The dual basis spans the vector space of linear transformations.
$begingroup$
Proof attempt:
Let $T in L(V,Bbb{F})$ where $L(V,Bbb{F})$ is the space of linear functionals of $V$ and where $Bbb{F}$ is some field. Let $mathscr{F}={Phi_1,cdots Phi_n}$ be the dual basis of, $B_v={v_1,cdots, v_n}$, the basis of $V$. Note $T(v)=T(sum_{i=1}^{n}c_i v_i)=sum_{i=1}^{n}c_i T(v_i)$ such that $forall i, T(v_i)inBbb{F}$. Since $T(v_i)inBbb{F}$ implies $exists lambda_iinBbb{F}, T(v_i)=lambda_i=lambda_icdot 1=lambda_icdotPhi(v_i)$. Then $T(v)=sum_{i=1}^{n}(c_ilambda_i)Phi(v_i)$ and $T=sum_{i=1}^n(c_ilambda_i)Phi_i$, as desired.
linear-algebra proof-verification linear-transformations dual-spaces
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
708715
$endgroup$
add a comment |
$begingroup$
Proof attempt:
Let $T in L(V,Bbb{F})$ where $L(V,Bbb{F})$ is the space of linear functionals of $V$ and where $Bbb{F}$ is some field. Let $mathscr{F}={Phi_1,cdots Phi_n}$ be the dual basis of, $B_v={v_1,cdots, v_n}$, the basis of $V$. Note $T(v)=T(sum_{i=1}^{n}c_i v_i)=sum_{i=1}^{n}c_i T(v_i)$ such that $forall i, T(v_i)inBbb{F}$. Since $T(v_i)inBbb{F}$ implies $exists lambda_iinBbb{F}, T(v_i)=lambda_i=lambda_icdot 1=lambda_icdotPhi(v_i)$. Then $T(v)=sum_{i=1}^{n}(c_ilambda_i)Phi(v_i)$ and $T=sum_{i=1}^n(c_ilambda_i)Phi_i$, as desired.
linear-algebra proof-verification linear-transformations dual-spaces
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
708715
$endgroup$
add a comment |
$begingroup$
Proof attempt:
Let $T in L(V,Bbb{F})$ where $L(V,Bbb{F})$ is the space of linear functionals of $V$ and where $Bbb{F}$ is some field. Let $mathscr{F}={Phi_1,cdots Phi_n}$ be the dual basis of, $B_v={v_1,cdots, v_n}$, the basis of $V$. Note $T(v)=T(sum_{i=1}^{n}c_i v_i)=sum_{i=1}^{n}c_i T(v_i)$ such that $forall i, T(v_i)inBbb{F}$. Since $T(v_i)inBbb{F}$ implies $exists lambda_iinBbb{F}, T(v_i)=lambda_i=lambda_icdot 1=lambda_icdotPhi(v_i)$. Then $T(v)=sum_{i=1}^{n}(c_ilambda_i)Phi(v_i)$ and $T=sum_{i=1}^n(c_ilambda_i)Phi_i$, as desired.
linear-algebra proof-verification linear-transformations dual-spaces
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
708715
$endgroup$
Proof attempt:
Let $T in L(V,Bbb{F})$ where $L(V,Bbb{F})$ is the space of linear functionals of $V$ and where $Bbb{F}$ is some field. Let $mathscr{F}={Phi_1,cdots Phi_n}$ be the dual basis of, $B_v={v_1,cdots, v_n}$, the basis of $V$. Note $T(v)=T(sum_{i=1}^{n}c_i v_i)=sum_{i=1}^{n}c_i T(v_i)$ such that $forall i, T(v_i)inBbb{F}$. Since $T(v_i)inBbb{F}$ implies $exists lambda_iinBbb{F}, T(v_i)=lambda_i=lambda_icdot 1=lambda_icdotPhi(v_i)$. Then $T(v)=sum_{i=1}^{n}(c_ilambda_i)Phi(v_i)$ and $T=sum_{i=1}^n(c_ilambda_i)Phi_i$, as desired.
linear-algebra proof-verification linear-transformations dual-spaces
linear-algebra proof-verification linear-transformations dual-spaces
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
708715
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
708715
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
708715
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
708715
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
708715
708715
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
7,8631730
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050919%2fthe-dual-basis-spans-the-vector-space-of-linear-transformations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
7,8631730
$endgroup$
add a comment |
$begingroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
7,8631730
$endgroup$
add a comment |
$begingroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
7,8631730
$endgroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
7,8631730
answered Dec 24 '18 at 4:21
Guido A.Guido A.
7,8631730
answered Dec 24 '18 at 4:21
Guido A.Guido A.
7,8631730
answered Dec 24 '18 at 4:21
Guido A.Guido A.
7,8631730
7,8631730
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050919%2fthe-dual-basis-spans-the-vector-space-of-linear-transformations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
