
714214
714214
2
Instead of<and>it is better to uselangleandrangle.
– Jendrik Stelzner
Jul 25 at 13:35
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 25 at 13:39
1
Please post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta.
– Shaun
Jul 25 at 13:39
add a comment |Â
2
Instead of<and>it is better to uselangleandrangle.
– Jendrik Stelzner
Jul 25 at 13:35
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 25 at 13:39
1
Please post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta.
– Shaun
Jul 25 at 13:39
2
2
Instead of
< and > it is better to use langle and rangle.– Jendrik Stelzner
Jul 25 at 13:35
Instead of
< and > it is better to use langle and rangle.– Jendrik Stelzner
Jul 25 at 13:35
1
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 25 at 13:39
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 25 at 13:39
1
1
Please post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta.
– Shaun
Jul 25 at 13:39
Please post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta.
– Shaun
Jul 25 at 13:39
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
A change of basis is not necessary, just take $G = S^-1 F^T S$:
$$langle Fx,yrangle = x^TF^TSy = x^TSS^-1F^TSy = langle x,S^-1F^TSyrangle.$$
answered Jul 25 at 14:29
LinAlg
5,4111319
add a comment |Â
up vote
2
down vote
1) There is a basis s.t. $S$ is congruent to $I$ (i.e. there is a matrix $P$ invertible s.t. $S=P^TP$. Let denote $A=P^-1FP$ the matrix of $f$ in the new basis.
Then $$langle APx,Pyrangle= x^TP^TA^TPy=langle Px,A^TPyrangle,$$
and thus $F^T$ is the matrix of $g$ in the new basis.
$$F=PAP^-1quad textandquad G=PA^TP^-1.$$
2) You have that $$leftlangle u(x),frightrangle_F,F^*=leftlangle x,u^T(f)rightrangle_E,E^*.$$
answered Jul 25 at 13:30
Surb
36.3k84274
1
As said above, @Surb, please use$langle$and$rangle$for $langle$ and $rangle$, respectively, instead of $<$ and $>$.
– Shaun
Jul 25 at 13:43
There's no comma in $langle x^TP^TF^TPyrangle$.
– Shaun
Jul 25 at 13:46
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
A change of basis is not necessary, just take $G = S^-1 F^T S$:
$$langle Fx,yrangle = x^TF^TSy = x^TSS^-1F^TSy = langle x,S^-1F^TSyrangle.$$
answered Jul 25 at 14:29
LinAlg
5,4111319
add a comment |Â
up vote
3
down vote
accepted
A change of basis is not necessary, just take $G = S^-1 F^T S$:
$$langle Fx,yrangle = x^TF^TSy = x^TSS^-1F^TSy = langle x,S^-1F^TSyrangle.$$
answered Jul 25 at 14:29
LinAlg
5,4111319
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
A change of basis is not necessary, just take $G = S^-1 F^T S$:
$$langle Fx,yrangle = x^TF^TSy = x^TSS^-1F^TSy = langle x,S^-1F^TSyrangle.$$
answered Jul 25 at 14:29
LinAlg
5,4111319
A change of basis is not necessary, just take $G = S^-1 F^T S$:
$$langle Fx,yrangle = x^TF^TSy = x^TSS^-1F^TSy = langle x,S^-1F^TSyrangle.$$
answered Jul 25 at 14:29
LinAlg
5,4111319
answered Jul 25 at 14:29
LinAlg
5,4111319
answered Jul 25 at 14:29
LinAlg
5,4111319
answered Jul 25 at 14:29
LinAlg
5,4111319
5,4111319
add a comment |Â
add a comment |Â
up vote
2
down vote
1) There is a basis s.t. $S$ is congruent to $I$ (i.e. there is a matrix $P$ invertible s.t. $S=P^TP$. Let denote $A=P^-1FP$ the matrix of $f$ in the new basis.
Then $$langle APx,Pyrangle= x^TP^TA^TPy=langle Px,A^TPyrangle,$$
and thus $F^T$ is the matrix of $g$ in the new basis.
$$F=PAP^-1quad textandquad G=PA^TP^-1.$$
2) You have that $$leftlangle u(x),frightrangle_F,F^*=leftlangle x,u^T(f)rightrangle_E,E^*.$$
answered Jul 25 at 13:30
Surb
36.3k84274
1
As said above, @Surb, please use$langle$and$rangle$for $langle$ and $rangle$, respectively, instead of $<$ and $>$.
– Shaun
Jul 25 at 13:43
There's no comma in $langle x^TP^TF^TPyrangle$.
– Shaun
Jul 25 at 13:46
add a comment |Â
up vote
2
down vote
1) There is a basis s.t. $S$ is congruent to $I$ (i.e. there is a matrix $P$ invertible s.t. $S=P^TP$. Let denote $A=P^-1FP$ the matrix of $f$ in the new basis.
Then $$langle APx,Pyrangle= x^TP^TA^TPy=langle Px,A^TPyrangle,$$
and thus $F^T$ is the matrix of $g$ in the new basis.
$$F=PAP^-1quad textandquad G=PA^TP^-1.$$
2) You have that $$leftlangle u(x),frightrangle_F,F^*=leftlangle x,u^T(f)rightrangle_E,E^*.$$
answered Jul 25 at 13:30
Surb
36.3k84274
1
As said above, @Surb, please use$langle$and$rangle$for $langle$ and $rangle$, respectively, instead of $<$ and $>$.
– Shaun
Jul 25 at 13:43
There's no comma in $langle x^TP^TF^TPyrangle$.
– Shaun
Jul 25 at 13:46
add a comment |Â
up vote
2
down vote
up vote
2
down vote
1) There is a basis s.t. $S$ is congruent to $I$ (i.e. there is a matrix $P$ invertible s.t. $S=P^TP$. Let denote $A=P^-1FP$ the matrix of $f$ in the new basis.
Then $$langle APx,Pyrangle= x^TP^TA^TPy=langle Px,A^TPyrangle,$$
and thus $F^T$ is the matrix of $g$ in the new basis.
$$F=PAP^-1quad textandquad G=PA^TP^-1.$$
2) You have that $$leftlangle u(x),frightrangle_F,F^*=leftlangle x,u^T(f)rightrangle_E,E^*.$$
answered Jul 25 at 13:30
Surb
36.3k84274
1) There is a basis s.t. $S$ is congruent to $I$ (i.e. there is a matrix $P$ invertible s.t. $S=P^TP$. Let denote $A=P^-1FP$ the matrix of $f$ in the new basis.
Then $$langle APx,Pyrangle= x^TP^TA^TPy=langle Px,A^TPyrangle,$$
and thus $F^T$ is the matrix of $g$ in the new basis.
$$F=PAP^-1quad textandquad G=PA^TP^-1.$$
2) You have that $$leftlangle u(x),frightrangle_F,F^*=leftlangle x,u^T(f)rightrangle_E,E^*.$$
answered Jul 25 at 13:30
Surb
36.3k84274
edited Jul 25 at 14:03
answered Jul 25 at 13:30
Surb
36.3k84274
answered Jul 25 at 13:30
Surb
36.3k84274
answered Jul 25 at 13:30
Surb
36.3k84274
36.3k84274
1
As said above, @Surb, please use$langle$and$rangle$for $langle$ and $rangle$, respectively, instead of $<$ and $>$.
– Shaun
Jul 25 at 13:43
There's no comma in $langle x^TP^TF^TPyrangle$.
– Shaun
Jul 25 at 13:46
add a comment |Â
1
As said above, @Surb, please use$langle$and$rangle$for $langle$ and $rangle$, respectively, instead of $<$ and $>$.
– Shaun
Jul 25 at 13:43
There's no comma in $langle x^TP^TF^TPyrangle$.
– Shaun
Jul 25 at 13:46
1
1
As said above, @Surb, please use
$langle$ and $rangle$ for $langle$ and $rangle$, respectively, instead of $<$ and $>$.– Shaun
Jul 25 at 13:43
As said above, @Surb, please use
$langle$ and $rangle$ for $langle$ and $rangle$, respectively, instead of $<$ and $>$.– Shaun
Jul 25 at 13:43
There's no comma in $langle x^TP^TF^TPyrangle$.
– Shaun
Jul 25 at 13:46
There's no comma in $langle x^TP^TF^TPyrangle$.
– Shaun
Jul 25 at 13:46
add a comment |Â
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2
Instead of
<and>it is better to uselangleandrangle.– Jendrik Stelzner
Jul 25 at 13:35
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 25 at 13:39
1
Please post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta.
– Shaun
Jul 25 at 13:39