Proving existence and uniqueness of the adjoint of $f$
$begingroup$
Let $f:Eto F$ be a linear transformation between euclidean spaces with finite dimension.
We define the following isomorphism:$quad Psi_E:Eto E^*quadtextrm{with}quadPsi_E(u)=langle u,·rangle$
Prove that there is only one linear transformation $f':Fto E$ such that
$$langle f(u),vrangle=langle u,f'(v)ranglequadforall uin E, vin F$$
where $f'$ is the adjoint of $f$.
My textbook gives the following proof:
Given $vin F$, we define the linear map $quad Etomathbb{R}quadtextrm{with}quad umapstolangle f(u),vrangle$.
Thus, $exists! f'(v)in E $ such that,
$$Psi_E(f'(v))(u)=langle f(u),vranglequadforall uin E$$
$$i.e.quadlangle u,f'(v)rangle=langle f(u),vranglequadforall uin E$$
So we have proven the existence and uniqueness of $f'$.
I have found other ways to prove the statement, but I don't clearly understand my textbook proof.
Why can we define the linear map with $umapstolangle f(u),vrangle$? Why does $Psi_E(f'(v))(u)equivlangle u,f'(v)rangle$?
Could you give me some clues? Thanks in advance!
linear-algebra linear-transformations
asked Dec 29 '18 at 13:04
GibbsGibbs
137111
$endgroup$
add a comment |
$begingroup$
Let $f:Eto F$ be a linear transformation between euclidean spaces with finite dimension.
We define the following isomorphism:$quad Psi_E:Eto E^*quadtextrm{with}quadPsi_E(u)=langle u,·rangle$
Prove that there is only one linear transformation $f':Fto E$ such that
$$langle f(u),vrangle=langle u,f'(v)ranglequadforall uin E, vin F$$
where $f'$ is the adjoint of $f$.
My textbook gives the following proof:
Given $vin F$, we define the linear map $quad Etomathbb{R}quadtextrm{with}quad umapstolangle f(u),vrangle$.
Thus, $exists! f'(v)in E $ such that,
$$Psi_E(f'(v))(u)=langle f(u),vranglequadforall uin E$$
$$i.e.quadlangle u,f'(v)rangle=langle f(u),vranglequadforall uin E$$
So we have proven the existence and uniqueness of $f'$.
I have found other ways to prove the statement, but I don't clearly understand my textbook proof.
Why can we define the linear map with $umapstolangle f(u),vrangle$? Why does $Psi_E(f'(v))(u)equivlangle u,f'(v)rangle$?
Could you give me some clues? Thanks in advance!
linear-algebra linear-transformations
asked Dec 29 '18 at 13:04
GibbsGibbs
137111
$endgroup$
add a comment |
$begingroup$
Let $f:Eto F$ be a linear transformation between euclidean spaces with finite dimension.
We define the following isomorphism:$quad Psi_E:Eto E^*quadtextrm{with}quadPsi_E(u)=langle u,·rangle$
Prove that there is only one linear transformation $f':Fto E$ such that
$$langle f(u),vrangle=langle u,f'(v)ranglequadforall uin E, vin F$$
where $f'$ is the adjoint of $f$.
My textbook gives the following proof:
Given $vin F$, we define the linear map $quad Etomathbb{R}quadtextrm{with}quad umapstolangle f(u),vrangle$.
Thus, $exists! f'(v)in E $ such that,
$$Psi_E(f'(v))(u)=langle f(u),vranglequadforall uin E$$
$$i.e.quadlangle u,f'(v)rangle=langle f(u),vranglequadforall uin E$$
So we have proven the existence and uniqueness of $f'$.
I have found other ways to prove the statement, but I don't clearly understand my textbook proof.
Why can we define the linear map with $umapstolangle f(u),vrangle$? Why does $Psi_E(f'(v))(u)equivlangle u,f'(v)rangle$?
Could you give me some clues? Thanks in advance!
linear-algebra linear-transformations
asked Dec 29 '18 at 13:04
GibbsGibbs
137111
$endgroup$
Let $f:Eto F$ be a linear transformation between euclidean spaces with finite dimension.
We define the following isomorphism:$quad Psi_E:Eto E^*quadtextrm{with}quadPsi_E(u)=langle u,·rangle$
Prove that there is only one linear transformation $f':Fto E$ such that
$$langle f(u),vrangle=langle u,f'(v)ranglequadforall uin E, vin F$$
where $f'$ is the adjoint of $f$.
My textbook gives the following proof:
Given $vin F$, we define the linear map $quad Etomathbb{R}quadtextrm{with}quad umapstolangle f(u),vrangle$.
Thus, $exists! f'(v)in E $ such that,
$$Psi_E(f'(v))(u)=langle f(u),vranglequadforall uin E$$
$$i.e.quadlangle u,f'(v)rangle=langle f(u),vranglequadforall uin E$$
So we have proven the existence and uniqueness of $f'$.
I have found other ways to prove the statement, but I don't clearly understand my textbook proof.
Why can we define the linear map with $umapstolangle f(u),vrangle$? Why does $Psi_E(f'(v))(u)equivlangle u,f'(v)rangle$?
Could you give me some clues? Thanks in advance!
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Dec 29 '18 at 13:04
GibbsGibbs
137111
asked Dec 29 '18 at 13:04
GibbsGibbs
137111
edited Dec 29 '18 at 14:44
Gibbs
asked Dec 29 '18 at 13:04
GibbsGibbs
137111
asked Dec 29 '18 at 13:04
GibbsGibbs
137111
asked Dec 29 '18 at 13:04
GibbsGibbs
137111
137111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are given $E$ and $F$ to be Euclidean spaces, with inner products $langlecdot,cdotrangle_E :Erightarrow mathbb{R}$ and $langlecdot,cdotrangle_F :Frightarrow mathbb{R}$. Note that the linear map $T:u mapsto langle f(u), v rangle_F$ is uniquely defined by $vin F$ and $f$. Since $T$ is a linear functional on $E$, hence there exist a unique vector $w$ in $E$, such that $T(u)=langle u, wrangle_E$, i.e. $$langle f(u), v rangle_F=langle u, wrangle_E, $$let's call $w=f'(v)$, since this vector is associated to $v$, while defining $T$. It is required be to check that $f':Frightarrow E$ is a linear map. And, then we have $$langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
I guess the notation, you have provided should be $Psi_E:Erightarrow E^*$ with $Psi_E(u)=langle cdot, urangle_E$, we get $$Psi_E(f'(v))(u)= langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
answered Dec 29 '18 at 14:21
DèöDèö
19017
$endgroup$
$begingroup$
Almost understood. Only one thing: Could you explain a little bit more in detail why there exists a unique vector $omega$ in $E$ such that $T(u)=langle u,wrangle_E$?
$endgroup$
– Gibbs
Dec 29 '18 at 14:54
1
$begingroup$
Let ${beta_1,...,beta_n}$ be an orthonormal basis for $E$. If we take $w=sum T(beta_i)beta_i$ and let $T_w(u) = langle u, w rangle_E$. Then on each basis vector, we can see $T_w=T$, and hence on the whole space $E$. It is not difficult to prove the uniqueness of $w$.
$endgroup$
– Dèö
Dec 29 '18 at 15:06
add a comment |
$begingroup$
In the last two equations he is using that $Psi_E$ is an isomorphism. "Canceling" the $v$ you get that $f'(u)=f(u)$ for all $uin E$.
answered Dec 29 '18 at 13:09
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
You are given $E$ and $F$ to be Euclidean spaces, with inner products $langlecdot,cdotrangle_E :Erightarrow mathbb{R}$ and $langlecdot,cdotrangle_F :Frightarrow mathbb{R}$. Note that the linear map $T:u mapsto langle f(u), v rangle_F$ is uniquely defined by $vin F$ and $f$. Since $T$ is a linear functional on $E$, hence there exist a unique vector $w$ in $E$, such that $T(u)=langle u, wrangle_E$, i.e. $$langle f(u), v rangle_F=langle u, wrangle_E, $$let's call $w=f'(v)$, since this vector is associated to $v$, while defining $T$. It is required be to check that $f':Frightarrow E$ is a linear map. And, then we have $$langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
I guess the notation, you have provided should be $Psi_E:Erightarrow E^*$ with $Psi_E(u)=langle cdot, urangle_E$, we get $$Psi_E(f'(v))(u)= langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
answered Dec 29 '18 at 14:21
DèöDèö
19017
$endgroup$
$begingroup$
Almost understood. Only one thing: Could you explain a little bit more in detail why there exists a unique vector $omega$ in $E$ such that $T(u)=langle u,wrangle_E$?
$endgroup$
– Gibbs
Dec 29 '18 at 14:54
1
$begingroup$
Let ${beta_1,...,beta_n}$ be an orthonormal basis for $E$. If we take $w=sum T(beta_i)beta_i$ and let $T_w(u) = langle u, w rangle_E$. Then on each basis vector, we can see $T_w=T$, and hence on the whole space $E$. It is not difficult to prove the uniqueness of $w$.
$endgroup$
– Dèö
Dec 29 '18 at 15:06
add a comment |
$begingroup$
You are given $E$ and $F$ to be Euclidean spaces, with inner products $langlecdot,cdotrangle_E :Erightarrow mathbb{R}$ and $langlecdot,cdotrangle_F :Frightarrow mathbb{R}$. Note that the linear map $T:u mapsto langle f(u), v rangle_F$ is uniquely defined by $vin F$ and $f$. Since $T$ is a linear functional on $E$, hence there exist a unique vector $w$ in $E$, such that $T(u)=langle u, wrangle_E$, i.e. $$langle f(u), v rangle_F=langle u, wrangle_E, $$let's call $w=f'(v)$, since this vector is associated to $v$, while defining $T$. It is required be to check that $f':Frightarrow E$ is a linear map. And, then we have $$langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
I guess the notation, you have provided should be $Psi_E:Erightarrow E^*$ with $Psi_E(u)=langle cdot, urangle_E$, we get $$Psi_E(f'(v))(u)= langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
answered Dec 29 '18 at 14:21
DèöDèö
19017
$endgroup$
$begingroup$
Almost understood. Only one thing: Could you explain a little bit more in detail why there exists a unique vector $omega$ in $E$ such that $T(u)=langle u,wrangle_E$?
$endgroup$
– Gibbs
Dec 29 '18 at 14:54
1
$begingroup$
Let ${beta_1,...,beta_n}$ be an orthonormal basis for $E$. If we take $w=sum T(beta_i)beta_i$ and let $T_w(u) = langle u, w rangle_E$. Then on each basis vector, we can see $T_w=T$, and hence on the whole space $E$. It is not difficult to prove the uniqueness of $w$.
$endgroup$
– Dèö
Dec 29 '18 at 15:06
add a comment |
$begingroup$
You are given $E$ and $F$ to be Euclidean spaces, with inner products $langlecdot,cdotrangle_E :Erightarrow mathbb{R}$ and $langlecdot,cdotrangle_F :Frightarrow mathbb{R}$. Note that the linear map $T:u mapsto langle f(u), v rangle_F$ is uniquely defined by $vin F$ and $f$. Since $T$ is a linear functional on $E$, hence there exist a unique vector $w$ in $E$, such that $T(u)=langle u, wrangle_E$, i.e. $$langle f(u), v rangle_F=langle u, wrangle_E, $$let's call $w=f'(v)$, since this vector is associated to $v$, while defining $T$. It is required be to check that $f':Frightarrow E$ is a linear map. And, then we have $$langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
I guess the notation, you have provided should be $Psi_E:Erightarrow E^*$ with $Psi_E(u)=langle cdot, urangle_E$, we get $$Psi_E(f'(v))(u)= langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
answered Dec 29 '18 at 14:21
DèöDèö
19017
$endgroup$
You are given $E$ and $F$ to be Euclidean spaces, with inner products $langlecdot,cdotrangle_E :Erightarrow mathbb{R}$ and $langlecdot,cdotrangle_F :Frightarrow mathbb{R}$. Note that the linear map $T:u mapsto langle f(u), v rangle_F$ is uniquely defined by $vin F$ and $f$. Since $T$ is a linear functional on $E$, hence there exist a unique vector $w$ in $E$, such that $T(u)=langle u, wrangle_E$, i.e. $$langle f(u), v rangle_F=langle u, wrangle_E, $$let's call $w=f'(v)$, since this vector is associated to $v$, while defining $T$. It is required be to check that $f':Frightarrow E$ is a linear map. And, then we have $$langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
I guess the notation, you have provided should be $Psi_E:Erightarrow E^*$ with $Psi_E(u)=langle cdot, urangle_E$, we get $$Psi_E(f'(v))(u)= langle
u, f'(v)rangle_E = langle f(u), v rangle_F.$$
answered Dec 29 '18 at 14:21
DèöDèö
19017
answered Dec 29 '18 at 14:21
DèöDèö
19017
answered Dec 29 '18 at 14:21
DèöDèö
19017
answered Dec 29 '18 at 14:21
DèöDèö
19017
19017
$begingroup$
Almost understood. Only one thing: Could you explain a little bit more in detail why there exists a unique vector $omega$ in $E$ such that $T(u)=langle u,wrangle_E$?
$endgroup$
– Gibbs
Dec 29 '18 at 14:54
1
$begingroup$
Let ${beta_1,...,beta_n}$ be an orthonormal basis for $E$. If we take $w=sum T(beta_i)beta_i$ and let $T_w(u) = langle u, w rangle_E$. Then on each basis vector, we can see $T_w=T$, and hence on the whole space $E$. It is not difficult to prove the uniqueness of $w$.
$endgroup$
– Dèö
Dec 29 '18 at 15:06
add a comment |
$begingroup$
Almost understood. Only one thing: Could you explain a little bit more in detail why there exists a unique vector $omega$ in $E$ such that $T(u)=langle u,wrangle_E$?
$endgroup$
– Gibbs
Dec 29 '18 at 14:54
1
$begingroup$
Let ${beta_1,...,beta_n}$ be an orthonormal basis for $E$. If we take $w=sum T(beta_i)beta_i$ and let $T_w(u) = langle u, w rangle_E$. Then on each basis vector, we can see $T_w=T$, and hence on the whole space $E$. It is not difficult to prove the uniqueness of $w$.
$endgroup$
– Dèö
Dec 29 '18 at 15:06
$begingroup$
Almost understood. Only one thing: Could you explain a little bit more in detail why there exists a unique vector $omega$ in $E$ such that $T(u)=langle u,wrangle_E$?
$endgroup$
– Gibbs
Dec 29 '18 at 14:54
$begingroup$
Almost understood. Only one thing: Could you explain a little bit more in detail why there exists a unique vector $omega$ in $E$ such that $T(u)=langle u,wrangle_E$?
$endgroup$
– Gibbs
Dec 29 '18 at 14:54
1
1
$begingroup$
Let ${beta_1,...,beta_n}$ be an orthonormal basis for $E$. If we take $w=sum T(beta_i)beta_i$ and let $T_w(u) = langle u, w rangle_E$. Then on each basis vector, we can see $T_w=T$, and hence on the whole space $E$. It is not difficult to prove the uniqueness of $w$.
$endgroup$
– Dèö
Dec 29 '18 at 15:06
$begingroup$
Let ${beta_1,...,beta_n}$ be an orthonormal basis for $E$. If we take $w=sum T(beta_i)beta_i$ and let $T_w(u) = langle u, w rangle_E$. Then on each basis vector, we can see $T_w=T$, and hence on the whole space $E$. It is not difficult to prove the uniqueness of $w$.
$endgroup$
– Dèö
Dec 29 '18 at 15:06
add a comment |
$begingroup$
In the last two equations he is using that $Psi_E$ is an isomorphism. "Canceling" the $v$ you get that $f'(u)=f(u)$ for all $uin E$.
answered Dec 29 '18 at 13:09
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
In the last two equations he is using that $Psi_E$ is an isomorphism. "Canceling" the $v$ you get that $f'(u)=f(u)$ for all $uin E$.
answered Dec 29 '18 at 13:09
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
In the last two equations he is using that $Psi_E$ is an isomorphism. "Canceling" the $v$ you get that $f'(u)=f(u)$ for all $uin E$.
answered Dec 29 '18 at 13:09
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
In the last two equations he is using that $Psi_E$ is an isomorphism. "Canceling" the $v$ you get that $f'(u)=f(u)$ for all $uin E$.
answered Dec 29 '18 at 13:09
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Dec 29 '18 at 13:09
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Dec 29 '18 at 13:09
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Dec 29 '18 at 13:09
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
802110
add a comment |
add a comment |
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