Mixing
What is common between adjoint operator and transpose of the matrix?
$begingroup$
I am confused.
What is Connection between Adjoint Operator and transpose of the Matrix?
I will be very grateful if someone can help me to clarify it.
functional-analysis
asked Mar 22 '14 at 16:08
learnerlearner
3781616
$endgroup$
add a comment |
$begingroup$
I am confused.
What is Connection between Adjoint Operator and transpose of the Matrix?
I will be very grateful if someone can help me to clarify it.
functional-analysis
asked Mar 22 '14 at 16:08
learnerlearner
3781616
$endgroup$
add a comment |
$begingroup$
I am confused.
What is Connection between Adjoint Operator and transpose of the Matrix?
I will be very grateful if someone can help me to clarify it.
functional-analysis
asked Mar 22 '14 at 16:08
learnerlearner
3781616
$endgroup$
I am confused.
What is Connection between Adjoint Operator and transpose of the Matrix?
I will be very grateful if someone can help me to clarify it.
functional-analysis
functional-analysis
asked Mar 22 '14 at 16:08
learnerlearner
3781616
asked Mar 22 '14 at 16:08
learnerlearner
3781616
asked Mar 22 '14 at 16:08
learnerlearner
3781616
asked Mar 22 '14 at 16:08
learnerlearner
3781616
asked Mar 22 '14 at 16:08
learnerlearner
3781616
3781616
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $f$ an endomorphism in an euclidean space $(E,langle,rangle)$ and let $A$ the matrix that represents $f$ relative to an orthonormal basis $mathcal B$. Let $x,yin E$ and $X=[x]_{mathcal B}, Y=[y]_{mathcal B}$ and $A^*$ the matrix of the adjoint $f^*$ of $f$ in $mathcal B$ then
$$langle f(x),yrangle=langle x,f^*(y)rangleiff(AX)^TY=X^TA^TY=X^TA^*Y$$
and the last equality imply:
$$A^*=A^T$$
$endgroup$
add a comment |
$begingroup$
A linear operator $L : Xrightarrow X$ on a linear space $X$ over a field $mathcal{F}$ does not require the notion of a basis in order to be defined. Such an $L$ is a function with the property that
$$
L(alpha x+beta y ) =alpha L(x)+beta L(y),;;; x,y in X,
;;alpha,beta in mathcal{F}.
$$
If $X$ is finite-dimensional, and if ${ v_{j}}_{j=1}^{N}$ is a basis for $X$, then you end up with a matrix representation of $L$ given by
$$
[Lx]_{v} = [L]_{v}[x]_{v}
$$
Then you can define a new linear transformation $M_{v} : Xrightarrow X$ such that
$$
[M_{v}x]_{v} = [L]_{v}^{T}[x]_{v},
$$
where $[L]_{v}^{T}$ is the transpose of $[L]$. A subscript of $v$ on the operator $M_{v}$ is necessary because the resulting linear operator definitely depends on the choice of basis; it is not uniquely determined by the linear operator $L$.
The adjoint of $L$ is also something which also requires a subscript, and it requires further specialization to the field $mathcal{F}$ of real or complex numbers. For your situation, assume $mathcal{F}$ is the field of real numbers. The general concept of adjoint in this case requires you to consider the dual space $X^{star}$ consisting of linear functions $x^{star} : Xrightarrowmathbb{R}$. The adjoint $L^{star} : X^{star}rightarrow X^{star}$ maps $x^{star}$ to $x^{star}circ L$. That is, $L^{star}x^{star}=x^{star}circ L$.
A connection between adjoint and transpose is found when choosing a basis ${ v_{n}^{star}}_{n=1}^{N}$ for $X^{star}$ such that $v_{j}^{star}(v_{k})=delta_{j,k}$. This is the so-called dual basis of ${ v_{j}}_{j=1}^{k}$ for $X^{star}$. In this basis, notice that $L^{star}v_{j}^{star}$ is defined by its action on the basis ${ v_{j}}_{j=1}^{k}$, which is easily found from the definition of $L^{star}$.
If $[L]_{v}=[l_{j,k}]$, then $Lv_{k}=l_{1,k}v_{1}+cdots+l_{N,k}v_{N}$ and
$$
(L^{star}v_{j}^{star})(v_{k})=(v_{j}^{star}circ L)(v_{k})=v_{j}^{star}(Lv_{k})=v_{j}^{star}(l_{1,k}v_{1}+cdots l_{N,k}v_{N})=l_{j,k}.
$$
In other words, $L^{star}v_{j}^{star}$ is determined by its action on the basis ${ v_{j}^{star}}$ to be
$$
L^{star}v_{j}=l_{j,1}v_{1}^{star}+cdots+l_{j,N}v_{N}^{star}.
$$
So the representation of $L^{star}$ with respect to ${ v_{j}^{star}}$ is the transpose of the matrix representation of $L$ with respect to ${ v_{j}}$. That is,
$$
[L^{star}]_{v^{star}}=[L]_{v}^{T}.
$$
But this holds only if $v^{star}$ is the dual basis of $v$.
If you have a real inner-product, then the dual basis ${ e_{n}^{star}}_{n=1}^{N}$ of an orthonormal basis ${ e_{n}}_{n=1}^{N}$ is identified with the original basis because $e_{j}^{star}(x)=(x,e_{j})$ does define $e_{j}^{star} in X^{star}$ such that $e_{j}^{star}(e_{k})=delta_{j,k}$.
answered Mar 22 '14 at 20:40
DisintegratingByPartsDisintegratingByParts
59.3k42580
$endgroup$
$begingroup$
It does not matter that the ground field is the field of real numbers. Everything holds over an arbitrary field.
$endgroup$
– user44400
Feb 8 '17 at 23:33
1
$begingroup$
@user44400 : You don't have inner products over arbitrary fields.
$endgroup$
– DisintegratingByParts
Feb 9 '17 at 0:17
1
$begingroup$
@DisintegratingByParts You can speak of nondegenerate (symmetric) bilinear forms over arbitrary fields, and this is all you really need to define an adjoint map.
$endgroup$
– ಠ_ಠ
Oct 15 '17 at 2:26
add a comment |
$begingroup$
Suppose a real operator $T:Bbb{R}^n to Bbb{R}^m$ is given by
$$
T(x) = Ax
$$
Where $x$ is a column vector and $A$ is an $m times n$ matrix. The adjoint of this operator will be given by
$$
T^*(y) = A^Ty
$$
answered Mar 22 '14 at 16:20
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $f$ an endomorphism in an euclidean space $(E,langle,rangle)$ and let $A$ the matrix that represents $f$ relative to an orthonormal basis $mathcal B$. Let $x,yin E$ and $X=[x]_{mathcal B}, Y=[y]_{mathcal B}$ and $A^*$ the matrix of the adjoint $f^*$ of $f$ in $mathcal B$ then
$$langle f(x),yrangle=langle x,f^*(y)rangleiff(AX)^TY=X^TA^TY=X^TA^*Y$$
and the last equality imply:
$$A^*=A^T$$
$endgroup$
add a comment |
$begingroup$
Let $f$ an endomorphism in an euclidean space $(E,langle,rangle)$ and let $A$ the matrix that represents $f$ relative to an orthonormal basis $mathcal B$. Let $x,yin E$ and $X=[x]_{mathcal B}, Y=[y]_{mathcal B}$ and $A^*$ the matrix of the adjoint $f^*$ of $f$ in $mathcal B$ then
$$langle f(x),yrangle=langle x,f^*(y)rangleiff(AX)^TY=X^TA^TY=X^TA^*Y$$
and the last equality imply:
$$A^*=A^T$$
$endgroup$
add a comment |
$begingroup$
Let $f$ an endomorphism in an euclidean space $(E,langle,rangle)$ and let $A$ the matrix that represents $f$ relative to an orthonormal basis $mathcal B$. Let $x,yin E$ and $X=[x]_{mathcal B}, Y=[y]_{mathcal B}$ and $A^*$ the matrix of the adjoint $f^*$ of $f$ in $mathcal B$ then
$$langle f(x),yrangle=langle x,f^*(y)rangleiff(AX)^TY=X^TA^TY=X^TA^*Y$$
and the last equality imply:
$$A^*=A^T$$
$endgroup$
Let $f$ an endomorphism in an euclidean space $(E,langle,rangle)$ and let $A$ the matrix that represents $f$ relative to an orthonormal basis $mathcal B$. Let $x,yin E$ and $X=[x]_{mathcal B}, Y=[y]_{mathcal B}$ and $A^*$ the matrix of the adjoint $f^*$ of $f$ in $mathcal B$ then
$$langle f(x),yrangle=langle x,f^*(y)rangleiff(AX)^TY=X^TA^TY=X^TA^*Y$$
and the last equality imply:
$$A^*=A^T$$
answered Mar 22 '14 at 16:26
user63181
add a comment |
add a comment |
$begingroup$
A linear operator $L : Xrightarrow X$ on a linear space $X$ over a field $mathcal{F}$ does not require the notion of a basis in order to be defined. Such an $L$ is a function with the property that
$$
L(alpha x+beta y ) =alpha L(x)+beta L(y),;;; x,y in X,
;;alpha,beta in mathcal{F}.
$$
If $X$ is finite-dimensional, and if ${ v_{j}}_{j=1}^{N}$ is a basis for $X$, then you end up with a matrix representation of $L$ given by
$$
[Lx]_{v} = [L]_{v}[x]_{v}
$$
Then you can define a new linear transformation $M_{v} : Xrightarrow X$ such that
$$
[M_{v}x]_{v} = [L]_{v}^{T}[x]_{v},
$$
where $[L]_{v}^{T}$ is the transpose of $[L]$. A subscript of $v$ on the operator $M_{v}$ is necessary because the resulting linear operator definitely depends on the choice of basis; it is not uniquely determined by the linear operator $L$.
The adjoint of $L$ is also something which also requires a subscript, and it requires further specialization to the field $mathcal{F}$ of real or complex numbers. For your situation, assume $mathcal{F}$ is the field of real numbers. The general concept of adjoint in this case requires you to consider the dual space $X^{star}$ consisting of linear functions $x^{star} : Xrightarrowmathbb{R}$. The adjoint $L^{star} : X^{star}rightarrow X^{star}$ maps $x^{star}$ to $x^{star}circ L$. That is, $L^{star}x^{star}=x^{star}circ L$.
A connection between adjoint and transpose is found when choosing a basis ${ v_{n}^{star}}_{n=1}^{N}$ for $X^{star}$ such that $v_{j}^{star}(v_{k})=delta_{j,k}$. This is the so-called dual basis of ${ v_{j}}_{j=1}^{k}$ for $X^{star}$. In this basis, notice that $L^{star}v_{j}^{star}$ is defined by its action on the basis ${ v_{j}}_{j=1}^{k}$, which is easily found from the definition of $L^{star}$.
If $[L]_{v}=[l_{j,k}]$, then $Lv_{k}=l_{1,k}v_{1}+cdots+l_{N,k}v_{N}$ and
$$
(L^{star}v_{j}^{star})(v_{k})=(v_{j}^{star}circ L)(v_{k})=v_{j}^{star}(Lv_{k})=v_{j}^{star}(l_{1,k}v_{1}+cdots l_{N,k}v_{N})=l_{j,k}.
$$
In other words, $L^{star}v_{j}^{star}$ is determined by its action on the basis ${ v_{j}^{star}}$ to be
$$
L^{star}v_{j}=l_{j,1}v_{1}^{star}+cdots+l_{j,N}v_{N}^{star}.
$$
So the representation of $L^{star}$ with respect to ${ v_{j}^{star}}$ is the transpose of the matrix representation of $L$ with respect to ${ v_{j}}$. That is,
$$
[L^{star}]_{v^{star}}=[L]_{v}^{T}.
$$
But this holds only if $v^{star}$ is the dual basis of $v$.
If you have a real inner-product, then the dual basis ${ e_{n}^{star}}_{n=1}^{N}$ of an orthonormal basis ${ e_{n}}_{n=1}^{N}$ is identified with the original basis because $e_{j}^{star}(x)=(x,e_{j})$ does define $e_{j}^{star} in X^{star}$ such that $e_{j}^{star}(e_{k})=delta_{j,k}$.
answered Mar 22 '14 at 20:40
DisintegratingByPartsDisintegratingByParts
59.3k42580
$endgroup$
$begingroup$
It does not matter that the ground field is the field of real numbers. Everything holds over an arbitrary field.
$endgroup$
– user44400
Feb 8 '17 at 23:33
1
$begingroup$
@user44400 : You don't have inner products over arbitrary fields.
$endgroup$
– DisintegratingByParts
Feb 9 '17 at 0:17
1
$begingroup$
@DisintegratingByParts You can speak of nondegenerate (symmetric) bilinear forms over arbitrary fields, and this is all you really need to define an adjoint map.
$endgroup$
– ಠ_ಠ
Oct 15 '17 at 2:26
add a comment |
$begingroup$
A linear operator $L : Xrightarrow X$ on a linear space $X$ over a field $mathcal{F}$ does not require the notion of a basis in order to be defined. Such an $L$ is a function with the property that
$$
L(alpha x+beta y ) =alpha L(x)+beta L(y),;;; x,y in X,
;;alpha,beta in mathcal{F}.
$$
If $X$ is finite-dimensional, and if ${ v_{j}}_{j=1}^{N}$ is a basis for $X$, then you end up with a matrix representation of $L$ given by
$$
[Lx]_{v} = [L]_{v}[x]_{v}
$$
Then you can define a new linear transformation $M_{v} : Xrightarrow X$ such that
$$
[M_{v}x]_{v} = [L]_{v}^{T}[x]_{v},
$$
where $[L]_{v}^{T}$ is the transpose of $[L]$. A subscript of $v$ on the operator $M_{v}$ is necessary because the resulting linear operator definitely depends on the choice of basis; it is not uniquely determined by the linear operator $L$.
The adjoint of $L$ is also something which also requires a subscript, and it requires further specialization to the field $mathcal{F}$ of real or complex numbers. For your situation, assume $mathcal{F}$ is the field of real numbers. The general concept of adjoint in this case requires you to consider the dual space $X^{star}$ consisting of linear functions $x^{star} : Xrightarrowmathbb{R}$. The adjoint $L^{star} : X^{star}rightarrow X^{star}$ maps $x^{star}$ to $x^{star}circ L$. That is, $L^{star}x^{star}=x^{star}circ L$.
A connection between adjoint and transpose is found when choosing a basis ${ v_{n}^{star}}_{n=1}^{N}$ for $X^{star}$ such that $v_{j}^{star}(v_{k})=delta_{j,k}$. This is the so-called dual basis of ${ v_{j}}_{j=1}^{k}$ for $X^{star}$. In this basis, notice that $L^{star}v_{j}^{star}$ is defined by its action on the basis ${ v_{j}}_{j=1}^{k}$, which is easily found from the definition of $L^{star}$.
If $[L]_{v}=[l_{j,k}]$, then $Lv_{k}=l_{1,k}v_{1}+cdots+l_{N,k}v_{N}$ and
$$
(L^{star}v_{j}^{star})(v_{k})=(v_{j}^{star}circ L)(v_{k})=v_{j}^{star}(Lv_{k})=v_{j}^{star}(l_{1,k}v_{1}+cdots l_{N,k}v_{N})=l_{j,k}.
$$
In other words, $L^{star}v_{j}^{star}$ is determined by its action on the basis ${ v_{j}^{star}}$ to be
$$
L^{star}v_{j}=l_{j,1}v_{1}^{star}+cdots+l_{j,N}v_{N}^{star}.
$$
So the representation of $L^{star}$ with respect to ${ v_{j}^{star}}$ is the transpose of the matrix representation of $L$ with respect to ${ v_{j}}$. That is,
$$
[L^{star}]_{v^{star}}=[L]_{v}^{T}.
$$
But this holds only if $v^{star}$ is the dual basis of $v$.
If you have a real inner-product, then the dual basis ${ e_{n}^{star}}_{n=1}^{N}$ of an orthonormal basis ${ e_{n}}_{n=1}^{N}$ is identified with the original basis because $e_{j}^{star}(x)=(x,e_{j})$ does define $e_{j}^{star} in X^{star}$ such that $e_{j}^{star}(e_{k})=delta_{j,k}$.
answered Mar 22 '14 at 20:40
DisintegratingByPartsDisintegratingByParts
59.3k42580
$endgroup$
$begingroup$
It does not matter that the ground field is the field of real numbers. Everything holds over an arbitrary field.
$endgroup$
– user44400
Feb 8 '17 at 23:33
1
$begingroup$
@user44400 : You don't have inner products over arbitrary fields.
$endgroup$
– DisintegratingByParts
Feb 9 '17 at 0:17
1
$begingroup$
@DisintegratingByParts You can speak of nondegenerate (symmetric) bilinear forms over arbitrary fields, and this is all you really need to define an adjoint map.
$endgroup$
– ಠ_ಠ
Oct 15 '17 at 2:26
add a comment |
$begingroup$
A linear operator $L : Xrightarrow X$ on a linear space $X$ over a field $mathcal{F}$ does not require the notion of a basis in order to be defined. Such an $L$ is a function with the property that
$$
L(alpha x+beta y ) =alpha L(x)+beta L(y),;;; x,y in X,
;;alpha,beta in mathcal{F}.
$$
If $X$ is finite-dimensional, and if ${ v_{j}}_{j=1}^{N}$ is a basis for $X$, then you end up with a matrix representation of $L$ given by
$$
[Lx]_{v} = [L]_{v}[x]_{v}
$$
Then you can define a new linear transformation $M_{v} : Xrightarrow X$ such that
$$
[M_{v}x]_{v} = [L]_{v}^{T}[x]_{v},
$$
where $[L]_{v}^{T}$ is the transpose of $[L]$. A subscript of $v$ on the operator $M_{v}$ is necessary because the resulting linear operator definitely depends on the choice of basis; it is not uniquely determined by the linear operator $L$.
The adjoint of $L$ is also something which also requires a subscript, and it requires further specialization to the field $mathcal{F}$ of real or complex numbers. For your situation, assume $mathcal{F}$ is the field of real numbers. The general concept of adjoint in this case requires you to consider the dual space $X^{star}$ consisting of linear functions $x^{star} : Xrightarrowmathbb{R}$. The adjoint $L^{star} : X^{star}rightarrow X^{star}$ maps $x^{star}$ to $x^{star}circ L$. That is, $L^{star}x^{star}=x^{star}circ L$.
A connection between adjoint and transpose is found when choosing a basis ${ v_{n}^{star}}_{n=1}^{N}$ for $X^{star}$ such that $v_{j}^{star}(v_{k})=delta_{j,k}$. This is the so-called dual basis of ${ v_{j}}_{j=1}^{k}$ for $X^{star}$. In this basis, notice that $L^{star}v_{j}^{star}$ is defined by its action on the basis ${ v_{j}}_{j=1}^{k}$, which is easily found from the definition of $L^{star}$.
If $[L]_{v}=[l_{j,k}]$, then $Lv_{k}=l_{1,k}v_{1}+cdots+l_{N,k}v_{N}$ and
$$
(L^{star}v_{j}^{star})(v_{k})=(v_{j}^{star}circ L)(v_{k})=v_{j}^{star}(Lv_{k})=v_{j}^{star}(l_{1,k}v_{1}+cdots l_{N,k}v_{N})=l_{j,k}.
$$
In other words, $L^{star}v_{j}^{star}$ is determined by its action on the basis ${ v_{j}^{star}}$ to be
$$
L^{star}v_{j}=l_{j,1}v_{1}^{star}+cdots+l_{j,N}v_{N}^{star}.
$$
So the representation of $L^{star}$ with respect to ${ v_{j}^{star}}$ is the transpose of the matrix representation of $L$ with respect to ${ v_{j}}$. That is,
$$
[L^{star}]_{v^{star}}=[L]_{v}^{T}.
$$
But this holds only if $v^{star}$ is the dual basis of $v$.
If you have a real inner-product, then the dual basis ${ e_{n}^{star}}_{n=1}^{N}$ of an orthonormal basis ${ e_{n}}_{n=1}^{N}$ is identified with the original basis because $e_{j}^{star}(x)=(x,e_{j})$ does define $e_{j}^{star} in X^{star}$ such that $e_{j}^{star}(e_{k})=delta_{j,k}$.
answered Mar 22 '14 at 20:40
DisintegratingByPartsDisintegratingByParts
59.3k42580
$endgroup$
A linear operator $L : Xrightarrow X$ on a linear space $X$ over a field $mathcal{F}$ does not require the notion of a basis in order to be defined. Such an $L$ is a function with the property that
$$
L(alpha x+beta y ) =alpha L(x)+beta L(y),;;; x,y in X,
;;alpha,beta in mathcal{F}.
$$
If $X$ is finite-dimensional, and if ${ v_{j}}_{j=1}^{N}$ is a basis for $X$, then you end up with a matrix representation of $L$ given by
$$
[Lx]_{v} = [L]_{v}[x]_{v}
$$
Then you can define a new linear transformation $M_{v} : Xrightarrow X$ such that
$$
[M_{v}x]_{v} = [L]_{v}^{T}[x]_{v},
$$
where $[L]_{v}^{T}$ is the transpose of $[L]$. A subscript of $v$ on the operator $M_{v}$ is necessary because the resulting linear operator definitely depends on the choice of basis; it is not uniquely determined by the linear operator $L$.
The adjoint of $L$ is also something which also requires a subscript, and it requires further specialization to the field $mathcal{F}$ of real or complex numbers. For your situation, assume $mathcal{F}$ is the field of real numbers. The general concept of adjoint in this case requires you to consider the dual space $X^{star}$ consisting of linear functions $x^{star} : Xrightarrowmathbb{R}$. The adjoint $L^{star} : X^{star}rightarrow X^{star}$ maps $x^{star}$ to $x^{star}circ L$. That is, $L^{star}x^{star}=x^{star}circ L$.
A connection between adjoint and transpose is found when choosing a basis ${ v_{n}^{star}}_{n=1}^{N}$ for $X^{star}$ such that $v_{j}^{star}(v_{k})=delta_{j,k}$. This is the so-called dual basis of ${ v_{j}}_{j=1}^{k}$ for $X^{star}$. In this basis, notice that $L^{star}v_{j}^{star}$ is defined by its action on the basis ${ v_{j}}_{j=1}^{k}$, which is easily found from the definition of $L^{star}$.
If $[L]_{v}=[l_{j,k}]$, then $Lv_{k}=l_{1,k}v_{1}+cdots+l_{N,k}v_{N}$ and
$$
(L^{star}v_{j}^{star})(v_{k})=(v_{j}^{star}circ L)(v_{k})=v_{j}^{star}(Lv_{k})=v_{j}^{star}(l_{1,k}v_{1}+cdots l_{N,k}v_{N})=l_{j,k}.
$$
In other words, $L^{star}v_{j}^{star}$ is determined by its action on the basis ${ v_{j}^{star}}$ to be
$$
L^{star}v_{j}=l_{j,1}v_{1}^{star}+cdots+l_{j,N}v_{N}^{star}.
$$
So the representation of $L^{star}$ with respect to ${ v_{j}^{star}}$ is the transpose of the matrix representation of $L$ with respect to ${ v_{j}}$. That is,
$$
[L^{star}]_{v^{star}}=[L]_{v}^{T}.
$$
But this holds only if $v^{star}$ is the dual basis of $v$.
If you have a real inner-product, then the dual basis ${ e_{n}^{star}}_{n=1}^{N}$ of an orthonormal basis ${ e_{n}}_{n=1}^{N}$ is identified with the original basis because $e_{j}^{star}(x)=(x,e_{j})$ does define $e_{j}^{star} in X^{star}$ such that $e_{j}^{star}(e_{k})=delta_{j,k}$.
answered Mar 22 '14 at 20:40
DisintegratingByPartsDisintegratingByParts
59.3k42580
edited Mar 22 '14 at 20:55
answered Mar 22 '14 at 20:40
DisintegratingByPartsDisintegratingByParts
59.3k42580
answered Mar 22 '14 at 20:40
DisintegratingByPartsDisintegratingByParts
59.3k42580
answered Mar 22 '14 at 20:40
DisintegratingByPartsDisintegratingByParts
59.3k42580
59.3k42580
$begingroup$
It does not matter that the ground field is the field of real numbers. Everything holds over an arbitrary field.
$endgroup$
– user44400
Feb 8 '17 at 23:33
1
$begingroup$
@user44400 : You don't have inner products over arbitrary fields.
$endgroup$
– DisintegratingByParts
Feb 9 '17 at 0:17
1
$begingroup$
@DisintegratingByParts You can speak of nondegenerate (symmetric) bilinear forms over arbitrary fields, and this is all you really need to define an adjoint map.
$endgroup$
– ಠ_ಠ
Oct 15 '17 at 2:26
add a comment |
$begingroup$
It does not matter that the ground field is the field of real numbers. Everything holds over an arbitrary field.
$endgroup$
– user44400
Feb 8 '17 at 23:33
1
$begingroup$
@user44400 : You don't have inner products over arbitrary fields.
$endgroup$
– DisintegratingByParts
Feb 9 '17 at 0:17
1
$begingroup$
@DisintegratingByParts You can speak of nondegenerate (symmetric) bilinear forms over arbitrary fields, and this is all you really need to define an adjoint map.
$endgroup$
– ಠ_ಠ
Oct 15 '17 at 2:26
$begingroup$
It does not matter that the ground field is the field of real numbers. Everything holds over an arbitrary field.
$endgroup$
– user44400
Feb 8 '17 at 23:33
$begingroup$
It does not matter that the ground field is the field of real numbers. Everything holds over an arbitrary field.
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– user44400
Feb 8 '17 at 23:33
1
1
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@user44400 : You don't have inner products over arbitrary fields.
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– DisintegratingByParts
Feb 9 '17 at 0:17
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@user44400 : You don't have inner products over arbitrary fields.
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– DisintegratingByParts
Feb 9 '17 at 0:17
1
1
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@DisintegratingByParts You can speak of nondegenerate (symmetric) bilinear forms over arbitrary fields, and this is all you really need to define an adjoint map.
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– ಠ_ಠ
Oct 15 '17 at 2:26
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@DisintegratingByParts You can speak of nondegenerate (symmetric) bilinear forms over arbitrary fields, and this is all you really need to define an adjoint map.
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– ಠ_ಠ
Oct 15 '17 at 2:26
add a comment |
$begingroup$
Suppose a real operator $T:Bbb{R}^n to Bbb{R}^m$ is given by
$$
T(x) = Ax
$$
Where $x$ is a column vector and $A$ is an $m times n$ matrix. The adjoint of this operator will be given by
$$
T^*(y) = A^Ty
$$
answered Mar 22 '14 at 16:20
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
add a comment |
$begingroup$
Suppose a real operator $T:Bbb{R}^n to Bbb{R}^m$ is given by
$$
T(x) = Ax
$$
Where $x$ is a column vector and $A$ is an $m times n$ matrix. The adjoint of this operator will be given by
$$
T^*(y) = A^Ty
$$
answered Mar 22 '14 at 16:20
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
add a comment |
$begingroup$
Suppose a real operator $T:Bbb{R}^n to Bbb{R}^m$ is given by
$$
T(x) = Ax
$$
Where $x$ is a column vector and $A$ is an $m times n$ matrix. The adjoint of this operator will be given by
$$
T^*(y) = A^Ty
$$
answered Mar 22 '14 at 16:20
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
Suppose a real operator $T:Bbb{R}^n to Bbb{R}^m$ is given by
$$
T(x) = Ax
$$
Where $x$ is a column vector and $A$ is an $m times n$ matrix. The adjoint of this operator will be given by
$$
T^*(y) = A^Ty
$$
answered Mar 22 '14 at 16:20
OmnomnomnomOmnomnomnom
128k791183
answered Mar 22 '14 at 16:20
OmnomnomnomOmnomnomnom
128k791183
answered Mar 22 '14 at 16:20
OmnomnomnomOmnomnomnom
128k791183
answered Mar 22 '14 at 16:20
OmnomnomnomOmnomnomnom
128k791183
128k791183
add a comment |
add a comment |
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