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Definition of the adjoint of a linear application
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Let $E$ and $F$ finite vector space. I know that if $uin mathcal L(E,F)$ we define the adjoint $u^*$ of $u$ as $u^*(f)=f(u)$, i.e. $u^*in mathcal L(F^*,E^*)$. It's written in my book that $$left<u(x),fright>_F,F^*=left<x,u^*(f)right>_E,E^*,$$
for all $fin F^*$ and all $xin E$.
I would like to make a similarity with the construction of the adjoint in a inner space. I recall that if $(E,left<cdot ,cdot right>)$ is a inner space of finite dimension, and $Tin mathcal L(E)$, then for $vin E$ fixed, $$varphi (u)=left<T(u),vright>,$$
is a linear form, and thus, there is a unique $v(T)$ s.t. $$left< T(u),vright>=varphi(u)=left<u,v(T)right> .$$
We denote $v(T):= T^*v$. The application $T^*$ is well defined since for all $vin V$, there is a unique $win V$ s.t. $w=T^*v$.
Now, I'm trying to reproduce this construction.
First try If I define $$varphi(x):=left<u(x),fright>_F,F^*$$
then $varphiin E^*$. I don't know theorem in this general case that tels me that there is a unique $u^*(f)in E^*$ s.t. $varphi(x)=left<x,u^*(f)right>$.
Second try So I tried to use that bijection $$Psi :Fto F^**$$ defined as $$Psi(x)(f)=left<f,xright>:= f(x),quad fin F^*.$$
Unfortunately $xlongmapsto left<u(x),fright>$ is an element of $E^*$ and not of $E^**$, so I can't conclude.
So is there a similarity between the definition of the adjoint of $u$ when we are in inner space and when we are in more general spaces ?
linear-algebra adjoint-operators
asked 7 hours ago
Henri
62
add a comment |Â
up vote
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Let $E$ and $F$ finite vector space. I know that if $uin mathcal L(E,F)$ we define the adjoint $u^*$ of $u$ as $u^*(f)=f(u)$, i.e. $u^*in mathcal L(F^*,E^*)$. It's written in my book that $$left<u(x),fright>_F,F^*=left<x,u^*(f)right>_E,E^*,$$
for all $fin F^*$ and all $xin E$.
I would like to make a similarity with the construction of the adjoint in a inner space. I recall that if $(E,left<cdot ,cdot right>)$ is a inner space of finite dimension, and $Tin mathcal L(E)$, then for $vin E$ fixed, $$varphi (u)=left<T(u),vright>,$$
is a linear form, and thus, there is a unique $v(T)$ s.t. $$left< T(u),vright>=varphi(u)=left<u,v(T)right> .$$
We denote $v(T):= T^*v$. The application $T^*$ is well defined since for all $vin V$, there is a unique $win V$ s.t. $w=T^*v$.
Now, I'm trying to reproduce this construction.
First try If I define $$varphi(x):=left<u(x),fright>_F,F^*$$
then $varphiin E^*$. I don't know theorem in this general case that tels me that there is a unique $u^*(f)in E^*$ s.t. $varphi(x)=left<x,u^*(f)right>$.
Second try So I tried to use that bijection $$Psi :Fto F^**$$ defined as $$Psi(x)(f)=left<f,xright>:= f(x),quad fin F^*.$$
Unfortunately $xlongmapsto left<u(x),fright>$ is an element of $E^*$ and not of $E^**$, so I can't conclude.
So is there a similarity between the definition of the adjoint of $u$ when we are in inner space and when we are in more general spaces ?
linear-algebra adjoint-operators
asked 7 hours ago
Henri
62
I'm not totally sure, but I don't think that there is a corelation between the definition of an adjoint of an application in a inner space and the adjoint of $uin mathcal L(E,F)$. I think that the fact that in both case we call it adjoint come from the fact that in both case, the matrix of the adjoint is given by the transpose of the matrix of the application (i.e. if $uin L(E,F)$, $A=(u)_B_FB_E$ then $(u^*)_B_E^*B_F^*=A^t$ and if $uin L(E)$ and be orthonormal, then if $A=(u)_BB$ then $(u^*)_BB=A^t$).
– Surb
7 hours ago
To justify your formula, just use the definition $$left<u(x),fright>=f(u(x))=f(u)(x)=left<x,f(u)right>=left<x,u^*(f)right>.$$ But it would be better to wait for the answer of someone that knows better the subject.
– Surb
7 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $E$ and $F$ finite vector space. I know that if $uin mathcal L(E,F)$ we define the adjoint $u^*$ of $u$ as $u^*(f)=f(u)$, i.e. $u^*in mathcal L(F^*,E^*)$. It's written in my book that $$left<u(x),fright>_F,F^*=left<x,u^*(f)right>_E,E^*,$$
for all $fin F^*$ and all $xin E$.
I would like to make a similarity with the construction of the adjoint in a inner space. I recall that if $(E,left<cdot ,cdot right>)$ is a inner space of finite dimension, and $Tin mathcal L(E)$, then for $vin E$ fixed, $$varphi (u)=left<T(u),vright>,$$
is a linear form, and thus, there is a unique $v(T)$ s.t. $$left< T(u),vright>=varphi(u)=left<u,v(T)right> .$$
We denote $v(T):= T^*v$. The application $T^*$ is well defined since for all $vin V$, there is a unique $win V$ s.t. $w=T^*v$.
Now, I'm trying to reproduce this construction.
First try If I define $$varphi(x):=left<u(x),fright>_F,F^*$$
then $varphiin E^*$. I don't know theorem in this general case that tels me that there is a unique $u^*(f)in E^*$ s.t. $varphi(x)=left<x,u^*(f)right>$.
Second try So I tried to use that bijection $$Psi :Fto F^**$$ defined as $$Psi(x)(f)=left<f,xright>:= f(x),quad fin F^*.$$
Unfortunately $xlongmapsto left<u(x),fright>$ is an element of $E^*$ and not of $E^**$, so I can't conclude.
So is there a similarity between the definition of the adjoint of $u$ when we are in inner space and when we are in more general spaces ?
linear-algebra adjoint-operators
asked 7 hours ago
Henri
62
Let $E$ and $F$ finite vector space. I know that if $uin mathcal L(E,F)$ we define the adjoint $u^*$ of $u$ as $u^*(f)=f(u)$, i.e. $u^*in mathcal L(F^*,E^*)$. It's written in my book that $$left<u(x),fright>_F,F^*=left<x,u^*(f)right>_E,E^*,$$
for all $fin F^*$ and all $xin E$.
I would like to make a similarity with the construction of the adjoint in a inner space. I recall that if $(E,left<cdot ,cdot right>)$ is a inner space of finite dimension, and $Tin mathcal L(E)$, then for $vin E$ fixed, $$varphi (u)=left<T(u),vright>,$$
is a linear form, and thus, there is a unique $v(T)$ s.t. $$left< T(u),vright>=varphi(u)=left<u,v(T)right> .$$
We denote $v(T):= T^*v$. The application $T^*$ is well defined since for all $vin V$, there is a unique $win V$ s.t. $w=T^*v$.
Now, I'm trying to reproduce this construction.
First try If I define $$varphi(x):=left<u(x),fright>_F,F^*$$
then $varphiin E^*$. I don't know theorem in this general case that tels me that there is a unique $u^*(f)in E^*$ s.t. $varphi(x)=left<x,u^*(f)right>$.
Second try So I tried to use that bijection $$Psi :Fto F^**$$ defined as $$Psi(x)(f)=left<f,xright>:= f(x),quad fin F^*.$$
Unfortunately $xlongmapsto left<u(x),fright>$ is an element of $E^*$ and not of $E^**$, so I can't conclude.
So is there a similarity between the definition of the adjoint of $u$ when we are in inner space and when we are in more general spaces ?
linear-algebra adjoint-operators
asked 7 hours ago
Henri
62
asked 7 hours ago
Henri
62
asked 7 hours ago
Henri
62
asked 7 hours ago
Henri
62
62
I'm not totally sure, but I don't think that there is a corelation between the definition of an adjoint of an application in a inner space and the adjoint of $uin mathcal L(E,F)$. I think that the fact that in both case we call it adjoint come from the fact that in both case, the matrix of the adjoint is given by the transpose of the matrix of the application (i.e. if $uin L(E,F)$, $A=(u)_B_FB_E$ then $(u^*)_B_E^*B_F^*=A^t$ and if $uin L(E)$ and be orthonormal, then if $A=(u)_BB$ then $(u^*)_BB=A^t$).
– Surb
7 hours ago
To justify your formula, just use the definition $$left<u(x),fright>=f(u(x))=f(u)(x)=left<x,f(u)right>=left<x,u^*(f)right>.$$ But it would be better to wait for the answer of someone that knows better the subject.
– Surb
7 hours ago
add a comment |Â
I'm not totally sure, but I don't think that there is a corelation between the definition of an adjoint of an application in a inner space and the adjoint of $uin mathcal L(E,F)$. I think that the fact that in both case we call it adjoint come from the fact that in both case, the matrix of the adjoint is given by the transpose of the matrix of the application (i.e. if $uin L(E,F)$, $A=(u)_B_FB_E$ then $(u^*)_B_E^*B_F^*=A^t$ and if $uin L(E)$ and be orthonormal, then if $A=(u)_BB$ then $(u^*)_BB=A^t$).
– Surb
7 hours ago
To justify your formula, just use the definition $$left<u(x),fright>=f(u(x))=f(u)(x)=left<x,f(u)right>=left<x,u^*(f)right>.$$ But it would be better to wait for the answer of someone that knows better the subject.
– Surb
7 hours ago
I'm not totally sure, but I don't think that there is a corelation between the definition of an adjoint of an application in a inner space and the adjoint of $uin mathcal L(E,F)$. I think that the fact that in both case we call it adjoint come from the fact that in both case, the matrix of the adjoint is given by the transpose of the matrix of the application (i.e. if $uin L(E,F)$, $A=(u)_B_FB_E$ then $(u^*)_B_E^*B_F^*=A^t$ and if $uin L(E)$ and be orthonormal, then if $A=(u)_BB$ then $(u^*)_BB=A^t$).
– Surb
7 hours ago
I'm not totally sure, but I don't think that there is a corelation between the definition of an adjoint of an application in a inner space and the adjoint of $uin mathcal L(E,F)$. I think that the fact that in both case we call it adjoint come from the fact that in both case, the matrix of the adjoint is given by the transpose of the matrix of the application (i.e. if $uin L(E,F)$, $A=(u)_B_FB_E$ then $(u^*)_B_E^*B_F^*=A^t$ and if $uin L(E)$ and be orthonormal, then if $A=(u)_BB$ then $(u^*)_BB=A^t$).
– Surb
7 hours ago
To justify your formula, just use the definition $$left<u(x),fright>=f(u(x))=f(u)(x)=left<x,f(u)right>=left<x,u^*(f)right>.$$ But it would be better to wait for the answer of someone that knows better the subject.
– Surb
7 hours ago
To justify your formula, just use the definition $$left<u(x),fright>=f(u(x))=f(u)(x)=left<x,f(u)right>=left<x,u^*(f)right>.$$ But it would be better to wait for the answer of someone that knows better the subject.
– Surb
7 hours ago
add a comment |Â
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I'm not totally sure, but I don't think that there is a corelation between the definition of an adjoint of an application in a inner space and the adjoint of $uin mathcal L(E,F)$. I think that the fact that in both case we call it adjoint come from the fact that in both case, the matrix of the adjoint is given by the transpose of the matrix of the application (i.e. if $uin L(E,F)$, $A=(u)_B_FB_E$ then $(u^*)_B_E^*B_F^*=A^t$ and if $uin L(E)$ and be orthonormal, then if $A=(u)_BB$ then $(u^*)_BB=A^t$).
– Surb
7 hours ago
To justify your formula, just use the definition $$left<u(x),fright>=f(u(x))=f(u)(x)=left<x,f(u)right>=left<x,u^*(f)right>.$$ But it would be better to wait for the answer of someone that knows better the subject.
– Surb
7 hours ago